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Eletromagnetismo

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DUPL Quantity Units PHYSICAL CONSTANTS Best Experimental Symbol Value Approximate Value for Problem Work Permittivity of free space Fm eu Permeability of free space Hm jo Intrinsic impedance of free rjo space fl Speed of light in vacuum ms c Electron charge C e Electron mass Kg mQ Proton mass kg mp Neutron mass Kg ma Boltzmann constant JK K Avogadros number Kgmole N Plancks constant J s h Acceleration due to gravity g ms2 Universal contant of gravitation G m2Kg s2 Electronvolt J eV 8854 x 10 12 4i7 x 107 3766 2998 x 108 16030 x 10l9 91066 x 1031 167248 x 1027 16749 x 1027 138047 x 1023 60228 x 1026 6624 x 1034 981 6658 x 10 16030 x 1019 126 x 107 120ir 3 X 108 16 x 1019 91 x 1031 167 x 1027 167 x 10 27 138 x 1023 6 x 1026 662 x 1034 98 666 x 1011 16 x 1019 CONTENTS Preface xiii A Note to the Student xvi PART 1 VECTOR ANALYSIS 1 Vector Algebra 3 11 Introduction 3 112 A Preview of the Book 4 13 Scalars and Vectors 4 14 Unit Vector 5 115 Vector Addition and Subtraction 6 16 Position and Distance Vectors 7 17 Vector Multiplication 11 18 Components of a Vector 16 Summary 22 Review Questions 23 Problems 25 2 Coordinate Systems and Transformation 28 21 Introduction 28 22 Cartesian Coordinates x y z 29 23 Circular Cylindrical Coordinates p f z 24 Spherical Coordinates r d z 32 f25 ConstantCoordinate Surfaces 41 Summary 46 Review Questions 47 Problems 49 29 VII Contents 3 Vector Calculus 53 31 Introduction 53 32 Differential Length Area and Volume 53 33 Line Surface and Volume Integrals 60 34 Del Operator 63 35 Gradient of a Scalar 65 36 Divergence of a Vector and Divergence Theorem 69 37 Curl of a Vector and Stokess Theorem 75 38 Laplacian of a Scalar 83 t39 Classification of Vector Fields 86 Summary 89 Review Questions 90 Problems 93 PART 2 ELECTROSTATICS 4 Electrostatic Fields 103 41 Introduction 103 42 Coulombs Law and Field Intensity 104 43 Electric Fields due to Continuous Charge Distributions 111 44 Electric Flux Density 122 45 Gausss LawMaxwells Equation 124 46 Applications of Gausss Law 126 47 Electric Potential 133 48 Relationship between E and VMaxwells Equation 139 49 An Electric Dipole and Flux Lines 142 410 Energy Density in Electrostatic Fields 146 Summary 150 Review Questions 153 Problems 155 5 Electric Fields in Material Space 161 51 Introduction 161 52 Properties of Materials 161 53 Convection and Conduction Currents 162 54 Conductors 165 55 Polarization in Dielectrics 171 56 Dielectric Constant and Strength 774 f 57 Linear Isotropic and Homogeneous Dielectrics 175 58 Continuity Equation and Relaxation Time 180 CONTENTS IX 59 Boundary Conditions 182 Summary 191 Review Questions 192 Problems 194 6 Electrostatic BoundaryValue Problems 199 61 Introduction 199 62 Poissons and Laplaces Equations 199 f 63 Uniqueness Theorem 201 64 General Procedure for Solving Poissons or Laplaces Equation 202 65 Resistance and Capacitance 223 66 Method of Images 240 Summary 246 Review Questions 247 Problems 249 PART 3 MAGNETOSTATICS 7 Magnetostatic Fields 261 71 Introduction 261 72 BiotSavarts Law 263 73 Amperes Circuit LawMaxwells Equation 273 74 Applications of Amperes Law 274 75 Magnetic Flux DensityMaxwells Equation 281 76 Maxwells Equations for Static EM Fields 283 77 Magnetic Scalar and Vector Potentials 284 f 78 Derivation of BiotSavarts Law and Amperes Law 290 Summary 292 Review Questions 293 Problems 296 8 Magnetic Forces Materials and Devices 304 81 Introduction 304 82 Forces due to Magnetic Fields 304 83 Magnetic Torque and Moment 316 84 A Magnetic Dipole 318 85 Magnetization in Materials 323 f 86 Classification of Magnetic Materials 327 87 Magnetic Boundary Conditions 330 88 Inductors and Inductances 336 Contents 89 Magnetic Energy 339 f810 Magnetic Circuits 347 1811 Force on Magnetic Materials Summary 354 Review Questions 356 Problems 358 349 PART 4 WAVES AND APPLICATIONS 9 Maxwells Equations 369 91 Introduction 369 92 Faradays Law 370 93 Transformer and Motional EMFs 372 94 Displacement Current 381 95 Maxwells Equations in Final Forms 384 t96 TimeVarying Potentials 387 97 TimeHarmonic Fields 389 Summary 400 Review Questions 407 Problems 404 10 Electromagnetic Wave Propagation 410 101 Introduction 410 tl02 Waves in General 411 103 Wave Propagation in Lossy Dielectrics 417 104 Plane Waves in Lossless Dielectrics 423 105 Plane Waves in Free Space 423 106 Plane Waves in Good Conductors 425 107 Power and the Poynting Vector 435 108 Reflection of a Plane Wave at Normal Incidence 440 f 109 Reflection of a Plane Wave at Oblique Incidence 451 Summary 462 Review Questions 464 Problems 466 11 Transmission Lines 473 111 Introduction 473 112 Transmission Line Parameters 474 113 Transmission Line Equations 477 114 Input Impedance SWR and Power 484 115 The Smith Chart 492 CONTENTS XI 116 Some Applications of Transmission Lines 505 f 117 Transients on Transmission Lines 512 1118 Microstrip Transmission Lines 524 Summary 528 Review Questions 530 Problems 533 12 Waveguides 542 121 Introduction 542 122 Rectangular Waveguides 543 123 Transverse Magnetic TM Modes 547 12A Transverse Electric TE Modes 552 125 Wave Propagation in the Guide 563 126 Power Transmission and Attenuation 565 tl27 Waveguide Current and Mode Excitation 569 128 Waveguide Resonators 575 Summary 581 Review Questions 582 Problems 583 13 Antennas 588 131 Introduction 588 132 Hertzian Dipole 590 133 HalfWave Dipole Antenna 594 134 QuarterWave Monopole Antenna 598 135 Small Loop Antenna 599 136 Antenna Characteristics 604 137 Antenna Arrays 612 1138 Effective Area and the Friis Equation 62 tl39 The Radar Equation 625 Summary 629 Review Questions 630 Problems 632 14 Modern Topics 638 141 Introduction 638 142 Microwaves 638 143 Electromagnetic Interference and Compatibility 144 Optical Fiber 649 Summary 656 Review Questions 656 Problems 658 644 ii Contents 15 Numerical Methods 660 151 tl52 153 154 155 Introduction 660 Field Plotting 667 The Finite Difference Method 669 The Moment Method 683 The Finite Element Method 694 Summary 713 Review Questions 714 Problems 76 Appendix A Mathematical Formulas 727 Appendix B Material Constants 737 Appendix C Answers to OddNumbered Problems 740 Index 763 PREFACE The fundamental objectives of the book remains the same as in the first editionto present electromagnetic EM concepts in a clearer and more interesting manner than earlier texts This objective is achieved in the following ways 1 To avoid complicating matters by covering EM and mathematical concepts simul taneously vector analysis is covered at the beginning of the text and applied gradually This approach avoids breaking in repeatedly with more background on vector analysis thereby creating discontinuity in the flow of thought It also separates mathematical theo rems from physical concepts and makes it easier for the student to grasp the generality of those theorems 2 Each chapter starts with a brief introduction that serves as a guide to the whole chapter and also links the chapter to the rest of the book The introduction helps students see the need for the chapter and how the chapter relates to the previous chapter Key points are emphasized to draw the readers attention to them A brief summary of the major con cepts is provided toward the end of the chapter 3 To ensure that students clearly understand important points key terms are defined and highlighted Essential formulas are boxed to help students identify them 4 Each chapter includes a reasonable amount of examples with solutions Since the examples are part of the text they are clearly explained without asking the reader to fill in missing steps Thoroughly workedout examples give students confidence to solve prob lems themselves and to learn to apply concepts which is an integral part of engineering ed ucation Each illustrative example is followed by a problem in the form of a Practice Exer cise with the answer provided 5 At the end of each chapter are ten review questions in the form of multiplechoice objective items It has been found that openended questions although intended to be thought provoking are ignored by most students Objective review questions with answers immediately following them provide encouragement for students to do the problems and gain immediate feedback A large number of problems are provided are presented in the same order as the mate rial in the main text Problems of intermediate difficulty are identified by a single asterisk the most difficult problems are marked with a double asterisk Enough problems are pro XIII iv Preface vided to allow the instructor to choose some as examples and assign some as homework problems Answers to oddnumbered problems are provided in Appendix C 6 Since most practical applications involve timevarying fields six chapters are devoted to such fields However static fields are given proper emphasis because they are special cases of dynamic fields Ignorance of electrostatics is no longer acceptable because there are large industries such as copier and computer peripheral manufacturing that rely on a clear understanding of electrostatics 7 The last chapter covers numerical methods with practical applications and com puter programs This chapter is of paramount importance because most practical problems are solvable only by using numerical techniques 8 Over 130 illustrative examples and 400 figures are given in the text Some addi tional learning aids such as basic mathematical formulas and identities are included in the Appendix Another guide is a special note to students which follows this preface In this edition a new chapter on modern topics such as microwaves electromagnetic interference and compatibility and fiber optics has been added Also the Fortran codes in previous editions have been converted to Matlab codes because it was felt that students are more familiar with Matlab than with Fortran Although this book is intended to be selfexplanatory and useful for selfinstruction the personal contact that is always needed in teaching is not forgotten The actual choice o1 course topics as well as emphasis depends on the preference of the individual instructor For example the instructor who feels that too much space is devoted to vector analysis o static fields may skip some of the materials however the students may use them as refer ence Also having covered Chapters 1 to 3 it is possible to explore Chapters 9 to 15 In structors who disagree with the vectorcalculusfirst approach may proceed with Chapter 1 and 2 then skip to Chapter 4 and refer to Chapter 3 as needed Enough material i covered for twosemester courses If the text is to be covered in one semester some sec tions may be skipped explained briefly or assigned as homework Sections marked wit the dagger sign t may be in this category A suggested schedule for a fourhour semester coverage is on page xv Acknowledgments I would like to thank Peter Gordon and the editorial and production staff of Oxford Un versity Press for a job well done This edition has benefited from the insightful commeni of the following reviewers Leo C Kempel Michigan State University Andrew Diene University of California Davis George W Hanson University of WisconsinMilwaukei Samir ElGhazaly Arizona State University and Sadasiva M Rao Auburn University am greatly indebted to Raymond Garcia Jerry Sagliocca and Dr Oladega Soriyan f helping with the solutions manual and to Dr Saroj Biswas for helping with Matlab I a grateful to Temple University for granting me a leave in Fall 1998 during which I was ab to work on the revision of this book I owe special thanks to Dr Keya Sadeghipour de of the College of Engineering and Dr John Helferty chairman of the Department of Ele trical and Computer Engineering for their constant support As always particular than PREFACE xv Suggested Schedule Chapter Title Approximate Number of Hours 1 Vector Algebra 2 Coordinate Systems and Transformation 3 Vector Calculus 4 Electrostatic Fields 5 Electric Fields in Material Space 6 Electrostatic BoundaryValue Problems 7 Magnetostatic Fields 8 Magnetic Forces Materials and Devices 9 Maxwells Equations 10 Electromagnetic Wave Propagation 11 Transmission Lines 12 Waveguides 13 Antennas 14 Modern Topics 15 Numerical Methods Exams TOTAL 2 2 4 6 4 5 4 6 4 5 5 4 5 3 6 4 60 go to my wife Chris and our daughters Ann and Joyce for the patience prayers and full support As usual I welcome your comments suggestions and corrections Matthew N O Sadiku A NOTE TO THE STUDENT Electromagnetic theory is generally regarded by most students as one of the most difficult courses in physics or the electrical engineering curriculum But this misconception may be proved wrong if you take some precautions From experience the following ideas are pro vided to help you perform to the best of your ability with the aid of this textbook 1 Pay particular attention to Part I on Vector Analysis the mathematical tool for this course Without a clear understanding of this section you may have problems with the rest of the book 2 Do not attempt to memorize too many formulas Memorize only the basic ones which are usually boxed and try to derive others from these Try to understand how for mulas are related Obviously there is nothing like a general formula for solving all prob lems Each formula has some limitations due to the assumptions made in obtaining it Be aware of those assumptions and use the formula accordingly 3 Try to identify the key words or terms in a given definition or law Knowing the meaning of these key words is essential for proper application of the definition or law 4 Attempt to solve as many problems as you can Practice is the best way to gain skill The best way to understand the formulas and assimilate the material is by solving problems It is recommended that you solve at least the problems in the Practice Exercise immediately following each illustrative example Sketch a diagram illustrating the problem before attempting to solve it mathematically Sketching the diagram not only makes the problem easier to solve it also helps you understand the problem by simplifying and organizing your thinking process Note that unless otherwise stated all distances are in meters For example 2 1 5 actually means 2 m 1 m 5 m A list of the powers of ten and Greek letters commonly used throughout this text is provided in the tables located on the inside cover Important formulas in calculus vectors and complex analysis are provided in Appendix A Answers to oddnumbered problems are in Appendix C XVI PART 1 VECTOR ANALYSIS Chapter 7 T V Sf VECTOR ALGEBRA One thing I have learned in a long life that all our science measured against reality is primitive and childlikeand yet is the most precious thing we have ALBERT EINSTEIN 11 INTRODUCTION Electromagnetics EM may be regarded as the study of the interactions between electric charges at rest and in motion It entails the analysis synthesis physical interpretation and application of electric and magnetic fields Kkctioniiiniutics kYli is a branch of physics or electrical engineering in which electric and magnetic phenomena are studied EM principles find applications in various allied disciplines such as microwaves an tennas electric machines satellite communications bioelectromagnetics plasmas nuclear research fiber optics electromagnetic interference and compatibility electromechanical energy conversion radar meteorology and remote sensing12 In physical medicine for example EM power either in the form of shortwaves or microwaves is used to heat deep tissues and to stimulate certain physiological responses in order to relieve certain patho logical conditions EM fields are used in induction heaters for melting forging annealing surface hardening and soldering operations Dielectric heating equipment uses shortwaves to join or seal thin sheets of plastic materials EM energy offers many new and exciting possibilities in agriculture It is used for example to change vegetable taste by reducing acidity EM devices include transformers electric relays radioTV telephone electric motors transmission lines waveguides antennas optical fibers radars and lasers The design of these devices requires thorough knowledge of the laws and principles of EM For numerous applications of electrostatics see J M Crowley Fundamentals of Applied Electro statics New York John Wiley Sons 1986 2For other areas of applications of EM see for example D Teplitz ed Electromagnetism Paths to Research New York Plenum Press 1982 4 Vector Algebra 12 A PREVIEW OF THE BOOK The subject of electromagnetic phenomena in this book can be summarized in Maxwells equations VD pv 11 V B 0 12 V X E 13 dt V X H J 14 dt where V the vector differential operator D the electric flux density B the magnetic flux density E the electric field intensity H the magnetic field intensity pv the volume charge density and J the current density Maxwell based these equations on previously known results both experimental and theo retical A quick look at these equations shows that we shall be dealing with vector quanti ties It is consequently logical that we spend some time in Part I examining the mathemat ical tools required for this course The derivation of eqs 11 to 14 for timeinvariant conditions and the physical significance of the quantities D B E H J and pv will be our aim in Parts II and III In Part IV we shall reexamine the equations for timevarying situa tions and apply them in our study of practical EM devices 13 SCALARS AND VECTORS Vector analysis is a mathematical tool with which electromagnetic EM concepts are most conveniently expressed and best comprehended We must first learn its rules and tech niques before we can confidently apply it Since most students taking this course have little exposure to vector analysis considerable attention is given to it in this and the next two chapters3 This chapter introduces the basic concepts of vector algebra in Cartesian coordi nates only The next chapter builds on this and extends to other coordinate systems A quantity can be either a scalar or a vector Indicates sections that may be skipped explained briefly or assigned as homework if the text is covered in one semester 3The reader who feels no need for review of vector algebra can skip to the next chapter I 14 UNIT VECTOR A scalar is a quantity that has only magnitude Quantities such as time mass distance temperature entropy electric potential and popu lation are scalars A vector is a quantity that has both magnitude and direction Vector quantities include velocity force displacement and electric field intensity Another class of physical quantities is called tensors of which scalars and vectors are special cases For most of the time we shall be concerned with scalars and vectors4 To distinguish between a scalar and a vector it is customary to represent a vector by a letter with an arrow on top of it such as A and B or by a letter in boldface type such as A and B A scalar is represented simply by a lettereg A B U and V EM theory is essentially a study of some particular fields A field is a function that specifies a particular quantity everywhere in a region If the quantity is scalar or vector the field is said to be a scalar or vector field Exam ples of scalar fields are temperature distribution in a building sound intensity in a theater electric potential in a region and refractive index of a stratified medium The gravitational force on a body in space and the velocity of raindrops in the atmosphere are examples of vector fields 14 UNIT VECTOR A vector A has both magnitude and direction The magnitude of A is a scalar written as A or A A unit vector aA along A is defined as a vector whose magnitude is unity ie 1 and its direction is along A that is 15 16 17 Note that aA 1 Thus we may write A as A AaA which completely specifies A in terms of its magnitude A and its direction aA A vector A in Cartesian or rectangular coordinates may be represented as Ax Ay Az or Ayay Azaz 4For an elementary treatment of tensors see for example A I Borisenko and I E Tarapor Vector and Tensor Analysis with Application Englewood Cliffs NJ PrenticeHall 1968 Vector Algebra H 1 y a b Figure 11 a Unit vectors ax ay and az b components of A along ax a and az where Ax A r and Az are called the components of A in the x y and z directions respec tively ax aT and az are unit vectors in the x y and z directions respectively For example ax is a dimensionless vector of magnitude one in the direction of the increase of the xaxis The unit vectors ax a and az are illustrated in Figure 11 a and the components of A along the coordinate axes are shown in Figure 11 b The magnitude of vector A is given by A VA2 X Al A and the unit vector along A is given by Axax Azaz VATATAI 18 19 15 VECTOR ADDITION AND SUBTRACTION Two vectors A and B can be added together to give another vector C that is C A B 110 The vector addition is carried out component by component Thus if A Ax Ay Az and B BxByBz C Ax Bxax Ay Byay Az Bzaz Vector subtraction is similarly carried out as D A B A B Ax Bxax Ay Byay Az Bzaz lH 112 B a 16 POSITION AND DISTANCE VECTORS b Figure 12 Vector addition C A B a parallelogram rule b headtotail rule Figure 13 Vector subtraction D A B a parallelogram rule b headtotail fA rule a b Graphically vector addition and subtraction are obtained by either the parallelogram rule or the headtotail rule as portrayed in Figures 12 and 13 respectively The three basic laws of algebra obeyed by any giveny vectors A B and C are sum marized as follows Law Addition Multiplication Commutative A B B A kA Ak Associative A B C A B C k A kA Distributive kA B kA ZfcB where k and are scalars Multiplication of a vector with another vector will be discussed in Section 17 16 POSITION AND DISTANCE VECTORS A point P in Cartesian coordinates may be represented by x y z The position vector r or radius vector of point P is as he directed silancc from the origin lo P ie r P OP xax yay 113 8 Vector Algebra 45 I I 111 I A I Figure 14 Illustration of position vector rP 3a 4a 5az Figure 15 Distance vector rPG The position vector of point P is useful in defining its position in space Point 3 4 5 for example and its position vector 3ax 4a 5az are shown in Figure 14 The distance vector is ihc displacement from one point to another If two points P and Q are given by xP yP zp and xe yQ ZQ the distance vector or separation vector is the displacement from P to Q as shown in Figure 15 that is rPQ rQ rP xQ xPax yQ yPy zQ zPaz 114 The difference between a point P and a vector A should be noted Though both P and A may be represented in the same manner as x y z and Ax Ay Az respectively the point P is not a vector only its position vector i is a vector Vector A may depend on point P however For example if A 2xyat y2ay xz2az and P is 2 14 then A at P would be 4a ay 32a A vector field is said to be constant or uniform if it does not depend on space variables x y and z For example vector B 3a 2a 10az is a uniform vector while vector A 2xyax y2ay xz2az is not uniform because B is the same everywhere whereas A varies from point to point EXAMPLE 11 If A 10ax 4ay 6azandB 2x av find a the component of A along ay b the magnitude of 3A B c a unit vector along A 2B 16 POSITION AND DISTANCE VECTORS Solution a The component of A along ay is Ay 4 b 3A B 310 4 6 2 1 0 301218 2 10 281318 Hence 3A B V282 132 182 VT277 3574 c Let C A 2B 10 4 6 4 2 0 14 2 6 A unit vector along C is 1426 or Vl4 2 22 62 ac 091 3ax 01302a 03906az Note that ac 1 as expected PRACTICE EXERCISE 11 Given vectors A ax 3a and B 5ax 2av 6a determine a A B b 5A B c The component of A along av d A unit vector parallel to 3A 4 B Answer a 7 b 0 2 21 c 0 d 09117 02279 03419 Points P and Q are located at 0 2 4 and 3 1 5 Calculate a The position vector P b The distance vector from P to Q c The distance between P and Q d A vector parallel to PQ with magntude of 10 10 Vector Algebra Solution a i 0ax 2av 4az 2a 4az b rPQ rQ i 3 1 5 0 2 4 3 1 1 or 3 a x ay az c Since rPQ is the distance vector from P to Q the distance between P and Q is the mag nitude of this vector that is Alternatively d ie V 9 1 1 3317 d VxQ xPf yQ yPf zQ zPf V 9 T T 3317 d Let the required vector be A then A AaA where A 10 is the magnitude of A Since A is parallel to PQ it must have the same unit vector as rPQ or rQP Hence rPQ and 311 3317 A I 0 3 9045a 3015a 3015az PRACTICE EXERCISE 12 Given points Pl 3 5 Q2 4 6 and R0 3 8 find a the position vectors of P and R b the distance vector rQR c the distance between Q and R Answer a ax 3ay 5az 3a 33 b 2a ay 2az EXAMPLE 13 A river flows southeast at 10 kmhr and a boat flows upon it with its bow pointed in the di rection of travel A man walks upon the deck at 2 kmhr in a direction to the right and per pendicular to the direction of the boats movement Find the velocity of the man with respect to the earth Solution Consider Figure 16 as illustrating the problem The velocity of the boat is ub 10cos 45 ax sin 45 a 7071a 7071a kmhr w 17 VECTOR MULTIPLICATION 11 Figure 16 For Example 13 The velocity of the man with respect to the boat relative velocity is um 2cos 45 ax sin 45 a 1414a 1414a kmhr Thus the absolute velocity of the man is uab um uh 5657a 8485ay u j 102563 that is 102 kmhr at 563 south of east PRACTICE EXERCISE 13 An airplane has a ground speed of 350 kmhr in the direction due west If there is a wind blowing northwest at 40 kmhr calculate the true air speed and heading of the airplane Answer 3793 kmhr 4275 north of west 17 VECTOR MULTIPLICATION When two vectors A and B are multiplied the result is either a scalar or a vector depend ing on how they are multiplied Thus there are two types of vector multiplication 1 Scalar or dot product A B 2 Vector or cross product A X B 12 HI Vector Algebra Multiplication of three vectors A B and C can result in either 3 Scalar triple product A B X C or 4 Vector triple product A X B X C A Dot Product The dot product of two vectors A and B wrilten as A B is defined geometrically as the product of the magnitudes of A and B and the cosine of the angle between them Thus A B AB cos I 115 where 6AB is the smaller angle between A and B The result of A B is called either the scalar product because it is scalar or the dot product due to the dot sign If A Ax Ay Az and B Bx By Bz then A B AXBX AyBy AZBZ 116 which is obtained by multiplying A and B component by component Two vectors A and B are said to be orthogonal or perpendicular with each other if A B 0 Note that dot product obeys the following i Commutative law A B B A 117 ii Distributive law A B C A B A C 118 A A A2 A2 119 iii Also note that ax ay ay az az ax 0 120a ax ax ay ay az az 1 120b It is easy to prove the identities in eqs 117 to 120 by applying eq 115 or 116 17 VECTOR MULTIPLICATION H 13 B Cross Product The cross product of two vectors A ind B written as A X B is a vector quantity whose magnitude is ihe area of the parallclopiped formed by A and It see Figure 17 and is in the direction of advance of a righthanded screw as A is turned into B Thus A X B AB sin 6ABan 121 where an is a unit vector normal to the plane containing A and B The direction of an is taken as the direction of the right thumb when the fingers of the right hand rotate from A to B as shown in Figure 18a Alternatively the direction of an is taken as that of the advance of a righthanded screw as A is turned into B as shown in Figure 18b The vector multiplication of eq 121 is called cross product due to the cross sign it is also called vector product because the result is a vector If A Ax B Bx By Bz then A X B ax Ax Bx av Ay By az K Bz AzByax AZBX AxBzay AxBy AyBxaz Ay Az and 122a 122b which is obtained by crossing terms in cyclic permutation hence the name cross product Figure 17 The cross product of A and B is a vector with magnitude equal to the area of the parallelogram and direction as indicated 14 H Vector Algebra A X B AX B A a b Figure 18 Direction of A X B and an using a righthand rule b righthanded screw rule Note that the cross product has the following basic properties i It is not commutative It is anticommutative A X B B X A A X B B X A ii It is not associative A X B X C h A X B X C iii It is distributive iv A X B C A X B A X C A X A 0 Also note that ax X ay az a X az ax az X ax ay 123a 123b 124 125 126 127 which are obtained in cyclic permutation and illustrated in Figure 19 The identities in eqs 125 to 127 are easily verified using eq 121 or 122 It should be noted that in ob taining an we have used the righthand or righthanded screw rule because we want to be consistent with our coordinate system illustrated in Figure 11 which is righthanded A righthanded coordinate system is one in which the righthand rule is satisfied that is ax X ay az is obeyed In a lefthanded system we follow the lefthand or lefthanded 17 VECTOR MULTIPLICATION 15 a b Figure 19 Cross product using cyclic permutation a moving clockwise leads to positive results b moving counterclockwise leads to negative results screw rule and ax X ay az is satisfied Throughout this book we shall stick to right handed coordinate systems Just as multiplication of two vectors gives a scalar or vector result multiplication of three vectors A B and C gives a scalar or vector result depending on how the vectors are multiplied Thus we have scalar or vector triple product C Scalar Triple Product Given three vectors A B and C we define the scalar triple product as A B X C B C X A C A X B 128 obtained in cyclic permutation If A Ax Ay Az B Bx By Bz and C Cx Cy Cz then A B X C is the volume of a parallelepiped having A B and C as edges and is easily obtained by finding the determinant of the 3 X 3 matrix formed by A B and C that is A B X C Bx By Bz Cy C 129 Since the result of this vector multiplication is scalar eq 128 or 129 is called the scalar triple product D Vector Triple Product For vectors A B and C we define the vector tiple product as A X B X C BA C CA B 130 16 Vector Algebra obtained using the baccab rule It should be noted that but A BC AB C A BC CA B 131 132 18 COMPONENTS OF A VECTOR A direct application of vector product is its use in determining the projection or compo nent of a vector in a given direction The projection can be scalar or vector Given a vector A we define the scalar component AB of A along vector B as see Figure 110a AB A cos 6AB A aB cos 6AB or AR A afl 133 The vector component AB of A along B is simply the scalar component in eq 133 multi plied by a unit vector along B that is AB ABaB A 134 Both the scalar and vector components of A are illustrated in Figure 110 Notice from Figure 110b that the vector can be resolved into two orthogonal components one com ponent AB parallel to B another A As perpendicular to B In fact our Cartesian repre sentation of a vector is essentially resolving the vector into three mutually orthogonal com ponents as in Figure llb We have considered addition subtraction and multiplication of vectors However di vision of vectors AB has not been considered because it is undefined except when A and B are parallel so that A kB where k is a constant Differentiation and integration of vectors will be considered in Chapter 3 B B a Figure 110 Components of A along B a scalar component AB b vector component AB EXAMPLE 14 18 COMPONENTS OF A VECTOR 17 Given vectors A 3ax 4ay az and B 2ay 5az find the angle between A and B Solution The angle dAB can be found by using either dot product or cross product A B 3 4 1 0 2 5 0 8 5 3 Alternatively A V3 2 42 I2 V26 COS BAR B VO2 22 52 V29 A B 3 I AII BI V2629 01092 9AR cos1 01092 8373 A X B 3 4 az 1 0 2 5 20 2ax 0 15ay 6 0az 22156 A X B 152 62 V745 sin 6AB A X Bj V745 26X29 0994 dAB cos1 0994 8373 PRACTICE EXERCISE 14 If A ax 3az and B 5a 2ay 6a find 6AB Answer 1206 EXAMPLE 15 Three field quantities are given by P 2ax a Q 2a ay 2az R 2ax 33 az Determine a P Q X P Q b Q R X P 18 Vector Algebra c P Q X R d sin0eR e P X Q X R f A unit vector perpendicular to both Q and R g The component of P along Q Solution a P Q X P Q P X P Q Q X P Q P X P P X Q Q X P Q X Q O Q X P Q X P O 2Q X P 2 ay a 2 1 2 2 0 1 21 0 ax 24 2 ay 20 2 az 2ar 12av 4a b The only way Q R X P makes sense is Q RX P 212 2 2 ay 3 0 a 1 1 2 12 3 4 6 6 4 12 14 Alternatively Q R X P 2 1 2 2 3 1 2 0 1 To find the determinant of a 3 X 3 matrix we repeat the first two rows and cross multiply when the cross multiplication is from right to left the result should be negated as shown below This technique of finding a determinant applies only to a 3 X 3 matrix Hence Q RXP 6021202 14 as obtained before 18 COMPONENTS OF A VECTOR 19 c From eq 128 or d P Q X R Q R X P 14 P Q X R 2 0 1 5 2 4 10 0 4 14 QXR IQIIRI 12 45 V5 3V14 V14 05976 e P X Q X R 2 0 1 X 5 2 4 2 3 4 Alternatively using the baccab rule P X Q X R QP R RP Q 2 1 24 0 1 2 3 14 0 2 2 3 4 f A unit vector perpendicular to both Q and R is given by Q X R 52 4 3 QXR 0745 0298 0596 Note that a l a Q 0 a R Any of these can be used to check a g The component of P along Q is cos 6PQaQ PQ P aGae 4 P QQ IQI2 2 9 4 1 4 04444ar 02222av 04444a7 PRACTICE EXERCISE 15 Let E 3av 4a and F 4a 10av 5ar a Find the component of E along F b Determine a unit vector perpendicular to both E and F Answer a 02837 07092 03546 b 09398 02734 0205 20 Vector Algebra FXAMPIF 1 f Derive the cosine formula and the sine formula a2 b2 c2 2bc cos A sin A sin B sin C a b c using dot product and cross product respectively Solution Consider a triangle as shown in Figure 111 From the figure we notice that a b c 0 that is b c a Hence a2 a a b c b c b b c c 2 b c a2 b2 c2 2bc cos A where A is the angle between b and c The area of a triangle is half of the product of its height and base Hence la X b lb X c lc X al ab sin C be sin A ca sin B Dividing through by abc gives sin A sin B sin C Figure 111 For Example 16 18 COMPONENTS OF A VECTOR 21 PRACTICE EXERCISE 16 Show that vectors a 4 0 1 b 13 4 and c 5 3 3 form the sides of a triangle Is this a right angle triangle Calculate the area of the triangle Answer Yes 105 EXAMPLE 17 Show that points Ptf 2 4 P2 1 2 and P 33 0 8 all lie on a straight line Deter mine the shortest distance between the line and point P43 1 0 Solution The distance vector fptp2 is given by Similarly rPJP2 rp2 TpP3 Tp31 rPtP4 rP4 rP P X rP rP 11 4 3 8 3 2 p a 4 8 00 2 1 08 2 10 3 a 1 2 0 5 6 2 4 5 2 4 12 5 2 4 4 az 6 12 showing that the angle between riP2 and rPiPi is zero sin 6 0 This implies that Ph P2 and P3 lie on a straight line Alternatively the vector equation of the straight line is easily determined from Figure 112a For any point P on the line joining P and P2 where X is a constant Hence the position vector r of the point P must satisfy i i Mp2 rP that is i i i2 i 5 2 4 X4 1 6 i 5 4X 2 X 4 6X This is the vector equation of the straight line joining Px and P2 If P3 is on this line the po sition vector of F3 must satisfy the equation r3 does satisfy the equation when X 2 22 Vector Algebra a Figure 112 For Example 17 The shortest distance between the line and point P43 1 0 is the perpendicular dis tance from the point to the line From Figure 112b it is clear that d rPiPt sin 6 rPp4 X aPp2 312 2426 53 Any point on the line may be used as a reference point Thus instead of using P as a ref erence point we could use P3 so that d sin PRACTICE EXERCISE 17 If P is 12 3 and P2 is 4 05 find a The distance PP2 b The vector equation of the line PP2 c The shortest distance between the line PP2 and point P37 12 Answer a 9644 b 1 5Xax 21 X av 8X 3 a c 82 SUMMARY 1 A field is a function that specifies a quantity in space For example Ax y z is a vector field whereas Vx y z is a scalar field 2 A vector A is uniquely specified by its magnitude and a unit vector along it that is A AaA REVIEW QUESTIONS M 23 3 Multiplying two vectors A and B results in either a scalar A B AB cos 6AB or a vector A X B AB sin 9ABan Multiplying three vectors A B and C yields a scalar A B X C or a vector A X B X C 4 The scalar projection or component of vector A onto B is AB A aB whereas vector projection of A onto B is AB ABaB 11 Identify which of the following quantities is not a vector a force b momentum c ac celeration d work e weight 12 Which of the following is not a scalar field a Displacement of a mosquito in space b Light intensity in a drawing room c Temperature distribution in your classroom d Atmospheric pressure in a given region e Humidity of a city 13 The rectangular coordinate systems shown in Figure 113 are righthanded except 14 Which of these is correct a A X A A2 b A X B B X A 0 c A B C B C A d axay az e ak ax ay where ak is a unit vector a y d e Figure 113 For Review Question 13 c y f 24 H Vector Algebra 15 Which of the following identities is not valid a ab c ab be b a X b c a X b a X c c a b b a d c a X b b a X c e aA aB cos dAB 16 Which of the following statements are meaningless a A B 2A 0 b A B 5 2A c AA B 2 0 d A A B B 0 17 Let F 2ax 63 10a2 and G ax Gyay 5az If F and G have the same unit vector Gy is a 6 d 0 b 3 e 6 18 Given that A ax aay az and B xax ay az if A and B are normal to each other a is a 2 d 1 b 12 e 2 c 0 19 The component of 6ax 2a 3az along 3ax 4a is a 12ax 9ay 3az b 30a 40a c 107 d 2 e 10 110 Given A 6ax 3ay 2az the projection of A along ay is a 12 b 4 c 3 d 7 e 12 Answers Lid 12a 13be 14b 15a 16bc 17b 18b 19d 110c PROBLEMS PROBLEMS 11 Find the unit vector along the line joining point 2 4 4 to point 3 2 2 25 12 Let A 2a 53 3az B 3a 4ay and C ax ay az a Determine A 2B b Calculate A 5C c For what values of k is kB 2 d Find A X BA B 13 If A ay 3az C 3ax 5av 7az determine a A 2B C b C 4A B 2A 3B d A C B2 e B X A C 14 If the position vectors of points T and S are 3a 23 az and Aax 4 6ay 2ax re spectively find a the coordinates of T and S b the distance vector from T to S c the distance between T and S 15 If Aay 6az A 5ax B x C 8ax 2a find the values of a and 3 such that aA 0B C is parallel to the yaxis 16 Given vectors A aax ay Aaz o a x T p3y O3Z C 5ax 2ay 7a determine a 3 and 7 such that the vectors are mutually orthogonal 17 a Show that A B2 A X B2 AB2 az X ax a aY X a a a X ay a ay X az 26 Vector Algebra 18 Given that P 2ax Ay 2az Q 4a 3ay 2a2 C ax ay 2az find a P Q R b P Q X R c Q X P R d P X Q Q X R e P X Q X Q X R f cos 6PR g sin 6PQ 19 Given vectors T 2ax 6ay 3az and 8 34 2ay az find a the scalar projec tion of T on S b the vector projection of S on T c the smaller angle between T and S 110 If A ax 6ay 5az andB ax 2ay 3ax find a the scalar projections of A on B b the vector projection of B on A c the unit vector perpendicular to the plane containing A and B 111 Calculate the angles that vector H 3ax 5ay 8az makes with the xy and zaxes 112 Find the triple scalar product of P Q and R given that P 2ax ay az Q a ay az and R 2a 3az 113 Simplify the following expressions a A X A X B b A X A X A X B 114 Show that the dot and cross in the triple scalar product may be interchanged ie A B X C A X B C 115 Points Pil 2 3 P25 2 0 and P32 7 3 form a triangle in space Calculate the area of the triangle 116 The vertices of a triangle are located at 4 1 3 2 5 4 and 016 Find the three angles of the triangle 117 Points P Q and R are located at 1 4 8 2 1 3 and 1 2 3 respectively Determine a the distance between P and Q b the distance vector from P to R c the angle between QP and QR d the area of triangle PQR e the perimeter of triangle PQR 118 If r is the position vector of the point x y z and A is a constant vector show that a r A A 0 is the equation of a constant plane b r A r 0 is the equation of a sphere Single asterisks indicate problems of intermediate difficulty PROBLEMS Figure 114 For Problem 120 27 c Also show that the result of part a is of the form Ax By Cz D 0 where D A2 B2 C2 and that of part b is of the form x2 y2 z2 r2 119 a Prove that P cos 0ix sin 6xay and Q cos 82ax sin 02ay are unit vectors in the xyplane respectively making angles i and 82 with the xaxis b By means of dot product obtain the formula for cos02 i By similarly formulat ing P and Q obtain the formula for cos02 i c If 6 is the angle between P and Q find P Q in terms of 6 120 Consider a rigid body rotating with a constant angular velocity w radians per second about a fixed axis through O as in Figure 114 Let r be the distance vector from O to P the position of a particle in the body The velocity u of the body at P is u dw r sin 6 co or u o X r If the rigid body is rotating with 3 radians per second about an axis parallel to ax 2ay 2az and passing through point 2 3 1 determine the velocity of the body at 1 34 121 Given A x2yax yzay yz2az determine a The magnitude of A at point T2 13 b The distance vector from T to 5 if S is 56 units away from T and in the same direction as A at T c The position vector of S 122 E and F are vector fields given by E 2xa ay yzaz and F xyax y2ay xyzaz Determine a Eatl2 3 b The component of E along F at 1 2 3 c A vector perpendicular to both E and F at 0 13 whose magnitude is unity Chapter 2 COORDINATE SYSTEMS AND TRANSFORMATION Education makes a people easy to lead but difficult to drive easy to govern but impossible to enslave HENRY P BROUGHAM 21 INTRODUCTION In general the physical quantities we shall be dealing with in EM are functions of space and time In order to describe the spatial variations of the quantities we must be able to define all points uniquely in space in a suitable manner This requires using an appropriate coordinate system A point or vector can be represented in any curvilinear coordinate system which may be orthogonal or nonorthogonal An orthogonal system is one in which the coordinates arc mutually perpendicular Nonorthogonal systems are hard to work with and they are of little or no practical use Examples of orthogonal coordinate systems include the Cartesian or rectangular the cir cular cylindrical the spherical the elliptic cylindrical the parabolic cylindrical the conical the prolate spheroidal the oblate spheroidal and the ellipsoidal1 A considerable amount of work and time may be saved by choosing a coordinate system that best fits a given problem A hard problem in one coordi nate system may turn out to be easy in another system In this text we shall restrict ourselves to the three bestknown coordinate systems the Cartesian the circular cylindrical and the spherical Although we have considered the Cartesian system in Chapter 1 we shall consider it in detail in this chapter We should bear in mind that the concepts covered in Chapter 1 and demonstrated in Cartesian coordinates are equally applicable to other systems of coordinates For example the procedure for For an introductory treatment of these coordinate systems see M R Spigel Mathematical Hand book of Formulas and Tables New York McGrawHill 1968 pp 124130 28 23 CIRCULAR CYLINDRICAL COORDINATES R F Z 29 finding dot or cross product of two vectors in a cylindrical system is the same as that used in the Cartesian system in Chapter 1 Sometimes it is necessary to transform points and vectors from one coordinate system to another The techniques for doing this will be presented and illustrated with examples 22 CARTESIAN COORDINATES X Y Z As mentioned in Chapter 1 a point P can be represented as x y z as illustrated in Figure 11 The ranges of the coordinate variables x y and z are 00 X 00 00yo 21 00 I 00 A vector A in Cartesian otherwise known as rectangular coordinates can be written as AxAyAJ or A A Ayay Azaz 22 where ax ay and az are unit vectors along the x y and zdirections as shown in Figure 11 23 CIRCULAR CYLINDRICAL COORDINATES p cj z The circular cylindrical coordinate system is very convenient whenever we are dealing with problems having cylindrical symmetry A point P in cylindrical coordinates is represented as p j z and is as shown in Figure 21 Observe Figure 21 closely and note how we define each space variable p is the radius of the cylinder passing through P or the radial distance from the zaxis f called the Figure 21 Point P and unit vectors in the cylindrical coordinate system 30 Coordinate Systems and Transformation azimuthal angle is measured from the xaxis in the xyplane and z is the same as in the Cartesian system The ranges of the variables are 0 p 0 27T 00 Z 00 A vector A in cylindrical coordinates can be written as 23 Ap A Az or Apap 24 where ap a and az are unit vectors in the p and directions as illustrated in Figure 21 Note that a is not in degrees it assumes the unit vector of A For example if a force of 10 N acts on a particle in a circular motion the force may be represented as F lOa N In this case a0 is in newtons The magnitude of A is Al p 2x12 25 Notice that the unit vectors ap a and az are mutually perpendicular because our co ordinate system is orthogonal ap points in the direction of increasing p a in the direction of increasing 0 and az in the positive zdirection Thus a az az 1 a a7 a 0 np X aj a a X az a az X ap a 26a 26b 26c 26d 26e where eqs 26c to 26e are obtained in cyclic permutation see Figure 19 The relationships between the variables x y z of the Cartesian coordinate system and those of the cylindrical system p j z are easily obtained from Figure 22 as cj tan1 x z 27 or x p cos 0 y p sin z z 28 Whereas eq 27 is for transforming a point from Cartesian x y z to cylindrical p z coordinates eq 28 is for p 4 z x y z transformation 23 CIRCULAR CYLINDRICAL COORDINATES p 0 z 11 31 Figure 22 Relationship between x y z and P z The relationships between ax ay az and ap a a2 are obtained geometrically from Figure 23 or cos 0 ap sin ap cos jax sin sin a7 cos b Figure 23 Unit vector transformation a cylindrical components of ax b cylin drical components of a r 29 210 32 Coordinate Systems and Transformation Finally the relationships between Ax Ay Az and Ap A0 Az are obtained by simply substituting eq 29 into eq 22 and collecting terms Thus A Ax cos j Ay sin jap AX sin j Ay cos 0a 0 Azaz 211 or Ap Ax cos t Ay sin f A AX sin f Ay cos tj 212 In matrix form we have the transformation of vector A from AxAyAz to Ap A0 A as 213 A Az cos sin 0 0 sinj cos 0 0 0 0 1 Ax Ay Az The inverse of the transformation Ap A Az Ax Ay Az is obtained as Ax cos t sin 0 sin cos 0 0 0 1 A 214 or directly from eqs 24 and 210 Thus cos sin 4 0 sin j cos j 0 0 0 1 V A 215 An alternative way of obtaining eq 214 or 215 is using the dot product For example 216 A Ay Az a a p az ap a a 0 a y a 0 a z a 0 x az az az az A A A The derivation of this is left as an exercise 24 SPHERICAL COORDINATES r 0 The spherical coordinate system is most appropriate when dealing with problems having a degree of spherical symmetry A point P can be represented as r 6 4 and is illustrated in Figure 24 From Figure 24 we notice that r is defined as the distance from the origin to 24 SPHERICAL COORDINATES r e 33 point P or the radius of a sphere centered at the origin and passing through P 6 called the colatitude is the angle between the zaxis and the position vector of P and 4 is measured from the xaxis the same azimuthal angle in cylindrical coordinates According to these definitions the ranges of the variables are O 0 i r 217 0 f 2TT A vector A in spherical coordinates may be written as ArAeA or Ar Agae A 218 where an ae and 3A are unit vectors along the r B and directions The magnitude of A is A A2 r A2 e Aj112 219 The unit vectors an a and a are mutually orthogonal ar being directed along the radius or in the direction of increasing r ae in the direction of increasing 6 and a0 in the di rection of increasing f Thus ar ar ae ar ae ae ar x ae a ae X a ar a 0 X ar a9 ar 0 220 Figure 24 Point P and unit vectors in spherical coordinates 34 Coordinate Systems and Transformation The space variables x y z in Cartesian coordinates can be related to variables r 0 p of a spherical coordinate system From Figure 25 it is easy to notice that Vx2 HZ2 0 tan z or x r sin 0 cos 0 y r sin 0 sin z r cos I 221 222 In eq 221 we have x y z r 0 point transformation and in eq 222 it is r 6 4 x y z point transformation The unit vectors ax ay a2 and ar ae a are related as follows or ax sin 0 cos 4 ar cos 0 cos as sin 83 sin 6 sin ar cos 6 sin 0 ae cos j az cos 6 ar sin 0 as ar sin 0 cos 0 a sin d sin ay c o s a cos 0 cos t ax cos 0 sin ay sin 223 224 sin cos ay Figure 25 Relationships between space variables x y z r 6 and p t z 24 SPHERICAL COORDINATES r e t 35 The components of vector A Ax Ay Az and A Ar Ae A are related by substitut ing eq 223 into eq 22 and collecting terms Thus A Ax sin 0 cos 4 Ay sin 0 sin 0 Az cos 0ar Ax cos 0 cos 0 Ay cos 0 sin 0 Az sin dae Ax sin 0 Ay cos Aa 225 and from this we obtain Ar A sin 0 cos t Ay sin 0 sin j Az cos 0 Ae Ax cos 0 cos 4 Ay cos 0 sin f Az sin A A sin Ay cos 0 226 A sin 6 cos 0 sin 0 sin 0 cos 0 cos 0 cos 0 cos 0 sin sin 0 sin 0 cos 4 0 X In matrix form the Ax Ay Az Ar Ae A vector transformation is performed accord ing to 227 The inverse transformation An Ae A Ax Ay Az is similarly obtained or we obtain it from eq 223 Thus 228 Alternatively we may obtain eqs 227 and 228 using the dot product For example AX Av sin sin cos 0 0 0 COS 0 sin 0 cos 0 cos cos 0 sin 0 sin0 sin cos T 0 Ar As Ar ar ax ar ay ar az 229 For the sake of completeness it may be instructive to obtain the point or vector trans formation relationships between cylindrical and spherical coordinates using Figures 25 and 26 where f is held constant since it is common to both systems This will be left as an exercise see Problem 29 Note that in point or vector transformation the point or vector has not changed it is only expressed differently Thus for example the magnitude of a vector will remain the same after the transformation and this may serve as a way of checking the result of the transformation The distance between two points is usually necessary in EM theory The distance d between two points with position vectors rl and r2 is generally given by dr2 230 36 Coordinate Systems and Transformation Figure 26 Unit vector transformations for cylindri cal and spherical coordinates or d2 x2 xf y2 yxf z2 zif Cartesian d2 p p2 2pp2 cos2 00 z2 Zf cylindrical 2rr2 cos d2 cos 0j sin 02 sin dx cos2 0i spherical d2 r r 2rr2 cos d2 cos 0j 231 232 233 EXAMPLE 21 Given point P2 6 3 and vector A yax x zay express P and A in cylindrical and spherical coordinates Evaluate A at P in the Cartesian cylindrical and spherical systems Solution At point P x 2 y 6 z 3 Hence p Vx 2 y2 V 4 36 632 4 tan1 tan1 10843 x 2 r Vx2 y2 z2 V4 36 9 7 6462 Vx2 y2 V40 d tan tan Z 5 Thus P2 6 3 P632 10843 3 P7 6462 10843 In the Cartesian system A at P is A 6ax a 24 SPHERICAL COORDINATES r e 37 For vector A Ax y Ay x z Az 0 Hence in the cylindrical system Ap A Az cos sin 4 t 0 sin cos 0 0 0 0 0 1 y X 0 z or Ap y cos j x z sin 0 A y sin 0 JC z cos A 0 But x p cos jy p sin 0 and substituting these yields A Ap A Az p cos 0 sin 0 p cos 0 z sin 0ap p sin 0 p cos 0 z cos AtP Hence V40 tan cos A sin 2 2 V40 V40 V40 V V40 40 6 2 V40 3 1 6 38 y p V40 V40 Similarly in the spherical system V40 V40 09487a 6OO8a0 Ar Ae A or sin 0 cos sin 9 sin 0 cos 0 cos 6 cos 0 cos 6 sin 0 sin 6 sin 0 cos 0 0 Ar y sin 0 cos j x zsin 6 sin 0 A9 y cos 0 cos 0 x zcos 0 sin A4 y sin j x zcos 0 x z 38 If Coordinate Systems and Transformation But x r sin 6 cos j y r sin 6 sin and z r cos 0 Substituting these yields r 8 rsin2 6 cos sin 0 sin 0 cos cos 6 sin 0 sin 4ar rsin 0 cos 6 sin cos 0 sin 0 cos cos 0 cos 6 sin AtP Hence r sin sin2 r 1 2 sin 6 cos 0 cos 8 cos tan 0 tan0 40 40 cos b V40 V40 cos t T sin 7 49 V40 V40 I VO 3 6 L 7 7 40 2 y 40 V40 V40 40 V 7 40 2 7 6 7 40 18 40 2 7 V40 38 a r i 7 7V40 40 08571a r 04066a9 6OO8a0 Note that A is the same in the three systems that is z Ar 0 6083 PRACTICE EXERCISE 21 a Convert points P 3 5 7X0 4 3 and S3 4 10 from Cartesian to cylindrical and spherical coordinates b Transform vector Q Vx2 to cylindrical and spherical coordinates c Evaluate Q at T in the three coordinate systems 24 SPHERICAL COORDINATES r e 39 Answer a P3162 7156 5 P5916 3231 7156 T4 270 3 T5 5313 270 55 2331 10 51118 15343 2331 b cos 4 ap sin j a z sin j az sin 9 sin 0 cos j r cos2 0 sin jar sin 0 cos 0 cos 0 r sin 0 sin ag sin 0 sin c 0 24az 0 24az 144ar 192a EXAMPLE 22 Express vector 10 B ar r cos 6 ae a in Cartesian and cylindrical coordinates Find B 3 4 0 and B 5 TT2 2 Solution Using eq 228 sin 0 cos sin 0 sin 4 cos 9 cos 0 cos sin cos 0 sin 0 cos t s i n 0 0 K r r cos I 1 or 10 Bx sin 0 cos j r cos 0 cos f sin 10 5 sin 0 sin j r cos 0 sin cos 10 5 7 cos 9 r cos 0 sin 0 r But r Vx 2 y2 z2 9 tan Hence and 6 tan x 40 H Coordinate Systems and Transformation Substituting all these gives loVx 2 y2 x Vx2 z2 z2x x2 y2 lOx xz x2 y2 z2 V 10W y2 y 2 2 2 y2 z2 f Vx2 y2 z 2 x2 y2 z2 y lOy B7 x2 y2 z2 V y2x2 y2 z2 lOz xA X x2 y2 zVx2 y2 x2 y2 z2 B B A Byay Bzaz where Bx By and Bz are as given above At 3 4 0 x 3 y 4 and z 0 so Thus 0 0 0 B For spherical to cylindrical vector transformation see Problem 29 sin cos 6 0 0 0 1 or cos d sin0 0 10 2 sin 6 r cos H r cos 10 7 cos d r r sin 6 cos 6 But r V p z zl and 6 tan V y2 7 25 CONSTANTCOORDINATE SURFACES 41 Thus sin V 7T7 cos C 7 T5 P Z P Hence B 10p At 5 TT2 2 p 5 0 TT2 and z 2 so B 50 4 29 V29V p Z lOz 2 z2 VTT 20 10 29 V29 2467ap Note that at 3 4 0 Bxyz BpAz Br 00 2907 This may be used to check the correctness of the result whenever possible PRACTICE EXERCISE 22 Express the following vectors in Cartesian coordinates a A pz sin 0 ap 3p cos 0 a p cos 0 sin 0 a b B r2 ar sin 6 a 1 Answer a A xyz 3xy ar zj 3x ay xy a j b B yixz y2 zz xay 25 CONSTANTCOORDINATE SURFACES Surfaces in Cartesian cylindrical or spherical coordinate systems are easily generated by keeping one of the coordinate variables constant and allowing the other two to vary In the 42 M Coordinate Systems and Transformation Cartesian system if we keep x constant and allow y and z to vary an infinite plane is gen erated Thus we could have infinite planes x constant y constant z constant 234 which are perpendicular to the x y and zaxes respectively as shown in Figure 27 The intersection of two planes is a line For example x constant y constant 235 is the line RPQ parallel to the zaxis The intersection of three planes is a point For example x constant y constant z constant 236 is the point Px y z Thus we may define point P as the intersection of three orthogonal infinite planes If P is 1 5 3 then P is the intersection of planes x 1 y 5 and z 3 Orthogonal surfaces in cylindrical coordinates can likewise be generated The sur faces p constant constant z constant 237 are illustrated in Figure 28 where it is easy to observe that p constant is a circular cylinder f constant is a semiinfinite plane with its edge along the zaxis and z constant is the same infinite plane as in a Cartesian system Where two surfaces meet is either a line or a circle Thus z constant p constant 238 z constant x constant Figure 27 Constant x y and z surfaces p constant z constant 25 CONSTANTCOORDINATE SURFACES H 43 Figure 28 Constant p j and z surfaces y p constant is a circle QPR of radius p whereas z constant j constant is a semiinfinite line A point is an intersection of the three surfaces in eq 237 Thus p 2 t 60 z 5 239 is the point P2 60 5 The orthogonal nature of the spherical coordinate system is evident by considering the three surfaces r constant 0 constant f constant 240 which are shown in Figure 29 where we notice that r constant is a sphere with its center at the origin 8 constant is a circular cone with the zaxis as its axis and the origin as its vertex 0 constant is the semiinfinite plane as in a cylindrical system A line is formed by the intersection of two surfaces For example r constant tj constant constant 241 Figure 29 Constant r 9 and j surfaces 44 Coordinate Systems and Transformation is a semicircle passing through Q and P The intersection of three surfaces gives a point Thus r 5 0 30 0 60 242 is the point P5 30 60 We notice that in general a point in threedimensional space can be identified as the intersection of three mutually orthogonal surfaces Also a unit normal vector to the surface n constant is an where n is x y z p r or 6 For example to plane 5 a unit normal vector is ax and to planed 20 a unit normal vector is a EXAMPLE 23 Two uniform vector fields are given by E 5a p 23 6az Calculate 3az and F ap a E X F b The vector component of E at P5 TT2 3 parallel to the line x 2 z 3 c The angle E makes with the surface z 3 at P Solution a E X F 5 10 1 2 6 60 6a 3 3Oa0 10 103 66 27 20 E X F V66 2 272 202 7406 b Line x 2 z 3 is parallel to the yaxis so the component of E parallel to the given line is E avav But at P5 TT2 3 Therefore sin t ap cos j a sin TT2 ap cos nil a a p E diy E apap or 5ay c Utilizing the fact that the zaxis is normal to the surface z 3 the angle between the zaxis and E as shown in Figure 210 can be found using the dot product E az E1 cos 6Ez 3 Vl34 cos Ez 3 cos oEz 134 02592 BEz 7498 Hence the angle between z 3 and E is 90 BEz 1502 J 25 CONSTANTCOORDINATE SURFACES Figure 210 For Example 23c 45 PRACTICE EXERCISE 23 Given the vector field H pz cos 0 ap e sin a p a At point 17r30 find a H a b H X a c The vector component of H normal to surface p 1 d The scalar component of H tangential to the plane z 0 Answer a 0433 b 05 ap c 0 ap d 05 EXAMPLE 24 Given a vector field D r sin 0 ar sin 6 cos 0 ae r2a determine a DatPQO 150 330 b The component of D tangential to the spherical surface r 10 at P c A unit vector at P perpendicular to D and tangential to the cone d 150 Solution a At P r 10 6 150 and 0 330 Hence D 10 sin 330 ar sin 150 cos 330 ae 100 a0 5 0043 100 46 SB Coordinate Systems and Transformation b Any vector D can always be resolved into two orthogonal components D D Dn where Dt is tangential to a given surface and Dn is normal to it In our case since ar is normal to the surface r 10 Hence Dn r sin 0 ar 5ar D D Dn 0043a c A vector at P perpendicular to D and tangential to the cone 0 1 5 0 is the same as the vector perpendicular to both D and ae Hence D X afl ar ae a 0 5 0043 100 0 1 0 1 0 0 a r 5a A unit vector along this is 100ar 53A a 09988ar 004993 VlOO2 52 PRACTICE EXERCISE 24 If A 3ar 2ae 6a0 and B 4a 33 determine a A B b A X B c The vector component of A along az at 1 TT3 5ir4 Answer a 6 b 3448 c 0116ar 0201a SUMMARY 1 The three common coordinate systems we shall use throughout the text are the Carte sian or rectangular the circular cylindrical and the spherical 2 A point P is represented as Px y z Pp j z and Pr 6 4 in the Cartesian cylin drical and spherical systems respectively A vector field A is represented as Ax Ay Az or Anx Ayay Azaz in the Cartesian system as Ap A Az or Apap Aa Azaz in the cylindrical system and as An Ae A or Aar Aeae Aa in the spherical system It is preferable that mathematical operations addition subtraction product etc be performed in the same coordinate system Thus point and vector transforma tions should be performed whenever necessary 3 Fixing one space variable defines a surface fixing two defines a line fixing three defines a point 4 A unit normal vector to surface n constant is an REVIEW QUESTIONS 47 PfVJEW QUtSTlONS 21 The ranges of d and as given by eq 217 are not the only possible ones The following are all alternative ranges of 6 and j except a 0 6 2TT 0 ct x b 0 6 2x 0 0 2x c TC 6ir 07r d ir2 0 TT2 0 0 2TT eO0Sx 7r07r f 7 T 0 7 r X 0 7 T 22 At Cartesian point 3 4 1 which of these is incorrect a p 5 b r Jlb c 6 tan1 d t t a n 1 23 Which of these is not valid at point 0 4 0 a b c d a a ar a az 4ay ay A unit normal vector t a b c d ar a a0 none of the above 25 At every point in space a 0 a 1 a True b False 26 If H 4afi 3a0 5az at 1 x2 0 the component of H parallel to surface p 1 is a 4ap b 5az c 3 a d 3a 5a2 e 5arf 3az 48 B Coordinate Systems and Transformation 27 Given G 20ar 50as 4Oa0 at 1 T2 TT6 the component of G perpendicular to surface 6 TT2 is a 20ar b c 0 d 20ar e 40a r 28 Where surfaces p 2 and z 1 intersect is a an infinite plane b a semiinfinite plane c a circle d a cylinder e a cone 29 Match the items in the left list with those in the right list Each answer can be used once more than once or not at all a 0 r4 b 2ir3 c JC 1 0 d r 10 e p 5 f p 3 A g p 10 z h r 4 i r 5 0 x3 f w2 5n73 1 TT6 TT3 i ii iii iv v vi vii viii ix x infinite plane semiinfinite plane circle semicircle straight line cone cylinder sphere cube point 210 A wedge is described by z 0 30 t 60 Which of the following is incorrect a The wedge lies in the x y plane b It is infinitely long c On the wedge 0 p d A unit normal to the wedge is az e The wedge includes neither the xaxis nor the yaxis Answers 21bf 22a 23c 24b 25b 26d 27b 28c 29avi bii ci dx evii fv giii hiv iiii 210b PROBLEMS 49 PROBLEMS 21 Express the following points in Cartesian coordinates aPl60 2 b G2 90 4 cR 45 210 d T4 TT2 TT6 22 Express the following points in cylindrical and spherical coordinates a Pl 4 3 b g3 0 5 c R2 6 0 23 a If V xz xy yz express V in cylindrical coordinates b If U x2 22 3z2 express U in spherical coordinates 24 Transform the following vectors to cylindrical and spherical coordinates a D x zay b E y2 x2ax xyzay x2 Z 2az 25 Convert the following vectors to cylindrical and spherical systems xax yay Aaz a F Vx2 b G x2 y2 xar Vx2 26 Express the following vectors in Cartesian coordinates a A pz2 lap pz cos j a0 b B 2r sin 6 cos j ar r cos 8 cos 6 ae r sin 4 27 Convert the following vectors to Cartesian coordinates a C z sin f ap p cos f a0 2pzaz sin d cos d b D ar r ae Vx2 28 Prove the following a ax ap cos a x a 0 s i n 33 sin 3y 3 0 COS 0 b ax ar sin 6 cos a as cos 0 cos By ar sin 0 sin 0 50 Coordinate Systems and Transformation ay ae cos 6 sin az ar cos 6 a as sin 6 29 a Show that point transformation between cylindrical and spherical coordinates is ob tained using p r sin 9 z r cos 9 4 4 b Show that vector transformation between cylindrical and spherical coordinates is ob tained using or Ar Ae A0 A sin cos 0 sin 0 cos ee e 9 0 cos 0 sin 1 0 cos 8 0 sin 8 9 9 0 1 0 Ar Hint Make use of Figures 25 and 26 210 a Express the vector field H xy2zax x2yzay in cylindrical and spherical coordinates b In both cylindrical and spherical coordinates determine H at 3 4 5 211 Let A p cos 9 ap pz2 sin j az a Transform A into rectangular coordinates and calculate its magnitude at point 3 4 0 b Transform A into spherical system and calculate its magnitude at point 3 4 0 212 The transformation Ap A0 Az Ax Ay Az in eq 215 is not complete Complete it by expressing cos 4 and sin f in terms of x y and z Do the same thing to the transforma tion Ar Ae A A x Ay Az in eq 228 213 In Practice Exercise 22 express A in spherical and B in cylindrical coordinates Evaluate A at 10 TT2 3TI74 and B at 2 TT6 1 PROBLEMS 51 214 Calculate the distance between the following pairs of points a 2 15 and 6 1 2 b 3 T2 1 and 5 3TT2 5 c 10 TT4 3TT4 and 5 x6 774 215 Describe the intersection of the following surfaces 10 a b c d e f X X r p r 2 2 10 g 60 y y e t z 0 5 1 30 40 10 90 216 At point 72 3 4 express az in the spherical system and ar in the rectangular system 217 Given vectors A 2a 4ay 10az and B 5a p a0 3az find a A BatP02 5 b The angle between A and B at P c The scalar component of A along B at P 218 Given that G x y2ax xzay z2 zyaz find the vector component of G along a0 at point P8 30 60 Your answer should be left in the Cartesian system 219 If J r sin 0 cos f ar cos 26 sin 4 ae tan In r a0 at T2 TT2 3 12 determine the vector component of J that is a Parallel to az b Normal to surface 4 37r2 c Tangential to the spherical surface r 2 d Parallel to the line y 2 z 0 220 Let H 5p sin f ap pz cos j a0 2paz At point P2 30 1 find a a unit vector along H b the component of H parallel to ax c the component of H normal to p 2 d the component of H tangential to j 30 52 11 Coordinate Systems and Transformation 221 Let and A pz2 lap pz cos t a o2z2az B r2 cos 0 ar 2r sin 0 a0 At r3 4 1 calculate a A and B b the vector component in cylindrical coordi nates of A along B at T c the unit vector in spherical coordinates perpendicular to both A and B at T 222 Another way of defining a point P in space is r a jS 7 where the variables are por trayed in Figure 211 Using this definition find r a 8 7 for the following points a 2 3 6 b 4 30 3 c 3 30 60 Hint r is the spherical r 0 a 0 7 2ir Figure 211 For Problem 222 223 A vector field in mixed coordinate variables is given by x cos 4 tyz x2 G az 1 j I a P p1 pz Express G completely in spherical system Chapter 3 VECTOR CALCULUS No man really becomes a fool until he stops asking questions CHARLES P STEINMETZ 31 INTRODUCTION Chapter 1 is mainly on vector addition subtraction and multiplication in Cartesian coordi nates and Chapter 2 extends all these to other coordinate systems This chapter deals with vector calculusintegration and differentiation of vectors The concepts introduced in this chapter provide a convenient language for expressing certain fundamental ideas in electromagnetics or mathematics in general A student may feel uneasy about these concepts at firstnot seeing what good they are Such a student is advised to concentrate simply on learning the mathematical techniques and to wait for their applications in subsequent chapters J2 DIFFERENTIAL LENGTH AREA AND VOLUME Differential elements in length area and volume are useful in vector calculus They are defined in the Cartesian cylindrical and spherical coordinate systems A Cartesian Coordinates From Figure 31 we notice that 1 Differential displacement is given by d dx ax dy ay dz az 31 53 54 Vector Calculus A Figure 31 Differential elements in the righthanded Cartesian coordinate system 2 Differential normal area is given by dS dy dz dxdz dzdy a av a and illustrated in Figure 32 3 Differential volume is given by dv dx dy dz 32 33 dy a a dz b ia z dy c Figure 32 Differential normal areas in Cartesian coordinates a dS dy dz a b dS dxdz ay c dS dx dy a 32 DIFFERENTIAL LENGTH AREA AND VOLUME 55 These differential elements are very important as they will be referred to again and again throughout the book The student is encouraged not to memorize them however but to learn to derive them from Figure 31 Notice from eqs 31 to 33 that d and dS are vectors whereas dv is a scalar Observe from Figure 31 that if we move from point P to Q or Q to P for example d dy ay because we are moving in the ydirection and if we move from Q to S or S to Q d dy ay dz az because we have to move dy along y dz along z and dx 0 no movement along x Similarly to move from D to Q would mean that dl dxax dyay dz az The way dS is denned is important The differential surface or area element dS may generally be defined as dS dSan 34 where dS is the area of the surface element and an is a unit vector normal to the surface dS and directed away from the volume if dS is part of the surface describing a volume If we consider surface ABCD in Figure 31 for example dS dydzax whereas for surface PQRS dS dy dz ax because an ax is normal to PQRS What we have to remember at all times about differential elements is d and how to get dS and dv from it Once d is remembered dS and dv can easily be found For example dS along ax can be obtained from d in eq 31 by multiplying the components of d along a and az that is dy dz ax Similarly dS along az is the product of the components of d along ax and ay that is dx dy az Also dv can be obtained from d as the product of the three components of dl that is dx dy dz The idea developed here for Cartesian coordinates will now be extended to other coordinate systems B Cylindrical Coordinates Notice from Figure 33 that in cylindrical coordinates differential elements can be found as follows 1 Differential displacement is given by dl dpap p dcj a 0 dz az 35 2 Differential normal area is given by dS p dj dz ap dp dz a p d4 dp az 36 and illustrated in Figure 34 3 Differential volume is given by dv p dp dcf dz 37 56 Vector Calculus dp 1 pdt dz z Figure 33 Differential elements in cylindrical coordinates As mentioned in the previous section on Cartesian coordinates we only need to re member dl dS and dv can easily be obtained from dl For example dS along az is the product of the components of dl along ap and a that is dp p df az Also dv is the product of the three components of dl that is dp p dj dz C Spherical Coordinates From Figure 35 we notice that in spherical coordinates 1 The differential displacement is dl drar rdd ae r sin 0 df a0 38 b c y Figure 34 Differential normal areas in cylindrical coordinates a dS pdj dz ap b dS dp dz a c dS p df dp az 32 DIFFERENTIAL LENGTH AREA AND VOLUME 57 pd r sin 6 dj Figure 35 Differential elements in the spherical coordinate system 2 The differential normal area is dS r2 sin 6 d6 df a r r sin 6 dr dj a r dr dd aA and illustrated in Figure 36 3 The differential volume is dv r2 sind drdd 39 310 r sin 01 rin a ar r sin 6dcf dr ae r b c w Figure 36 Differential normal areas in spherical coordinates a dS r2 sin 0 dO dj ar b dS r sin 0 dr dj a c dS rdr dd a 58 Vector Calculus Again we need to remember only dl from which dS and dv are easily obtained For example dS along ae is obtained as the product of the components of dl along ar and a that is dr r sin 6 d4 dv is the product of the three components of dl that is dr r dd r sin 6 dt EXAMPLE 31 Consider the object shown in Figure 37 Calculate a The distance BC b The distance CD c The surface area ABCD d The surface area ABO e The surface area A OFD f The volume ABDCFO Solution Although points A B C and D are given in Cartesian coordinates it is obvious that the object has cylindrical symmetry Hence we solve the problem in cylindrical coordinates The points are transformed from Cartesian to cylindrical coordinates as follows A500A500 50 5 0 5 5 0 C0 5 10 C 5 10 D50 1050 10 05 0 10 Figure 37 For Example 31 C0 5 10 5050 32 DIFFERENTIAL LENGTH AREA AND VOLUME 59 a Along BC dl dz hence BC dl dz 10 b Along CD dl pdf and p 5 so CTSIl CD p dj 5 c For ABCD dS pdt dz p 5 Hence ir2 r 10 area ABCD dS I pd ir2 25TT 5 x2 r10 25TT d For ABO dS pdj dp and z 0 so TTr2 r 5 TT2 5 p dcj dp dj p dp 625TT e For AOFD d5 dp dz and 0 0 so area AOFD dp dz 50 f For volume ABDCFO dv pdf dz dp Hence r5 rir2 rlO 10 ir2 5 v Lv p d0 dz dp dz J 625TT PRACTICE EXERCISE 31 Refer to Figure 326 disregard the differential lengths and imagine that the object is part of a spherical shell It may be described as 3 S r 5 60 0 90 45 4 60 where surface r 3 is the same as AEHD surface 0 60 is AFB and surface 45 is AfiCO Calculate a The distance DH b The distance FG c The surface area AEHD d The surface area ABDC e The volume of the object Answer a 07854 b 2618 c 1179 d 4189 e 4276 60 Vector Calculus 33 LINE SURFACE AND VOLUME INTEGRALS The familiar concept of integration will now be extended to cases when the integrand in volves a vector By a line we mean the path along a curve in space We shall use terms such as line curve and contour interchangeably The line integral A d is the integral of ihc tangential component of A along curve L Given a vector field A and a curve L we define the integral fb Adl 311 as the line integral of A around L see Figure 38 If the path of integration is a closed curve such as abca in Figure 38 eq 311 becomes a closed contour integral Adl 312 which is called the circulation of A around L Given a vector field A continuous in a region containing the smooth surface S we define the surface integral or thewx of A through S see Figure 39 as or simply 313 A cos OdS AandS Figure 38 Path of integration of vector field A pathi 33 LINE SURFACE AND VOLUME INTEGRALS 61 surface S Figure 39 The flux of a vector field A through surface S where at any point on S an is the unit normal to S For a closed surface defining a volume eq 313 becomes A dS s 314 which is referred to as the net outward flux of A from S Notice that a closed path defines an open surface whereas a closed surface defines a volume see Figures 311 and 316 We define the integral 315 as the volume integral of the scalar pv over the volume v The physical meaning of a line surface or volume integral depends on the nature of the physical quantity represented by A or pv Note that d dS and dv are all as defined in Section 32 Given that F x2ax xzy y2z calculate the circulation of F around the closed path shown in Figure 310 Solution The circulation of F around path L is given by F d h h V Fdl where the path is broken into segments numbered 1 to 4 as shown in Figure 310 For segment 1 v 0 z jc2ax dxax Notice that d is always taken as along ax so that the direction on segment 1 is taken care of by the limits of integration Thus d x2dx 62 Vector Calculus Figure 310 For Example 32 y For segment 2 x 0 z F y2az d dy ay F d 0 Hence F dl 0 For segment 3 y 1 F x2ax xzy az and dl dxax dz a2 so F d xldx dz But on 3 z that is Jx dz Hence l 3 F d xz 1 dx x 3 For segment 4 x 1 so F ax zay y2az and d dy ay dz az Hence Fdl zrfydz But on 4 z y that is dz rfy so F d 4 J 0 2 3 5 1 6 By putting all these together we obtain 34 DEL OPERATOR Figure 311 For Practice Exercise 32 63 PRACTICE EXERCISE 32 Calculate the circulation of A p cos j ap z sin a2 around the edge L of the wedge defined by 0 p 2 0 t 60 z 0 and shown in Figure 311 Answer 1 34 DEL OPERATOR The del operator written V is the vector differential operator In Cartesian coordinates 316 dx V a Ty a Tz a This vector differential operator otherwise known as the gradient operator is not a vector in itself but when it operates on a scalar function for example a vector ensues The oper ator is useful in denning 1 The gradient of a scalar V written as W 2 The divergence of a vector A written as V A 3 The curl of a vector A written as V X A 4 The Laplacian of a scalar V written as V V Each of these will be denned in detail in the subsequent sections Before we do that it is appropriate to obtain expressions for the del operator V in cylindrical and spherical coordinates This is easily done by using the transformation formulas of Section 23 and 24 64 Vector Calculus To obtain V in terms of p j and z we recall from eq 27 that1 Hence cos q dx dp tan 0 x sin j d p dq d d COS 4 d sin f 1 dy dp p 8t 317 318 Substituting eqs 317 and 318 into eq 316 and making use of eq 29 we obtain V in cylindrical coordinates as 319 V d 3 f dp 3 P d dp a d dz Similarly to obtain V in terms of r 6 and p we use Vx2 y2 z2 tan 0 tan d x to obtain d d cos 6 cos 4 d sin j d sin 6 cos cp 1 dx dr r 30 p dcp 8 d cos 0 sin 0 3 cos 0 5 sin 0 sin p 1 1 dy dr r 80 P dj d d sin 0 d cos 6 dz dr r 80 320 321 322 Substituting eqs 320 to 322 into eq 316 and using eq 223 results in V in spheri cal coordinates 1 d 1 323 Notice that in eqs 319 and 323 the unit vectors are placed to the right of the differen tial operators because the unit vectors depend on the angles A more general way of deriving V V A V X A W and V2V is using the curvilinear coordinates See for example M R Spiegel Vector Analysis and an Introduction to Tensor Analysis New York McGrawHill 1959 pp 135165 35 GRADIENT OF A SCALAR 65 35 GRADIENT OF A SCALAR The gradient of a scalar field V is a vccior thai represents both the magnitude and the direction of the maximum space rale of increase of V A mathematical expression for the gradient can be obtained by evaluating the difference in the field dV between points Pl and P2 of Figure 312 where V V2 and V3 are contours on which V is constant From calculus dV dV dV dx dy dz dx dy dz dV dV dV ax ay az dx ax dy ay dz az ay For convenience let Then dV dV dV x ay dx By dz dV G d G cos 6 dl 324 325 or dV Gcos6 dl 326 Figure 312 Gradient of a scalar y 66 Vector Calculus where d is the differential displacement from P to P2 and 6 is the angle between G and d From eq 326 we notice that dVdl is a maximum when 0 0 that is when d is in the direction of G Hence dV dl dn 327 where dVdn is the normal derivative Thus G has its magnitude and direction as those of the maximum rate of change of V By definition G is the gradient of V Therefore dV 3V dV grad V W ax av a s 5x dy y dz 328 By using eq 328 in conjunction with eqs 316 319 and 323 the gradient of V can be expressed in Cartesian cylindrical and spherical coordinates For Cartesian co ordinates Vy dV a dx dV ay av y dV a dz for cylindrical coordinates vy dV dp i dV a4 3f dV 1 a dz 329 and for spherical coordinates av i ay 1 ay ar a dr r 39 rsinfl df 330 The following computation formulas on gradient which are easily proved should be noted a Vv u vy vu b Vvu yvt uw c d yvc nVnxVV 331a 331b 331c 331d where U and V are scalars and n is an integer Also take note of the following fundamental properties of the gradient of a scalar field V 1 The magnitude of Vy equals the maximum rate of change in V per unit distance 2 Vy points in the direction of the maximum rate of change in V 3 Vy at any point is perpendicular to the constant V surface that passes through that point see points P and Q in Figure 312 35 GRADIENT OF A SCALAR 67 4 The projection or component of VV in the direction of a unit vector a is W a and is called the directional derivative of V along a This is the rate of change of V in the direction of a For example dVdl in eq 326 is the directional derivative of V along PiP2 in Figure 312 Thus the gradient of a scalar function V provides us with both the direction in which V changes most rapidly and the magnitude of the maximum directional derivative of V 5 If A VV V is said to be the scalar potential of A EXAMPLE 33 Find the gradient of the following scalar fields a V ez sin 2x cosh y b U p2z cos 2t c W lOrsin20cos Solution a dvt dx dV dV v 2e zcos 2x cosh y ax e zsin 2x sinh y ay e Jsin 2x cosh y az 1 dU dU P d0 dz a7 2pz cos 2f ap 2pz sin 2 a p cos 20 az aw i aw i aw dr a r 36 rsinfl 90 10 sin2 6 cos 0 ar 10 sin 26 cos 0 a 10 sin 0 sin 01 PRACTICE EXERCISE 33 Determine the gradient of the following scalar fields a U x2y xyz b V pz sin t z2 cos2 t p2 c cos 6 sin 0 In r r2j Answer a y2x zax xx zay xyaz b z sin 0 2pap z cos 0 sin p sin t 2z cos 2 jaz cos 0 sin i c 2r0 Jar cot 0 I cos q In r r cosec 0 sin 9 sin In r ae 68 i EXAMPLE 34 Vector Calculus Given W x2y2 xyz compute VW and the direction derivative dWdl in the direction 3ax 4ay 12az at 210 Solution dW dW dW VW a av a dx x dy y dz z 2xyz yzax 2xzy xzay xyaz At 2 10 VW Hence ay 2az PRACTICE EXERCISE 34 Given P xy yz xz find gradient 0 at point 12 3 and the directional deriv ative of P at the same point in the direction toward point 344 Answer 5ax 4a 3az 7 EXAMPLE 35 Find the angle at which line x y 2z intersects the ellipsoid x2 y2 2z2 10 Solution Let the line and the ellipsoid meet at angle j as shown in Figure 313 The line x y 2z can be represented by rX 2Xa 2X3 Xaz where X is a parameter Where the line and the ellipsoid meet 2X2 2X2 2X2 10 X 1 Taking X 1 for the moment the point of intersection is x y z 221 At this point r 2a 2a az ellipsoid Figure 313 For Example 35 plane of intersection of a line with an ellipsoid 36 DIVERGENCE OF A VECTOR AND DIVERGENCE THEOREM 69 The surface of the ellipsoid is defined by fxyzx2 y2 2z2W The gradient ofis Vf2xax 2yay 4zaz At 221 V 4ax 4ay 4ar Hence a unit vector normal to the ellipsoid at the point of intersection is v a a a V3 Taking the positive sign for the moment the angle between an and r is given by cos 6 an r 2 2 1 r V W 9 3V3 n p Hence j 7421 Because we had choices of or for X and an there are actually four possible angles given by sin i 53 V3 PRACTICE EXERCISE 35 Calculate the angle between the normals to the surfaces x y z 3 and x log z y2 4 at the point of intersection 1 21 Answer 734 36 DIVERGENCE OF A VECTOR AND DIVERGENCE THEOREM From Section 33 we have noticed that the net outflow of the flux of a vector field A from a closed surface S is obtained from the integral A dS We now define the divergence of A as the net outward flow of flux per unit volume over a closed incremental surface The divergence of A at a given point P is ihc outward lux per unii volume as the volume shrinks about P Hence div A V A lim Av0 AdS Av 332 70 Vector Calculus a b P c Figure 314 Illustration of the divergence of a vector field at P a positive divergence b negative divergence c zero divergence where Av is the volume enclosed by the closed surface S in which P is located Physically we may regard the divergence of the vector field A at a given point as a measure of how much the field diverges or emanates from that point Figure 314a shows that the diver gence of a vector field at point P is positive because the vector diverges or spreads out at P In Figure 314b a vector field has negative divergence or convergence at P and in Figure 314c a vector field has zero divergence at P The divergence of a vector field can also be viewed as simply the limit of the fields source strength per unit volume or source density it is positive at a source point in the field and negative at a sink point or zero where there is neither sink nor source We can obtain an expression for V A in Cartesian coordinates from the definition in eq 332 Suppose we wish to evaluate the divergence of a vector field A at point Pxoyo zo we let the point be enclosed by a differential volume as in Figure 315 The surface integral in eq 332 is obtained from A dS M A dS 333 S front back left right Aop bottorr A threedimensional Taylor series expansion of Ax about P is BAr Axx v z Axxo yo Zo dx z Zo dz yyo dy 334 higherorder terms For the front side x xo dx2 and dS dy dz ax Then dx dA A dS dy dz front dx xo yo zo 2 dx higherorder terms For the back side x x0 dx2 dS dy dzax Then dx dA L 2 dx back L A dS dydzl Axx0 yo zo higherorder terms 36 DIVERGENCE OF A VECTOR AND DIVERGENCE THEOREM 71 top side Figure 315 Evaluation of V A at point Px0 Jo Zo front side i dz dy Jx right side dA A dS I A dS dx dy dz Hence front back By taking similar steps we obtain and left A dS AdS dxdydz right dx dAy dy AdS A dS dx dy dz dz higherorder terms 335 higherorder terms 336 higherorder terms 337 top bottom Substituting eqs 335 to 337 into eq 333 and noting that Av dx dy dz we get A dS s AVO Av Mi dz 338 because the higherorder terms will vanish as Av 0 Thus the divergence of A at point Pxo yo zo in a Cartesian system is given by 339 Similar expressions for V A in other coordinate systems can be obtained directly from eq 332 or by transforming eq 339 into the appropriate coordinate system In cylindrical coordinates substituting eqs 215 317 and 318 into eq 339 yields VA 1 dA6 dA f P dp 340 72 Vector Calculus Substituting eqs 228 and 320 to 322 into eq 339 we obtain the divergence of A in spherical coordinates as 341 V A 1 d 2 dr A 1 r sin i e sin 0 H 1 rsind df Note the following properties of the divergence of a vector field 1 It produces a scalar field because scalar product is involved 2 The divergence of a scalar V div V makes no sense 3 V A B V A V B 4 V VA VV A A VV From the definition of the divergence of A in eq 332 it is not difficult to expect that 342 This is called the divergence theorem otherwise known as the GaussOstrogradsky theorem Hie divergence theorem stales thai Ihe total mil ward llux of a vector licld A through ihc closed surface V is ihe same as the volume integral of the divergence of A To prove the divergence theorem subdivide volume v into a large number of small cells If the Mi cell has volume Avk and is bounded by surface Sk AdS Avt 343 Since the outward flux to one cell is inward to some neighboring cells there is cancellation on every interior surface so the sum of the surface integrals over Sks is the same as the surface integral over the surface 5 Taking the limit of the righthand side of eq 343 and incorporating eq 332 gives AdS VAdv 344 which is the divergence theorem The theorem applies to any volume v bounded by the closed surface S such as that shown in Figure 316 provided that A and V A are continu Volume v 36 DIVERGENCE OF A VECTOR AND DIVERGENCE THEOREM M 73 Figure 316 Volume v enclosed by surface S Closed Surface S EXAMPLE 36 ous in the region With a little experience it will soon become apparent that volume inte grals are easier to evaluate than surface integrals For this reason to determine the flux of A through a closed surface we simply find the righthand side of eq 342 instead of the lefthand side of the equation Determine the divergence of these vector fields a P x2yz ax xz az b Q p sin 0 ap p2z a cos j az c T z cos d ar r sin 6 cos j a cos Solution a V P Px Pv Pz dx x dy y dz z dx x2yz dx 2xyz x dy b V Q pQp Q Qz P dp p Bj dz 1 S 1 d 2 d P sin 0 p z z cos t P dp p dt dz 2 sin cos j c V T 3 r2Tr r dr r sin 8 Te sin 6 B6 r sin 9 1 d 1 cos 6 r2 dr d o l a r sin 9 cos H cos 9 r sin 0 30 r sin 0 dd 2r sin 0 cos 6 cos t 0 74 Vector Calculus PRACTICE EXERCISE 36 Determine the divergence of the following vector fields and evaluate them at the specified points a A yzx 4xyy ja z at 1 2 3 b B pz sin t ap 3pz2 cos a at 5 TT2 1 c C 2r cos 0 cos j ar r y at 1 ir6 ir3 Answer a Ax 4 b 2 3zz sin f 1 c 6 cos 6 cos 2598 EXAMPLE 37 If Gr lOe 2zaP aj determine the flux of G out of the entire surface of the cylinder p l 0 z 1 Confirm the result using the divergence theorem Solution If P is the flux of G through the given surface shown in Figure 317 then where ft Vfc and Ys are the fluxes through the top bottom and sides curved surface of the cylinder as in Figure 317 For Ytz dS pdp dj az Hence 10e dp dt Figure 317 For Example 37 J 37 CURL OF A VECTOR AND STOKESS THEOREM For Yb z 0 and dS pdp djaz Hence G dS 1 0 7 T 75 10epdpdt 1027ry For Ys p 1 dS p dz dj a Hence I G dS Kte2zp2 dz d4 10l22ir 2 0 J 2z 2 10TT1 e2 Thus IOTT 10TT1 e2 0 Alternatively since S is a closed surface we can apply the divergence theorem But P GdS VGrfv P ap p 2l0e 2 z 20e2z 0 P dp showing that G has no source Hence V G dv 0 PRACTICE EXERCISE 37 Determine the flux of D p2 cos2 0 a z sin 0 a over the closed surface of the cylinder 0 j I p 4 Verify the divergence theorem for this case Answer Air 37 CURL OF A VECTOR AND STOKESS THEOREM In Section 33 we defined the circulation of a vector field A around a closed path L as the integral LA d 76 Vector Calculus The curl of A is an axial or rotational vector whose magnitude is the maximum cir culation of A per unit area as the area lends to zero and whose direction is the normal direction of the area when the area is oriented so as to make the circulation maximum That is curl A V X A lim A d 345 where the area AS is bounded by the curve L and an is the unit vector normal to the surface AS and is determined using the righthand rule To obtain an expression for V X A from the definition in eq 345 consider the dif ferential area in the zplane as in Figure 318 The line integral in eq 345 is obtained as A dl jAdl 346 JL Lb be Kd dj We expand the field components in a Taylor series expansion about the center point PXoyozo as in eq 334 and evaluate eq 346 On side ab d dyay and z zo dzli so ab dz A d dy Axo yo zo On side be d dz az and y yo dy2 so Adl dz be zxo yo Zo 2 dz dy dAz 2 dy On side cd d dy ay and z z0 dzJ2 so A d dy Ayxo yo zo y j 347 348 349 Figure 318 Contour used in evaluating the component of V X A at point Px0 y0 z0 2Because of its rotational nature some authors use rot A instead of curl A J 37 CURL OF A VECTOR AND STOKESS THEOREM 17 On side da d dz Skz and y yo dy2 so da A d dz Azxo yo Zo dydAz 2 dy 350 Substituting eqs 347 to 350 into eq 346 and noting that AS dy dz we have A d bA7 dAv lim AS or curl Ax dy dz dA7 dAy By dz The y and xcomponents of the curl of A can be found in the same way We obtain dA7 351 curl Ay curl Az dz dx AAj dAx dx dy 352a 352b The definition of V X A in eq 345 is independent of the coordinate system In Cartesian coordinates the curl of A is easily found using VX A A A A dx dy dz A A A nx riy nz 353 V X A dy ldAy dx dAJ dy J z aA zl dx yy 354 By transforming eq 354 using point and vector transformation techniques used in Chapter 2 we obtain the curl of A in cylindrical coordinates as ap p a A A dp d 78 Vector Calculus or 355 and in spherical coordinates as V X A ar r ae r sin 0 a dr dd d4 Ar rAe r sin 0 yL or V X A rsinflL 3 8 a r l 1 dAr 3M 0 r s i n 0 9 ar 1 i W M r ar ae a 356 Note the following properties of the curl 1 The curl of a vector field is another vector field 2 The curl of a scalar field V V X V makes no sense 3 VXA B V X A V X B 4 V X A X B AV B BV A B VA A VB 5 V X VA VV X A VV X A 6 The divergence of the curl of a vector field vanishes that is V V X A 0 7 The curl of the gradient of a scalar field vanishes that is V X VV 0 Other properties of the curl are in Appendix A The physical significance of the curl of a vector field is evident in eq 345 the curl provides the maximum value of the circulation of the field per unit area or circulation density and indicates the direction along which this maximum value occurs The curl of a vector field A at a point P may be regarded as a measure of the circulation or how much the field curls around P For example Figure 319a shows that the curl of a vector field around P is directed out of the page Figure 319b shows a vector field with zero curl Figure 319 Illustration of a curl a curl at P points out of the page b curl at P is zero a b 37 CURL OF A VECTOR AND STOKESS THEOREM 79 dS Figure 320 Determining the sense of d and dS involved in Stokess theorem Closed Path L Surfaces J Also from the definition of the curl of A in eq 345 we may expect that A d VXAiS 357 This is called Stokess theorem Stokess theorem suites that ihe circulation of a vcclor Meld A around a closed pain is equal lo the surface integral ollhe curl of A over the open surface S bounded by see Figure 320 provided that A and V X A are continuous on V The proof of Stokess theorem is similar to that of the divergence theorem The surface S is subdivided into a large number of cells as in Figure 321 If the Ath cell has surface area ASk and is bounded by path Lk Adl A d l Adl AS ASt 358 Figure 321 Illustration of Stokess theorem 80 Vector Calculus As shown in Figure 321 there is cancellation on every interior path so the sum of the line integrals around Lks is the same as the line integral around the bounding curve L There fore taking the limit of the righthand side of eq 358 as ASk 0 and incorporating eq 345 leads to A dl V X A dS which is Stokess theorem The direction of dl and dS in eq 357 must be chosen using the righthand rule or righthanded screw rule Using the righthand rule if we let the fingers point in the direc tion ofdl the thumb will indicate the direction of dS see Fig 320 Note that whereas the divergence theorem relates a surface integral to a volume integral Stokess theorem relates a line integral circulation to a surface integral EXAMPLE 38 Determine the curl of the vector fields in Example 36 Solution dPz dPy dPx dP fdPy dPx a V X P U i a V 3j 3z dz dx J dx dy a7 0 0 ax x2y z x2y zay x2zaz 0 x2z az y sin 0 p2 ap 0 08 3p2z p cos taz z sin t p3ap 3pz cos c V X T r sin 0 L 30 30 i r i d ar r sin I r cos 0 sin 0 r sin 0 cos 0 ar 30 30 1 3 cos0 3 r c o s sin 0 30 r2 3r dr r sin 0 cos 0 d COS0 cos 26 r sin 0 sin 0a H 0 cos 6ae r sin 0 1 2r sin 0 cos 0 H sin 9 5 r V r 2 cos 20 cos 0 rr sin 0 I ar Vrsin0 r 1 2 cos 0 sin 0 a0 37 CURL OF A VECTOR AND STOKESS THEOREM 81 PRACTICE EXERCISE 38 Determine the curl of the vector fields in Practice Exercise 36 and evaluate them at the specified points Answer a ax yay 4y zaz ax 2 1 la b 6pz cos 0 ap p sin 0 a 6z lz cos 0 a2 5a cot 6 I 3 c ar 2 cot 0 sin 0 1732 ar 45 ae 05 a0 2 sin 6 cos 0 a EXAMPLE 39 If A p cos 0 ap sin 0 a evaluate A d around the path shown in Figure 322 Confirm this using Stokess theorem Solution Let Adl I I A J 1 b c d where path L has been divided into segments ab be cd and da as in Figure 322 Along ab p 2 and d p dt a0 Hence kd 30 p sin 0 dj 2cos 0 60 30 V3 1 60 5 t I is 7 Xb 0 I 1 L r C 1 1 Figure 322 For Example 39 82 H Vector Calculus Along be t 30 and d dp ap Hence A d p cos dp cos 30 Along cd p 5 and dl p df a Hence 2lV3 60 p sin jdj 5 c o s f 60 fV30 30 Along da j 60 and dl dp ap Hence A dl p cos jdp cos 60 2 21 Putting all these together results in 5V3 5 21 2 2 4 V 3 1 4941 4 Using Stokess theorem because L is a closed path But dS pddpaz and 1 3A V X A a j 0 0 A dl V X A I dAo dAz ld aj I aT dAn Hence V X A 60 z5 60 sin i 30 J2 dp p sin 4 az 1 p sin fi p dp dt 5 cos 27 V3 1 4941 38 LAPLACIAN OF A SCALAR 83 PRACTICE EXERCISE 39 Use Stokess theorem to confirm your result in Practice Exercise 32 Answer 1 EXAMPLE 310 For a vector field A show explicitly that V V X A 0 that is the divergence of the curl of any vector field is zero Solution This vector identity along with the one in Practice Exercise 310 is very useful in EM For simplicity assume that A is in Cartesian coordinates dx dy dz dx dy dz dx dy dz A A A dAz dy dAy dz d fdAz dAy dx dy dz dAy dA dx dz y dx dy d dA v dA d2Az d2Av V dx dz d2Ax dx dx 2AV d 2 A dx dy dx dz 0 dy dx dy dz dz dx dz 3y because d2A7 d2A7 dx dy dy dx and so on PRACTICE EXERCISE 310 For a scalar field V show that V X W 0 that is the curl of the gradient of any scalar field vanishes Answer Proof 8 LAPLACIAN OF A SCALAR For practical reasons it is expedient to introduce a single operator which is the composite of gradient and divergence operators This operator is known as the Laplacian 84 Vector Calculus The Laplacian of u scalar field V written as V2V is the divergence of the gradient of V Thus in Cartesian coordinates Laplacian V V W V2V d dx d d dz 01 dx OX By dV dz a7 359 that is 360 Notice that the Laplacian of a scalar field is another scalar field The Laplacian of V in other coordinate systems can be obtained from eq 360 by transformation In cylindrical coordinates 361 and in spherical coordinates 1 d 2dV r2 dr dr J r2 1 sin d f V ou V dV dd r2 1 sin2 d2V dj 2 362 A scalar field V is said to be harmonic in a given region if its Laplacian vanishes in that region In other words if V 2V0 363 is satisfied in the region the solution for V in eq 363 is harmonic it is of the form of sine or cosine Equation 363 is called Laplaces equation Solving this equation will be our major task in Chapter 6 We have only considered the Laplacian of a scalar Since the Laplacian operator V2 is a scalar operator it is also possible to define the Laplacian of a vector A In this context V2A should not be viewed as the divergence of the gradient of A which makes no sense Rather V2A is defined as the gradient of the divergence of A minus the curl of the curl of A That is V2A VV A V X V X A 364 J 38 LAPLACIAN OF A SCALAR 85 This equation can be applied in finding V2A in any coordinate system In the Cartesian system and only in that system eq 364 becomes V2A 365 EXAMPLE 311 Find the Laplacian of the scalar fields of Example 33 that is a V ez sin 2x cosh y b U p2z cos 20 c W lOr sin2 0 cos 0 Solution The Laplacian in the Cartesian system can be found by taking the first derivative and later the second derivative d2V t2V r2V dx 5y dz 2ez cos 2x cosh y H e z cos 2x sinh y dx dy H e z sin 2x cosh y dz Aez sin 2xcoshy ez sin 2xcoshy ez sin 2xcoshy 2ez sin 2x cosh y 1 d2U d2U b P dp 2p2z cos 20 4p2z cos 20 0 P dp pz Az cos 20 Az cos 20 0 1 r2 sin 6 d r dr 1 r 10r2 sin2 0 cos 0 lOr sin2 0 cos 0 r2 sin2 0 d ndW d9 d2W r2sin20 dct2 1 f 10r sin 20 sin 6 cos 0 sin0 d6 20 sin2 0 cos 0 20r cos 20 sin 6 cos 0 r2 sin 0 lOr sin 26 cos 0 cos 0 10 cos 0 r2 sin 6 r 2 sin2 0 2 cos 20 2 cos2 0 1 1 2 cos 26 86 Vector Calculus PRACTICE EXERCISE 311 Determine the Laplacian of the scalar fields of Practice Exercise 33 that is a V x2y xyz b V pz sin 4 z2 cos2 l p2 c cos 0 sin 4 In r r2 j 2z2 1 Answer a 2y b 4 2 cos 0 5cos2 c z cos 0 sin 1 21nr cosec2 In r 6 39 CLASSIFICATION OF VECTOR FIELDS A vector field is uniquely characterized by its divergence and curl Neither the diver gence nor curl of a vector field is sufficient to completely describe the field All vector fields can be classified in terms of their vanishing or nonvanishing divergence or curl as follows a V A 0 V X A 0 b V A 0 V X A 0 c V A 0 V X A 0 d V A O V X A O Figure 323 illustrates typical fields in these four categories J 1 O I a b c d Figure 323 Typical fields with vanishing and nonvanishing divergence or curl a A kax V A 0 V X A 0 b A kr V A 3k V X A 0 c A k X r V A 0 V X A 2k d A k X r cr V A 3c V X A 2k 39 CLASSIFICATION OF VECTOR FIELDS 87 A vector field A is said to be solenoidal or divergenceless if V A 0 Such a field has neither source nor sink of flux From the divergence theorem A dS V A dv 0 366 Hence flux lines of A entering any closed surface must also leave it Examples of sole noidal fields are incompressible fluids magnetic fields and conduction current density under steady state conditions In general the field of curl F for any F is purely solenoidal because V V X F 0 as shown in Example 310 Thus a solenoidal field A can always be expressed in terms of another vector F that is if then VA 0 A JS 0 and F V X A 367 A vector field A is said to be irrotational or potential if V X A 0 That is a curlfree vector is irrotational3 From Stokess theorem V X A dS 4 A d 0 368 Thus in an irrotational field A the circulation of A around a closed path is identically zero This implies that the line integral of A is independent of the chosen path Therefore an ir rotational field is also known as a conservative field Examples of irrotational fields include the electrostatic field and the gravitational field In general the field of gradient V for any scalar VO is purely irrotational since see Practice Exercise 310 V X VV 0 369 Thus an irrotational field A can always be expressed in terms of a scalar field V that is V X A 0 A d 0 and one reason to use the term irrotational A vy 370 For this reason A may be called potential field and V the scalar potential of A The neg ative sign in eq 370 has been inserted for physical reasons that will become evident in Chapter 4 In fact curl was once known as rotation and curl A is written as rot A in some textbooks This is 88 Vector Calculus A vector A is uniquely prescribed within a region by its divergence and its curl If we let and V A P v V X A ps 371a 371b pv can be regarded as the source density of A and ps its circulation density Any vector A satisfying eq 371 with both pv and ps vanishing at infinity can be written as the sum of two vectors one irrotational zero curl the other solenoidal zero divergence This is called Helmholtz s theorem Thus we may write A V V V X B 372 If we let A W and As V X B it is evident from Example 310 and Practice Exer cise 310 that V X A 0 and V X As 0 showing that A is irrotational and As is sole noidal Finally it is evident from eqs 364 and 371 that any vector field has a Lapla cian that satisfies V2A Vpv V X ps 373 EXAMPLE 312 Show that the vector field A is conservative if A possesses one of these two properties a The line integral of the tangential component of A along a path extending from a point P to a point Q is independent of the path b The line integral of the tangential component of A around any closed path is zero Solution a If A is conservative V X A 0 so there exists a potential V such that dV dV dV A W 1 ar a a Hence Q Adl dx dV dx dy dV dz dV dz or dV dx dVdy dVdz dx ds dy ds dz ds J QdV Q ds dV ds J A d VP VQ I SUMMARY 89 showing that the line integral depends only on the end points of the curve Thus for a con Q servative field A d is simply the difference in potential at the end points b If the path is closed that is if P and Q coincide then Adl VP VP 0 PRACTICE EXERCISE 312 Show that B y z cos xzax xay x cos xz az is conservative without comput ing any integrals Answer Proof 1 The differential displacements in the Cartesian cylindrical and spherical systems are respectively d dxax dy ay dz az d dp ap p defy a dz az d dr ar r dd ae r sin 6 dj a Note that d is always taken to be in the positive direction the direction of the dis placement is taken care of by the limits of integration 2 The differential normal areas in the three systems are respectively dS dy dz six y dx dy az dS pdcj dz ap dp dz a p dp dj az dS r 2 sin 6 dd dj ar r sin 8 dr dj a r dr dd a Note that dS can be in the positive or negative direction depending on the surface under consideration 3 The differential volumes in the three systems are dv dxdy dz dv p dp dt dz dv r2 sin 6 dr dd d 90 M Vector Calculus 4 The line integral of vector A along a path L is given by JL A d If the path is closed the line integral becomes the circulation of A around L that is jL A d 5 The flux or surface integral of a vector A across a surface S is defined as Js A dS When the surface S is closed the surface integral becomes the net outward flux of A across S that is cf A dS 6 The volume integral of a scalar pv over a volume v is defined as pv dv 7 Vector differentiation is performed using the vector differential operator V The gradi ent of a scalar field V is denoted by V V the divergence of a vector field A by V A the curl of A by V X A and the Laplacian of V by V2V 8 The divergence theorem j5 A dS v V A dv relates a surface integral over a closed surface to a volume integral 9 Stokess theorem fL A d sV X A dS relates a line integral over a closed path to a surface integral 10 If Laplaces equation V2V 0 is satisfied by a scalar field V in a given region V is said to be harmonic in that region 11 A vector field is solenoidal if V A 0 it is irrotational or conservative if V XA 0 12 A summary of the vector calculus operations in the three coordinate systems is pro vided on the inside back cover of the text 13 The vector identities V V X A 0 and V X VV 0 are very useful in EM Other vector identities are in Appendix A 10 REVIEW QUESTIONS 31 Consider the differential volume of Figure 324 Match the items in the left column with those in the right column a d from A to B i dydzax b dlfromAtoD ii dxdzay c d from A to iii dx dy az d dS for face ABCD iv dxdyaz e dS for face AEHD v dxax f dS for face DCGH vi dy ay g dS for face ABFE vii dzaz 32 For the differential volume in Figure 325 match the items in the left list with those in the right list a d from to A i p d4 dz ap b dlfromBtoA ii dpdza c d from D to A iii p dp dj az d dS for face ABCD iv pdpdfaz e dS for face AEHD v dp ap f dS for face ABFE vi pdta g dS for face DCGH vii dz az REVIEW QUESTIONS F Figure 324 For Review Question 31 91 33 A differential volume in spherical coordinates is shown in Figure 326 For the volume element match the items in the left column with those in the right column a dlfromAtoD b lfromtoA c dlfmmAtoB d dS for face EFGH e dS for face AEHD f dS for face ABFE i r2 sin ii r sin drdjae iii r dr d6 a0 iv drar v rd6ae vi r sin dj 34 If r xax yay zaz the position vector of point x y z and r r which of the fol lowing is incorrect a Vr rlr b V r 1 c V2r r 6 d V X r 0 35 Which of the following is a meaningless combination a graddiv b divcurl c curl grad d curl grad e div curl F Figure 325 For Review Question 32 92 Vector Calculus F Figure 326 For Review Question 33 and also for Practice Exercise 31 36 Which of the following is zero a grad div b div grad c curl grad d curl curl 37 Given field A 3x2yz ax x3z ay x3y 2zaz it can be said that A is a Harmonic b Divergenceless c Solenoidal d Rotational e Conservative 38 The surface current density J in a rectangular waveguide is plotted in Figure 327 It is evident from the figure that J diverges at the top wall of the guide whereas it is diver genceless at the side wall a True b False 39 Stokess theorem is applicable only when a closed path exists and the vector field and its derivatives are continuous within the path a True b False c Not necessarily 310 If a vector field Q is solenoidal which of these is true a L Q d 0 b s Q dS 0 c V X Q 0 d V X Q 0 e V2Q 0 PROBLEMS H 93 Figure 327 For Review Question 38 u Answers 31avi bvii cv di eii fiv giii 32avi bv cvii dii ei fOv giii 33av bvi civ diii ei fii 34b 35c 36c 37e 38a 39a 310b Using the differential length dl find the length of each of the following curves a p 3 TT4 0 ir2 z constant b r 10 3OO06O c r 4 30 6 90 constant 32 Calculate the areas of the following surfaces using the differential surface area dS a p 2 0 z 5 ir3 vr2 b z 1 1 p 3 0 4 TT4 c r 10 TT4 6 2TT3 0 j 2ir d 0 r 4 60 6 90 t constant 33 Use the differential volume dv to determine the volumes of the following regions a 0 x 1 1 y 2 3 z 3 b 2 p 5 TI73 7T 1 z 4 c 1 r 3 TT2 0 2x3 TT6 x2 34 Given that ps x2 xy calculate SsPsdS over the region y x2 0 x 1 35 Given that H x ax y a evaluate H dl where L is along the curve y x from 36 Find the volume cut from the sphere radius r a by the cone 6 a Calculate the volume when a x3 and a TT2 94 Vector Calculus B 37 If the integral I F dl is regarded as the work done in moving a particle from A to B A find the work done by the force field F 2xy ax x2 z2 ay 3xz2 az on a particle that travels from A0 0 0 to B2 1 3 along a The segment 0 0 0 0 1 0 2 1 0 2 1 3 b The straight line 0 0 0 to 2 1 3 38 If H x yax x2 zyay 5yz az evaluate H dl along the contour of Figure 328 39 If V x yz evaluate p V dS where S is the surface of the cylindrical wedge defined by 0 4 7i72 0 z 2 and dS is normal to that surface 310 Let A 2xyax xzay yaz Evaluate J A dv over a a rectangular region 0 2 J 2 O S J 2 0 Z S 2 b a cylindrical region p 3 0 s j 5 c a spherical region r 4 311 The acceleration of a particle is given by a 24az ms2 The initial position of the particle is r 0 0 0 while its initial velocity is v 2a 5az ms a Find the position of the particle at time t 1 b Determine the velocity of the particle as a func tion of t 312 Find the gradient of the these scalar fields a U 4xz2 3yz b W 2pz2 1 cos t c H r2 cos 6 cos t Figure 328 For Problem 38 PROBLEMS 95 313 Determine the gradient of the following fields and compute its value at the specified point a V eax3y cos 5z 01 02 04 b T 5pe2z sin 2 ir3 0 314 Determine the unit vector normal to Sx y z x2 y2 z at point 1 30 315 The temperature in an auditorium is given by T x2 y2 z A mosquito located at 1 1 2 in the auditorium desires to fly in such a direction that it will get warm as soon as possible In what direction must it fly 316 Find the divergence and curl of the following vectors a A e ax sin xy ay cos2 xz az b B pz2 cos 4 ap z sin2 az c C r cos 6 ar sin 0 ae 2r2 sin 8 a 317 Evaluate V X A and V V X A if a A x2yax y2zay 2xzaz b A p2zap p 3pz sin 4 cos f c A ar 318 The heat flow vector H kWT where T is the temperature and k is the thermal conduc tivity Show that where KX x y T 50 sin cosh 2 2 then V H 0 319 a Prove that V VA VV A A VV where V is a scalar field and A is a vector field b Evaluate V VA when A 2xay 3yay 4zaz and V xyz 320 a Verify that V X VA VV X A VV X A where V and A are scalar and vector fields respectively b Evaluate V X VA when V r and A r cos 8 ar r sin 8 ae sin 8 cos a0 r 321 lfU xz x2y y V evaluate div grad U 96 Vector Calculus 322 Show that V In p V X jaz 323 Prove that V0 V X r V 0 sin 8J 324 Evaluate VV V W and V X VV if a V 3x2y xz b V pz cos 0 c V Ar2 cos d sin 0 325 If r xax yay zaz and T 2zyax xy2ay x2yzaz determine a V rT b r VT c VrrT d r Vr2 326 If r xax yay zaz is the position vector of point x y z r r and n is an integer show that a V rr n 3r b V X rr 0 327 If r and r are as defined in the previous problem prove that a V In r r b V2 In r 328 For each of the following scalar fields find V2V a Vi x3 y3 z3 b V2 pz2 sin 20 c V3 r cos 6 sin j 329 Find the Laplacian of the following scalar fields and compute the value at the specified point a U x3y2exz 1 11 b V p2zcos 4 sin 0 5 TT6 2 c W er sin 6 cos 0 1 TT3 TT6 330 If V x2y2z2 and A x2y ax xz3 ay y2z2 az find a V2V b V2A c grad div A d curl curl A J PROBLEMS 97 Figure 329 For Problem 331 331 Given that F x2y ax y ay find a fL F d where L is shown in Figure 329 b JsVxFfi where S is the area bounded by L c Is Stokess theorem satisfied 332 Let D 2pz p cos2 jaz Evaluate a DdS b V Ddv over the region defined by 0 s p 5 1 z 1 0 2TT 333 If F x y z2 1 a find F dS where S is defined by p 2 0 z 2 0 t 2TT 334 a Given that A xyax yzay xzaz evaluate cfs A dS where S is the surface of the cube defined b y O x l 0 v l 0 z 1 b Repeat part a if S remains the same but A yzax xzay xyaz 335 Verify the divergence theorem ArfS V Adv for each of the following cases a A xy2ax y3ay y2zaz and S is the surface of the cuboid defined by 0 x 1 0 y 10 z 1 b A 2pzap 3z sin 4 a Ap cos 4 az and S is the surface of the wedge 0 p 2 0 0 45 0 z 5 c A r2ar r sin 6 cos a and S is the surface of a quarter of a sphere defined by 0 r 3 0 t TT2 0 0 x2 98 Vector Calculus a Figure 330 For Problem 337 336 The moment of inertia about the zaxis of a rigid body is proportional to x2 y2 dx dy dz Express this as the flux of some vector field A through the surface of the body 337 Let A p sin ap p2 a0 Evaluate L A d given that a L is the contour of Figure 330a b L is the contour of Figure 33Ob 338 Calculate the total outward flux of vector F p2 sin 0 ap z cos f a0 pzz through the hollow cylinder defined b y 2 p 3 0 z 5 339 Find the flux of the curl of field T r cos 6 ar r sin 6 cos a cos 6 su r through the hemisphere r 4 z 0 340 A vector field is given by Q Vx2 x yax x yay Evaluate the following integrals a JL Q d where L is the circular edge of the volume in the form of an icecream cone shown in Figure 331 b Jj V X Q S where S is the top surface of the volume c Js2 V X Q dS where S2 is the slanting surface of the volume Double asterisks indicate problems of highest difficulty J PROBLEMS 99 Figure 331 Volume in form of icecream cone for Problem 340 y d JSl Q dS e fs2QdS f Jv V Q dv How do your results in parts a to f compare 341 A rigid body spins about a fixed axis through its center with angular velocity to If u is the velocity at any point in the body show that to 12 V X u 342 Let U and V be scalar fields show that UVVdl t VVUdl 343 Show that f i c rnx dr n 3 where r r and n are as defined in Problem 326 344 Given the vector field G 16xy a Is G irrotational or conservative T axy 0z3ax 3x2 is irrotational determine a 0 and y Find V T at 2 1 0 b Find the net flux of G over the cube 0 x y z 1 c Determine the circulation of G around the edge of the square z 0 0 x y 1 Assume anticlockwise direction 3xZ 2 yaz PART 2 ELECTROSTATICS Chapter 4 ELECTROSTATIC FIELDS Take risks if you win you will be happy if you lose you will be wise PETER KREEFT 41 INTRODUCTION Having mastered some essential mathematical tools needed for this course we are now prepared to study the basic concepts of EM We shall begin with those fundamental con cepts that are applicable to static or timeinvariant electric fields in free space or vacuum An electrostatic field is produced by a static charge distribution A typical example of such a field is found in a cathoderay tube Before we commence our study of electrostatics it might be helpful to examine briefly the importance of such a study Electrostatics is a fascinating subject that has grown up in diverse areas of application Electric power transmission Xray machines and lightning protection are associated with strong electric fields and will require a knowledge of elec trostatics to understand and design suitable equipment The devices used in solidstate electronics are based on electrostatics These include resistors capacitors and active devices such as bipolar and field effect transistors which are based on control of electron motion by electrostatic fields Almost all computer peripheral devices with the exception of magnetic memory are based on electrostatic fields Touch pads capacitance keyboards cathoderay tubes liquid crystal displays and electrostatic printers are typical examples In medical work diagnosis is often carried out with the aid of electrostatics as incorpo rated in electrocardiograms electroencephalograms and other recordings of organs with electrical activity including eyes ears and stomachs In industry electrostatics is applied in a variety of forms such as paint spraying electrodeposition electrochemical machining and separation of fine particles Electrostatics is used in agriculture to sort seeds direct sprays to plants measure the moisture content of crops spin cotton and speed baking of bread and smoking of meat12 For various applications of electrostatics see J M Crowley Fundamentals of Applied Electrostat ics New York John Wiley Sons 1986 A D Moore ed Electrostatics and Its Applications New York John Wiley Sons 1973 and C E Jowett Electrostatics in the Electronics Environment New York John Wiley Sons 1976 2An interesting story on the magic of electrostatics is found in B Bolton Electromagnetism and Its Applications London Van Nostrand 1980 p 2 103 104 Electrostatic Fields We begin our study of electrostatics by investigating the two fundamental laws gov erning electrostatic fields 1 Coulombs law and 2 Gausss law Both of these laws are based on experimental studies and they are interdependent Although Coulombs law is ap plicable in finding the electric field due to any charge configuration it is easier to use Gausss law when charge distribution is symmetrical Based on Coulombs law the concept of electric field intensity will be introduced and applied to cases involving point line surface and volume charges Special problems that can be solved with much effort using Coulombs law will be solved with ease by applying Gausss law Throughout our discussion in this chapter we will assume that the electric field is in a vacuum or free space Electric field in material space will be covered in the next chapter 42 COULOMBS LAW AND FIELD INTENSITY Coulombs law is an experimental law formulated in 1785 by the French colonel Charles Augustin de Coulomb It deals with the force a point charge exerts on another point charge By a point charge we mean a charge that is located on a body whose dimensions are much smaller than other relevant dimensions For example a collection of electric charges on a pinhead may be regarded as a point charge Charges are generally measured in coulombs C One coulomb is approximately equivalent to 6 X 1018 electrons it is a very large unit of charge because one electron charge e 16019 X 1019C Coulombs law states that the force between two point charges and Q2 is 1 Along the line joining them 2 Directly proportional to the product QtQ2 of the charges 3 Inversely proportional to the square of the distance R between them Expressed mathematically F R2 41 where k is the proportionality constant In SI units charges 2i and Q2 are in coulombs C the distance R is in meters m and the force F is in newtons N so that k 14TTS0 The constant so is known as the permittivity of free space in farads per meter and has the value 8854 X 1012 rFm 9 X 109 mF 47Tn 42 3Further details of experimental verification of Coulombs law can be found in W F Magie A Source Book in Physics Cambridge Harvard Univ Press 1963 pp 40820 42 COULOMBS LAW AND FIELD INTENSITY 105 Thus eq 41 becomes F QxQi 4irEoR2 43 If point charges Qy and Q2 are located at points having position vectors I and r2 then the force F12 on Q2 due to Qy shown in Figure 41 is given by 44 45a 45b 45c 46a 46b where Rl2 r2 R R12 aR2 R By substituting eq 45 into eq 44 we may write eq 44 as 12 or It is worthwhile to note that Q1Q2 r2 r 4xeor2 1 As shown in Figure 41 the force F2 on Qy due to Q2 is given by F2i F2aR21 F12aRi2 or F2 F 1 2 since 47 Figure 41 Coulomb vector force on point changes Qy and Q2 Origin 106 Electrostatic Fields a b c Figure 42 a b Like charges repel c unlike charges attract 2 Like charges charges of the same sign repel each other while unlike charges attract This is illustrated in Figure 42 3 The distance R between the charged bodies 2i and Q2 must be large compared with the linear dimensions of the bodies that is 2i and Q2 must be point charges 4 Qx and Q2 must be static at rest 5 The signs of Qx and Q2 must be taken into account in eq 44 If we have more than two point charges we can use the principle of superposition to determine the force on a particular charge The principle states that if there are N charges 2i 62 QN located respectively at points with position vectors r1 r2 r the resultant force F on a charge Q located at point r is the vector sum of the forces exerted on Q by each of the charges Qu Q2 QN Hence QQdX rn or eer r 47reor r 17r 4ireo 2 r r N 2 r2 r23 G r rr 48 We can now introduce the concept of electric field intensity The electric field intensity or electric field strength K is the force per unit charge when placed in the electric field Thus or simply F E lim 0o Q 49 E Q 410 The electric field intensity E is obviously in the direction of the force F and is measured in newtonscoulomb or voltsmeter The electric field intensity at point r due to a point charge located at r is readily obtained from eqs 46 and 410 as E Q r r r 411 42 COULOMBS LAW AND FIELD INTENSITY 107 For N point charges Qu Q2 QN located at r b r2 rN the electric field in tensity at point r is obtained from eqs 48 and 410 as or t 6ir 4xejr fepr r2 4irsor r2 A 2J 4TTO TX r rN 4TTO r 412 EXAMPLE 41 Point charges 1 mC and 2 mC are located at 3 2 1 and 1 14 respectively Calculate the electric force on a 10nC charge located at 0 3 1 and the electric field in tensity at that point Solution QQk rk Ai247reor rk Q 1030 3 1 321 21030 3 1 114 47re0 I 03 1 3 2 1 3 103 10 109 r 312 1 1 43 4TT 9 1 0 F i At that point 10 9 2143 9 1 4yu 1 16 9 32 36TT 23 12 286 14Vl4 26V26 3817ay 7506azmN E Q 6507 3817 7506 10 10 109 E 6507ax 3817a 7506azkVm PRACTICE EXERCISE 41 Point charges 5 nC and 2 nC are located at 20 4 and 30 5 respectively a Determine the force on a 1nC point charge located at 1 3 7 b Find the electric field E at 1 3 7 Answer a 1004a 1284a 14aznN b 1004ax 1284a14a2Vm 108 Electrostatic Fields EXAMPLE 42 Two point charges of equal mass m charge Q are suspended at a common point by two threads of negligible mass and length t Show that at equilibrium the inclination angle a of each thread to the vertical is given by Q 16x eomg sin a tan a If a is very small show that Solution Consider the system of charges as shown in Figure 43 where Fe is the electric or coulomb force T is the tension in each thread and mg is the weight of each charge At A or B T sin a Fe T cos a mg Hence But Hence or sin a Fe 1 Q2 cos a mg mg 4ireor r 2 sin a Q cos a I6irejng2 sin3 a Q2 I6irsomg2 sin2 a tan a as required When a is very small tan a a sin a Figure 43 Suspended charged particles for Example 42 42 COULOMBS LAW AND FIELD INTENSITY 109 and so or 23 Q2 I6wsomgtla a Ql 16ireomg PRACTICE EXERCISE 42 Three identical small spheres of mass m are suspended by threads of negligible masses and equal length from a common point A charge Q is divided equally between the spheres and they come to equilibrium at the corners of a horizontal equi lateral triangle whose sides are d Show that Q2 where g acceleration due to gravity Answer Proof r21l2 EXAMPLE 43 A practical application of electrostatics is in electrostatic separation of solids For example Florida phosphate ore consisting of small particles of quartz and phosphate rock can be separated into its components by applying a uniform electric field as in Figure 44 Assum ing zero initial velocity and displacement determine the separation between the particles after falling 80 cm Take E 500 kVm and Qlm 9 xCkg for both positively and neg atively charged particles Figure 44 Electrostatic separation of solids for Example 43 Phosphate Quartz 110 Electrostatic Fields Solution Ignoring the coulombic force between particles the electrostatic force is acting horizon tally while the gravitational force weight is acting vertically on the particles Thus or Integrating twice gives dt2 Q 2m c2 where C and c2 are integration constants Similarly or dt2 Integrating twice we get y Since the initial displacement is zero xt yf Also due to zero initial velocity dx dt dy dt l2gt2 0 0 0 n 0 0 0 ct c 4 c4 0 0 0 0 Thus QE 2 2m 43 ELECTRIC FIELDS DUE TO CONTINUOUS CHARGE DISTRIBUTIONS Wheny 8 0 cm 08 m f2 1633 and 111 x 12 X 9 X 106 X 5 X 105 X 01633 03673 m The separation between the particles is 2x 7347 cm PRACTICE EXERCISE 43 An ion rocket emits positive cesium ions from a wedgeshape electrode into the region described by y The electric field is E 400a 200a kVm The ions have single electronic charges e 16019 X 1019 C and mass m 222 X 1025 kg and travel in a vacuum with zero initial velocity If the emission is confined to 40 cm v 40 cm find the largest value of x which can be reached Answer 08 m 43 ELECTRIC FIELDS DUE TO CONTINUOUS CHARGE DISTRIBUTIONS So far we have only considered forces and electric fields due to point charges which are es sentially charges occupying very small physical space It is also possible to have continuous charge distribution along a line on a surface or in a volume as illustrated in Figure 45 It is customary to denote the line charge density surface charge density and volume charge density by pL in Cm ps in Cm2 and pv in Cm3 respectively These must not be confused with p without subscript used for radial distance in cylindrical coordinates The charge element dQ and the total charge Q due to these charge distributions are ob tained from Figure 45 as pLdl line charge 413a Point charge Line charge Ps Surface charge Volume charge Figure 45 Various charge distributions and charge elements 112 Electrostatic Fields dQ psdSQ psdS surface charge 4 dQ pv dv Q pv dv volume charge 413b 413c The electric field intensity due to each of the charge distributions pL ps and pv may be regarded as the summation of the field contributed by the numerous point charges making up the charge distribution Thus by replacing Q in eq 411 with charge element dQ pL dl ps dS or pv dv and integrating we get E E E PLdl 4jrsJt2 PsdS AweJi2 pvdv line charge surface charge volume charge 414 415 416 It should be noted that R2 and a vary as the integrals in eqs 413 to 416 are evaluated We shall now apply these formulas to some specific charge distributions A A Line Charge Consider a line charge with uniform charge density pL extending from A to B along the zaxis as shown in Figure 46 The charge element dQ associated with element dl dz of the line is dQ pLdl pL dz 0027 00 z dEz dE Figure 46 Evaluation of the E field due to i l i n e 43 ELECTRIC FIELDS DUE TO CONTINUOUS CHARGE DISTRIBUTIONS 113 and hence the total charge Q is Q 417 The electric field intensity E at an arbitrary point Px y z can be found using eq 414 It is important that we learn to derive and substitute each term in eqs 414 to 415 for a given charge distribution It is customary to denote the field point4 by x y z and the source point by x y z Thus from Figure 46 dl dz R x y Z 0 0 z xax yay z zaz or R pap z z az 2 R z zaz R2 R z zf 2132 Substituting all this into eq 414 we get PL E N 2132 4ireo J p2 Z z To evaluate this it is convenient that we define a au and a2 as in Figure 46 R p2 z zfm p sec a z OT p tan a dz p sec2 a da Hence eq 418 becomes pL ai p sec2 a cos a a sin a az da E 4iren p2 sec2 a PL cos a a sin a aj da Thus for a finite line charge E PL sin a2 sin aOa cos a2 cos aaz 418 419 420 4The field point is the point at which the field is to be evaluated 114 Electrostatic Fields As a special case for an infinite line charge point B is at 0 0 and A at 0 0 co so that al x2 a2 x2 the zcomponent vanishes and eq 420 becomes E PL 421 Bear in mind that eq 421 is obtained for an infite line charge along the zaxis so that p and ap have their usual meaning If the line is not along the zaxis p is the perpendicular distance from the line to the point of interest and ap is a unit vector along that distance di rected from the line charge to the field point B A Surface Charge Consider an infinite sheet of charge in the xyplane with uniform charge density ps The charge associated with an elemental area dS is dQ Ps dS and hence the total charge is Q PsdS 422 From eq 415 the contribution to the E field at point P0 0 h by the elemental surface 1 shown in Figure 47 is JE dQ 423 Figure 47 Evaluation of the E field due to an infinite sheet of charge 43 ELECTRIC FIELDS DUE TO CONTINUOUS CHARGE DISTRIBUTIONS From Figure 47 115 2 2112 R pap haz R R Lc fr n aR K Ps dS psp dj dp Substitution of these terms into eq 423 gives pspdjdppap haz 424 Due to the symmetry of the charge distribution for every element 1 there is a correspond ing element 2 whose contribution along ap cancels that of element 1 as illustrated in Figure 47 Thus the contributions to Ep add up to zero so that E has only zcomponent This can also be shown mathematically by replacing a with cos ax sin a r Integra tion of cos j or sin over 0 j 2ir gives zero Therefore E Ps hp dp dj 132 trH2V 425 that is E has only zcomponent if the charge is in the xyplane In general for an infinite sheet of charge 426 where an is a unit vector normal to the sheet From eq 425 or 426 we notice that the electric field is normal to the sheet and it is surprisingly independent of the distance between the sheet and the point of observation P In a parallel plate capacitor the electric field existing between the two plates having equal and opposite charges is given by C A Volume Charge Let the volume charge distribution with uniform charge density pv be as shown in Figure 48 The charge dQ associated with the elemental volume dv is dQ pv dv 116 Electrostatic Fields p dE z dE2 P0 0 z V dv at r S j Figure 48 Evaluation of the E field due to a volume charge distribution and hence the total charge in a sphere of radius a is Q pv dv pv dv 4ira3 428 The electric field dE at P0 0 z due to the elementary volume charge is E a s 4x S o 2 where aR cos a a sin a ap Due to the symmetry of the charge distribution the con tributions to Ex or Ey add up to zero We are left with only Ez given by Ez E az dE cos a Again we need to derive expressions for dv R2 and cos a dv r2 sin 6 dr dd dt Applying the cosine rule to Figure 48 we have R2 z2 r2 2zr cos B r2 z2 R2 2zR cos a dv c o s 429 430 w 43 ELECTRIC FIELDS DUE TO CONTINUOUS CHARGE DISTRIBUTIONS 117 It is convenient to evaluate the integral in eq 429 in terms of R and r Hence we express cos d cos a and sin 6 dd in terms of R and r that is cos a z 2 2 Z h R 2 2zR Yr 2 r2 R 2 2zr Differentiating eq 431b with respect to 0 keeping z and r fixed we obtain RdR sin 6 dd zr Substituting eqs 430 to 432 into eq 429 yields E 4xeo J o dt J r 0 J ra rzr dr r0 JR zr r ro i 1 1 IA 4r2dr 2 47reo z z V3 or 431a 431b 432 433 This result is obtained for E at P0 0 z Due to the symmetry of the charge distribution the electric field at Pr 9 j is readily obtained from eq 433 as E Q ar 434 which is identical to the electric field at the same point due to a point charge Q located at the origin or the center of the spherical charge distribution The reason for this will become obvious as we cover Gausss law in Section 45 EXAMPLE 44 A circular ring of radius a carries a uniform charge pL Cm and is placed on the xyplane with axis the same as the zaxis a Show that E0 0 h pLah 2eoh 2 a2132 z 118 Electrostatic Fields b What values of h gives the maximum value of E c If the total charge on the ring is Q find E as a 0 Solution a Consider the system as shown in Figure 49 Again the trick in finding E using eq 414 is deriving each term in the equation In this case dl a d4 R R R a haz a2 12112 R R or a R aap 2 R3 a2 h2 132 Hence E PL aap By symmetry the contributions along ap add up to zero This is evident from the fact that for every element dl there is a corresponding element diametrically opposite it that gives an equal but opposite dEp so that the two contributions cancel each other Thus we are left with the zcomponent That is pLahaz 4vsoh2 a2 132 dt pLahaz 2soh2 a2f2 as required Figure 49 Charged ring for Example 44 I 43 ELECTRIC FIELDS DUE TO CONTINUOUS CHARGE DISTRIBUTIONS 119 b dh 2eo For maximum E 0 which implies that dh h 2 a23 a2 2hz 0 or h c Since the charge is uniformly distributed the line charge density is Q so that Asa or in general PL E Qh 2132 Hz E si AKSJI2 4ireor which is the same as that of a point charge as one would expect PRACTICE EXERCISE 44 A circular disk of radius a is uniformly charged with ps Cm2 If the disk lies on the z 0 plane with its axis along the zaxis a Show that at point 0 0 h h b From this derive the E field due to an infinite sheet of charge on the z 0 plane c If a 3C h show that E is similar to the field due to a point charge Answer a Proof b a c Proof 2en 120 Electrostatic Fields EXAMPLE 45 I The finite sheet 0 x 1 0 y 1 on the z 0 plane has a charge density p s xyx2 y2 2532 nCm2 Find a The total charge on the sheet b The electric field at 0 0 5 c The force experienced by a 1 mC charge located at 0 0 5 Solution a g I psdS I xyx2 y2 2532 dx dy nC J Jo Jo Since x dx 12 dx2 we now integrate with respect to x2 or change variables x2 u so that x dx dull 1 Q j 1 f1 y2 2532 dx2 dy nC y2 25 2 dy 1 2 721 3315 nC b E r r 4ireor2 J 4reor r3 where r r 0 0 5 x y 0 x y 5 Hence E o Jo 10 y2 25fxax yay 5azdxdy 109 36TT yz 2532 r rl rl r 1 rl f l rl 9 x2 dx ydyax x dx y2dy ay 5 xdx y dy 6 6 4 1515 1125 Vm c F E 15 15 1125 mN 43 ELECTRIC FIELDS DUE TO CONTINUOUS CHARGE DISTRIBUTIONS 121 PRACTICE EXERCISE 45 A square plate described by 2 S A 2 2y2z 0 carries a charge 12 y mCm2 Find the total charge on the plate and the electric field intensity at 0 0 10 Answer 192 mC 1646 a MVm EXAMPLE 46 Planes x 2 and y 3 respectively carry charges 10 nCm2 and 15 nCm2 If the line x 0 z 2 carries charge lOx nCm calculate E at 1 1 1 due to the three charge distributions Solution Let E E E2 E3 where Ej E2 and E3 are respectively the contributions to E at point 1 1 1 due to the infinite sheet 1 infinite sheet 2 and infinite line 3 as shown in Figure 410a Applying eqs 426 and 421 gives v9 ar 1807rar 36r 15 10 10 9 270TT av 36TT y 3 l b Figure 410 For Example 46 a three charge distributions b finding p and ap on plane y 1 122 Electrostatic Fields and E3 PL 2ireop where ap not regular ap but with a similar meaning is a unit vector along LP perpendicu lar to the line charge and p is the length LP to be determined from Figure 410b Figure 410b results from Figure 410a if we consider plane y 1 on which E3 lies From Figure 410b the distance vector from L to P is R P VTo R Hence lOir 10 9 2 T T 10 9 10 Vio Vio a 3az 36TT 187rax 3a Thus by adding Eu E2 and E3 we obtain the total field as E 162Trax 270ira 54x3 Vm Note that to obtain ar ap or a which we always need for finding F or E we must go from the charge at position vector r to the field point at position vector r hence ar ap or an is a unit vector along r r Observe this carefully in Figures 46 to 410 PRACTICE EXERCISE 46 In Example 46 if the line x 0 z 2 is rotated through 90 about the point 0 2 2 so that it becomes x 0 y 2 find E at 1 1 1 Answer 2827a 5645a Vm 44 ELECTRIC FLUX DENSITY The flux due to the electric field E can be calculated using the general definition of flux in eq 313 For practical reasons however this quantity is not usually considered as the most useful flux in electrostatics Also eqs 411 to 416 show that the electric field in tensity is dependent on the medium in which the charge is placed free space in this chapter Suppose a new vector field D independent of the medium is defined by D eoE 435 II 44 ELECTRIC FLUX DENSITY H 123 We define electric flux f in terms of D using eq 313 namely DdS 436 In SI units one line of electric flux emanates from 1 C and terminates on 1 C There fore the electric flux is measured in coulombs Hence the vector field D is called the elec tric flux density and is measured in coulombs per square meter For historical reasons the electric flux density is also called electric displacement From eq 435 it is apparent that all the formulas derived for E from Coulombs law in Sections 42 and 43 can be used in calculating D except that we have to multiply those formulas by eo For example for an infinite sheet of charge eqs 426 and 435 give 437 438 Note from eqs 437 and 438 that D is a function of charge and position only it is in dependent of the medium and for a volume charge distribution eqs 416 and 435 give D Pvdv EXAMPLE 47 Determine D at 4 0 3 if there is a point charge 5TT mC at 4 0 0 and a line charge 3TT mCm along the yaxis Solution Let D DQ DL where D e and DL are flux densities due to the point charge and line charge respectively as shown in Figure 411 Q r r eoE Q 4mR 47rr r where r r 4 0 3 4 0 0 0 0 3 Hence Also In this case 3 DQ 5TT 100 0 3 4TT00 3 3 2 0138 az mCm 4 0 3 0 0 0 4 0 3 40 3 00 0 5 p 403 000 5 124 Electrostatic Fields 3rCm y Q SirC Figure 411 Flux density D due to a point charge and an infinite line charge Hence Thus 3TT 2TT25 3az 024ax 018azmCm2 D DG DL 240a 42a2 xCm2 PRACTICE EXERCISE 47 A point charge of 30 nC is located at the origin while plane y 3 carries charge 10nCm2 Find D at 04 3 Answer 5076a 00573az nCm2 45 GAUSSS LAWMAXWELLS EQUATION Gausss5 law constitutes one of the fundamental laws of electromagnetism Gausss law stales thai the loial electric Mux V through any closed surface is equal to the total charge enclosed by that surface Karl Friedrich Gauss 17771855 a German mathematician developed the divergence theorem of Section 36 popularly known by his name He was the first physicist to measure electric and mag netic quantities in absolute units For details on Gausss measurements see W F Magie A Source Book in Physics Cambridge Harvard Univ Press 1963 pp 519524 r if 45 GAUSSS LAWMAXWELLS EQUATION 125 Thus that is 439 T d D dS Total charge enclosed Q I pv dv 440 or 441 By applying divergence theorem to the middle term in eqs 441 D dS I V D dv Comparing the two volume integrals in eqs 441 and 442 results in V D 442 443 which is the first of the four Maxwells equations to be derived Equation 443 states that the volume charge density is the same as the divergence of the electric flux density This should not be surprising to us from the way we defined the divergence of a vector in eq 332 and from the fact that pv at a point is simply the charge per unit volume at that point Note that 1 Equations 441 and 443 are basically stating Gausss law in different ways eq 441 is the integral form whereas eq 443 is the differential or point form of Gausss law 2 Gausss law is an alternative statement of Coulombs law proper application of the divergence theorem to Coulombs law results in Gausss law 3 Gausss law provide an easy means of finding E or D for symmetrical charge dis tributions such as a point charge an infinite line charge an infinite cylindrical surface charge and a spherical distribution of charge A continuous charge distribution has rectan gular symmetry if it depends only on x or y or z cylindrical symmetry if it depends only on p or spherical symmetry if it depends only on r independent of 6 and j It must be stressed that whether the charge distribution is symmetric or not Gausss law always holds For example consider the charge distribution in Figure 412 where V and v2 are closed surfaces or volumes The total flux leaving vl is 10 5 5 nC because only 10 nC and 5 nC charges are enclosed by vj Although charges 20 nC and 15 nC outside Vi do contribute to the flux crossing v1 the net flux crossing vi according to Gausss law is irrespective of those charges outside vj Similarly the total flux leaving v2 is zero 126 Electrostatic Fields 20 nC Figure 412 Illustration of Gausss law flux leaving v is 5 nC and that 15 nC leaving v2 is 0 C because no charge is enclosed by v2 Thus we see that Gausss law f 2enciosed is still obeyed even though the charge distribution is not symmetric However we cannot use the law to determine E or D when the charge distribution is not symmetric we must resort to Coulombs law to determine E or D in that case 46 APPLICATIONS OF GAUSSS LAW The procedure for applying Gausss law to calculate the electric field involves first knowing whether symmetry exists Once symmetric charge distribution exists we con struct a mathematical closed surface known as a Gaussian surface The surface is chosen such that D is normal or tangential to the Gaussian surface When D is normal to the surface D dS D dS because D is constant on the surface When D is tangential to the surface D dS 0 Thus we must choose a surface that has some of the symmetry ex hibited by the charge distribution We shall now apply these basic ideas to the following cases A Point Charge Suppose a point charge Q is located at the origin To determine D at a point P it is easy to see that choosing a spherical surface containing P will satisfy symmetry conditions Thus a spherical surface centered at the origin is the Gaussian surface in this case and is shown in Figure 413 Figure 413 Gaussian surface about a point charge y Gaussian surface 46 APPLICATIONS OF GAUSSS LAW 127 Since D is everywhere normal to the Gaussian surface that is D Dn applying Gausss law V genciosed gives Q i D dS Dr dS Dr Aitr 444 where dS surface Thus LQ 0 r2 sin 6 dd dcf 4irr2 is the surface area of the Gaussian 44511 as expected from eqs 411 and 435 B Infinite Line Charge Suppose the infinite line of uniform charge pL Cm lies along the zaxis To determine D at a point P we choose a cylindrical surface containing P to satisfy symmetry condition as shown in Figure 414 D is constant on and normal to the cylindrical Gaussian surface that is D Dpap If we apply Gausss law to an arbitrary length of the line PJ Q Dp 2irp 446 where dS 2irp is the surface area of the Gaussian surface Note that J D dS evalu ated on the top and bottom surfaces of the cylinder is zero since D has no zcomponent that means that D is tangential to those surfaces Thus D as expected from eqs 421 and 435 2irp 447 Figure 414 Gaussian surface about an infinite line line charge PCm Gaussian surface 128 Electrostatic Fields C Infinite Sheet of Charge Consider the infinite sheet of uniform charge ps Cm2 lying on the z 0 plane To deter mine D at point P we choose a rectangular box that is cut symmetrically by the sheet of charge and has two of its faces parallel to the sheet as shown in Figure 415 As D is normal to the sheet D Dzaz and applying Gausss law gives Ps dS Q f D dS Dz dS dS op bottom 448 Note that D dS evaluated on the sides of the box is zero because D has no components along ax and ay If the top and bottom area of the box each has area A eq 448 becomes and thus PsA DZA A Ps 449 or 450 as expected from eq 425 D Uniformly Charged Sphere Consider a sphere of radius a with a uniform charge pv Cm3 To determine D everywhere we construct Gaussian surfaces for eases r a and r a separately Since the charge has spherical symmetry it is obvious that a spherical surface is an appropriate Gaussian surface Infinite sheet of charge ps Cm2 Figure 415 Gaussian surface about an infinite line sheet of charge Gaussian surface 46 APPLICATIONS OF GAUSSS LAW H 129 For r a the total charge enclosed by the spherical surface of radius r as shown in Figure 416 a is Gene Pvdv pAdv pA I I r2 sin 6 drdd dj 451 and V P D dS Dr dS Dr Dr4xr2 Hence TP 2enc gi y e s D r 4xr 2 rlsm6ded o 452 or 0 r a 453 For r a the Gaussian surface is shown in Figure 416b The charge enclosed by the surface is the entire charge in this case that is while G e n e p v d v p v d v p pv ira sinO drdd o cb D dS Dr4irr2 454 455 Gaussian surface II f I Figure 416 Gaussian surface for a uniformly charged sphere when a r a and b r a a b 130 M Electrostatic Fields IDI Figure 417 Sketch of D against r for a uniformly charged sphere just as in eq 452 Hence or r 3s a 456 Thus from eqs 453 and 456 D everywhere is given by D 0 r s a pvar r a 457 and D is as sketched in Figure 417 Notice from eqs 444 446 448 and 452 that the ability to take D out of the integral sign is the key to finding D using Gausss law In other words D must be constant on the Gaussian surface EXAMPLE 48 Given that D Zp cos20 az Cm2 calculate the charge density at d T4 3 and the total charge enclosed by the cylinder of radius 1 m with 2 z 2 m Solution pv V D z p cos2 dZ At 1 TT4 3 Pv 1 cos27r4 05 Cm3 The total charge enclosed by the cylinder can be found in two different ways 46 APPLICATIONS OF GAUSSS LAW 131 Method 1 This method is based directly on the definition of the total volume charge Q pv dv p cos j pdtdp dz dz 2 p2dp 4TT13 Method 2 Alternatively we can use Gausss law D y IP where f f and P6 are the flux through the sides the top surface and the bottom surface of the cylinder respectively see Figure 317 Since D does not have component along ap Ys 0 for dS pdj dp az so zp cos2 4 p dt dp 2 I p2dp I cos2 and for Wb dS p df dp az so i cos t p dj dp 2TT T x Thus 2 2 Jo c o s z 4 dt as obtained previously PRACTICE EXERCISE 48 If D 22 ZAX 4xyay xaz Cm2 find a The volume charge density at 1 0 3 b The flux through the cube defined b y 0 J t l 0 y l 0 z l c The total charge enclosed by the cube Answer a 4 Cm b 2 C c 2 C 132 Electrostatic Fields A charge distribution with spherical symmetry has density Pr rv r EXAMPLE 49 1 R r R Determine E everywhere Solution The charge distribution is similar to that in Figure 416 Since symmetry exists we can apply Gausss law to find E y a For r R SoPEdS gene P r r IT r 2ir eoEr 4 x r Qenc pv r sin 9 dt dB dr Jo Jo Jo i 2 Por PoTr4 4Trr dr R R or b For r R Qenc r rir r2ic 0 J0 pvr sin 6 dj dd dr I 4irr2dr 0 4wr2 dr Jo R R or PRACTICE EXERCISE 49 A charge distribution in free space has pv 2r nCm3 for 0 r 10 m and zero otherwise Determine E at r 2 m and r 12 m Answer 226ar Vm 3927ar kVm 47 ELECTRIC POTENTIAL 133 47 ELECTRIC POTENTIAL From our discussions in the preceding sections the electric field intensity E due to a charge distribution can be obtained from Coulombs law in general or from Gausss law when the charge distribution is symmetric Another way of obtaining E is from the electric scalar po tential V to be defined in this section In a sense this way of rinding E is easier because it is easier to handle scalars than vectors Suppose we wish to move a point charge Q from point A to point B in an electric field E as shown in Figure 418 From Coulombs law the force on Q is F QE so that the work done in displacing the charge by d is dW F d QE d 458 The negative sign indicates that the work is being done by an external agent Thus the total work done or the potential energy required in moving Q from A to B is 459 Dividing W by Q in eq 459 gives the potential energy per unit charge This quantity denoted by VAB is known as the potential difference between points A and B Thus 460 Note that 1 In determining VAB A is the initial point while B is the final point 2 If VAB is negative there is a loss in potential energy in moving Q from A to B this implies that the work is being done by the field However if VAB is positive there is a gain in potential energy in the movement an external agent performs the work 3 VAB is independent of the path taken to be shown a little later 4 VAB is measured in joules per coulomb commonly referred to as volts V Origin Figure 418 Displacement of point charge Q in an electrostatic field E 134 Electrostatic Fields As an example if the E field in Figure 418 is due to a point charge Q located at the origin then E so eq 460 becomes 4iren Q Q 2 rA 4ireor Q l d r r 461 462a or vAB vBvA 462b where VB and VA are the potentials or absolute potentials at B and A respectively Thus the potential difference VAB may be regarded as the potential at B with reference to A In problems involving point charges it is customary to choose infinity as reference that is we assume the potential at infinity is zero Thus if VA 0 as rA in eq 462 the po tential at any point rB r due to a point charge Q located at the origin is V Q 4irenr 463 Note from eq 462a that because E points in the radial direction any contribution from a displacement in the 6 or direction is wiped out by the dot product E d E cos 8 dl E dr Hence the potential difference VAB is independent of the path as asserted earlier The potential il an poim is the pulomial dittcrcntx helwecn thai poim and a chosen poinl in which the potential is em In other words by assuming zero potential at infinity the potential at a distance r from the point charge is the work done per unit charge by an external agent in transferring a test charge from infinity to that point Thus V E dl 464 If the point charge Q in eq 463 is not located at the origin but at a point whose po sition vector is r the potential Vx y z or simply Vr at r becomes Vr Q 4iren r r 465 47 ELECTRIC POTENTIAL 135 We have considered the electric potential due to a point charge The same basic ideas apply to other types of charge distribution because any charge distribution can be regarded as consisting of point charges The superposition principle which we applied to electric fields applies to potentials For n point charges Qu Q2 Qn located at points with po sition vectors r b r2 rn the potential at r is Vr Qn 47TE 4ireor r2 4ire0 r or r r point charges 466 For continuous charge distributions we replace Qk in eq 466 with charge element pL dl ps dS or pv dv and the summation becomes an integration so the potential at r becomes Vr Vr r 1 4irso 1 47Tfio 1 r r PsrdS 4TTS O r rr PvrdV r r line charge surface charge volume charge 467 468 469 where the primed coordinates are used customarily to denote source point location and the unprimed coordinates refer to field point the point at which Vis to be determined The following points should be noted 1 We recall that in obtainingqs 463 to 469 the zero potential reference point has been chosen arbitrarily to be at infinity If any other point is chosen as reference eq 465 for example becomes V Q C 470 where C is a constant that is determined at the chosen point of reference The same idea applies to eqs 463 to 469 2 The potential at a point can be determined in two ways depending on whether the charge distribution or E is known If the charge distribution is known we use one of eqs 465 to 470 depending on the charge distribution If E is known we simply use V E dl C The potential difference VAB can be found generally from fB vAB vBvA w 471 472 136 I EXAMPLE 410 Electrostatic Fields Two point charges 4 juC and 5 jtC are located at 2 1 3 and 0 4 2 respectively Find the potential at 1 0 1 assuming zero potential at infinity Solution Let 6i 4 Vr Q2 5 4vreo r r 4Treor r2 C0 If 0 Co 0 r r 1 0 1 2 1 3 1 1 2 V6 r r2 1 0 1 0 4 2 1 4 3 V26 Hence Vl0 1 4TT X 4 1OV6 V26J 36TT 9 X 103 1633 09806 5872 kV PRACTICE EXERCISE 410 If point charge 3 fiC is located at the origin in addition to the two charges of example 410 find the potential at 1 5 2 assuming Vo 0 Answer 1023 kV EXAMPLE 411 A point charge 5 nC is located at 3 4 0 while line y 1 z 1 carries uniform charge 2 nCm a If V 0 V at O0 0 0 find V at A5 0 1 b If V 100 V at 51 2 1 find Vat C2 5 3 c If V 5 V at O find VBC Solution Let the potential at any point be V VQ VL 47 ELECTRIC POTENTIAL 137 where VQ and VL are the contributions to V at that point due to the point charge and the line charge respectively For the point charge VQ SEd Q ar dr ar Q c 4xeor For the infinite line charge VL I E d PL 2rop dp ap PL 2irsc In p C2 Hence V PL 2ireo lnp Q 4irenr C where C Cx C2 constant p is the perpendicular distance from the line y 1 z 1 to the field point and r is the distance from the point charge to the field point a If V 0 at O0 0 0 and V at A5 0 1 is to be determined we must first determine the values of p and r at O and A Finding r is easy we use eq 231 To find p for any point x y z we utilize the fact that p is the perpendicular distance from x y z to line y 1 z 1 which is parallel to the xaxis Hence p is the distance between x y z and x 1 1 because the distance vector between the two points is perpendicular to ax Thus p x y z x 1 1 Vy I2 z I2 Applying this for p and eq 231 for r at points O and A we obtain P o 000 0l r o 000 3 pA 50 1 5 1 rA 50 1 3 Tl 5 1 9 Hence 0 PA 2 109 fl In J 1 ro r 5 109 2TT 10 9 1 4TT 10 9 36TT 36TT 0 V 3 6 In V 2 45 138 Electrostatic Fields or VA 36 In V 2 4 8477 V fnote H T I T g Stant C by Subtractig one another and that it does not matter which one is subtracted from which b If V 100 at 51 2 1 and Vat C2 5 3 is to be determined we find PB 1121 111 1 rB 1 2 1 3 4 0 V2T Pc K253 211 V20 rc 253 340 2xeo or 361n 50175 V Vc 49825 V 21 J c To find the potential difference between two points we do not need a potential refer ence if a common reference is assumed Vc VB 49825 100 50175 V as obtained in part b PRACTICE EXERCISE 411 A point charge of 5 nC is located at the origin If V 2 V at 0 6 8 find a The potential at A3 26 b The potential at B 5 7 c The potential difference VAB Answer a 3929 V b 2696 V c 1233 V 48 RELATIONSHIP BETWEEN E AND VMAXWELLS EQUATION 139 48 RELATIONSHIP BETWEEN E AND V MAXWELLS EQUATION As shown in the previous section the potential difference between points A and B is inde pendent of the path taken Hence VAB that is VBA VAB E d 0 or 473 This shows that the line integral of E along a closed path as shown in Figure 419 must be zero Physically this implies that no net work is done in moving a charge along a closed path in an electrostatic field Applying Stokess theorem to eq 473 gives E d V X E dS 0 or V X E 0 474 Any vector field that satisfies eq 473 or 474 is said to be conservative or irrotational as discussed in Section 38 Thus an electrostatic field is a conservative field Equation 473 or 474 is referred to as Maxwells equation the second Maxwells equation to be derived for static electric fields Equation 473 is the integral form and eq 474 is the differential form they both depict the conservative nature of an electrostatic field From the way we defined potential V E d it follows that dV Edl Ex dx Eydy Ez dz Figure 419 Conservative nature of an electrosta tic field 140 Electrostatic Fields But dV dV dV dV dx dy dz dx dy dz Comparing the two expressions for dV we obtain dV Ex dV dx E dy z dz Thus E V V 475 476 that is the electric field intensity is the gradient of V The negative sign shows that the di rection of E is opposite to the direction in which V increases E is directed from higher to lower levels of V Since the curl of the gradient of a scalar function is always zero V X V V 0 eq 474 obviously implies that E must be a gradient of some scalar func tion Thus eq 476 could have been obtained from eq 474 Equation 476 shows another way to obtain the E field apart from using Coulombs or Gausss law That is if the potential field V is known the E can be found using eq 476 One may wonder how one function V can possibly contain all the information that the three components of E carry The three components of E are not independent of one another They are explicitly interrelated by the condition V X E 0 What the poten tial formulation does is to exploit this feature to maximum advantage reducing a vector problem to a scalar one EXAMPLE 412 Given the potential V sin 6 cos 0 a Find the electric flux density D at 2 TT2 0 b Calculate the work done in moving a 10C charge from point Al 30 120 to B4 90 60 Solution a D eoE But 1 dV dr r dd rsind ari 20 10 r sin 0 cos p ar r cos d cos 10 Hr sin i r 48 RELATIONSHIP BETWEEN E AND VMAXWELLS EQUATION 141 At 2 TT2 0 20 D eoE r 2 6 TT2 0 0 so ar 0ae 0a I 25eoarCm2 221 arpCm2 b The work done can be found in two ways using either E or V Method 1 W or I E dl W Q E dl and because the electrostatic field is conservative the path of integration is immaterial Hence the work done in moving Q from Al 30 120 to 54 90 60 is the same as that in moving Q from A to A from A to B and from B to B where Al 30 i dl A4 30 120 drar 120 dl rddag B4 90 B4 90 60 120 r sin 6 df That is instead of moving Q directly from A and B it is moved from A A A B B B so that only one variable is changed at a time This makes the line integral a lot easier to evaluate Thus W Q 1 AA AB BB Edl 20 sin 6 cos jdr 30 t 120 90 1 0 cos 6 cos 30 60 10 sin rdd r sin 6 dt r 4 41 10 1 l6 S m e 75 K 32 32 16 10 cos 0 60 120 142 Electrostatic Fields or 45 W Q 28125 Method 2 Since Vis known this method is a lot easier WQ VA 10 j sin 90 cos 60 y sin 30 cos 120 1T6 28125 J as obtained before PRACTICE EXERCISE 412 Given that E 32 v a Aa Wm find the work done in moving a 2 MC charge from 0 5 0 to 2 1 0 by taking the path a 050 2 50 2 10 b y 5 3x Answer a 12 mJ b 12 mJ 49 AN ELECTRIC DIPOLE AND FLUX LINES An electric dipole is formed when two poim charges of equal magnitude but oppo site sign are separated by a small distance The importance of the field due to a dipole will be evident in the subsequent chapters Consider the dipole shown in Figure 420 The potential at point Pr 6 0 is given by r2 4TTEO 477 where r and r2 are the distances between P and Q and P and Q respectively If r dr2 r d cos 6 r2rx r2 and eq 477 becomes V Q dcosd 478 49 A N ELECTRIC DIPOLE AND FLUX LINES Hi 143 Figure 420 An electric dipole dcosd Since d cos 6 d ar where d daz if we define as the dipole moment eq 478 may be written as 479 480 Note that the dipole moment p is directed from Q to Q If the dipole center is not at the origin but at r eq 480 becomes Vr p r r 47renr r 481 The electric field due to the dipole with center at the origin shown in Figure 420 can be obtained readily from eqs 476 and 478 as E V V Qd cos 0 27Tnr3 ay l ay QJ sin 6 or E 2 cos 6 ar sin 6 ae 482 where p p Qd 144 Electrostatic Fields Notice that a point charge is a monopole and its electric field varies inversely as r2 while its potential field varies inversely as r see eqs 461 and 463 From eqs 480 and 482 we notice that the electric field due to a dipole varies inversely as r3 while its potential varies inversely as r2 The electric fields due to successive higherorder multi poles such as a quadrupole consisting of two dipoles or an octupole consisting of two quadrupoles vary inversely as r4 r5 r6 while their corresponding potentials vary in versely as r3 r4 r5 The idea of electric flux lines or electric lines of force as they are sometimes called was introduced by Michael Faraday 17911867 in his experimental investigation as a way of visualizing the electric field An electric flux line is an imaginary path or line drawn in such a way thai its direc tion at any poinl is the direction of Ihc electric field at that point In other words they are the lines to which the electric field density D is tangential at every point Any surface on which the potential is the same throughout is known as an equipoten tial surface The intersection of an equipotential surface and a plane results in a path or line known as an equipotential line No work is done in moving a charge from one point to another along an equipotential line or surface VA VB 0 and hence Edl 483 on the line or surface From eq 483 we may conclude that the lines of force or flux lines or the direction of E are always normal to equipotential surfaces Examples of equipotential surfaces for point charge and a dipole are shown in Figure 421 Note from these examples that the direction of E is everywhere normal to the equipotential flux line Figure 421 Equipotential surfaces for a a point charge and b an electric dipole 49 A N ELECTRIC DIPOLE AND FLUX LINES 145 lines We shall see the importance of equipotential surfaces when we discuss conducting bodies in electric fields it will suffice to say at this point that such bodies are equipoten tial volumes A typical application of field mapping flux lines and equipotential surfaces is found in the diagnosis of the human heart The human heart beats in response to an electric field potential difference across it The heart can be characterized as a dipole with the field map similar to that of Figure 42 lb Such a field map is useful in detecting abnormal heart po sition6 In Section 152 we will discuss a numerical technique for field mapping EXAMPLE 413 Two dipoles with dipole moments 5a z nCm and 9az nCm are located at points 0 0 2 and 0 0 3 respectively Find the potential at the origin Solution where Hence 5az tri 4ireor3 k p2 r2 r r 0 0 0 0 0 2 2az j 2 p2 9az r2 0 0 0 0 0 3 3az r2 jr2 3 10 27 10 9 L 2 3 3 3 109 36vr 2025 V PRACTICE EXERCISE 413 An electric dipole of 100 a pC m is located at the origin Find V and E at points a 00 10 b I7i73 TT2 Answer a 9 mV 18ar mVm b 045 V 09ar 07794a Vm 6For more information on this see R Plonsey Bioelectric Phenomena New York McGrawHill 1969 146 Electrostatic Fields 410 ENERGY DENSITY IN ELECTROSTATIC FIELDS To determine the energy present in an assembly of charges we must first determine the amount of work necessary to assemble them Suppose we wish to position three point charges Qx Q2 and Q3 in an initially empty space shown shaded in Figure 422 No work is required to transfer Qx from infinity to Pl because the space is initially charge free and there is no electric field from eq 459 W 0 The work done in transferring Q2 from infinity to P2 is equal to the product of Q2 and the potential V2i a t P2 due to Qx Similarly the work done in positioning Q3 at P3 is equal to Q3V32 V31 where V32 and V31 are the potentials at P3 due to Q2 and Qu respectively Hence the total work done in positioning the three charges is W3 0 Q2V2l If the charges were positioned in reverse order WE W3 W2 1 0 v32 V13 484 485 where V23 is the potential at P2 due to Q3 Vl2 and Vl3 are respectively the potentials at Pi due to Q2 and Q3 Adding eqs 484 and 485 gives 2WE V13 Q2V2l V23 Q3V3 Q2V2 Q3V3 V32 or Q2V2 Q3V3 486 where Vu V2 and V3 are total potentials at Pu P2 and P3 respectively In general if there are n point charges eq 486 becomes in joules 487 Figure 422 Assembling of charges 410 ENERGY DENSITY IN ELECTROSTATIC FIELDS 147 If instead of point charges the region has a continuous charge distribution the sum mation in eq 487 becomes integration that is WE pLVdl line charge W psV dS surface charge WE I pvV dv volume charge Since pv V D eq 490 can be further developed to yield WE jVDVdv But for any vector A and scalar V the identity V VA A VV VV A or V AV V VA A VV holds Applying the identity in eqs 492 to 491 we get WE V VD dv D VV dv 488 489 490 491 492 493 By applying divergence theorem to the first term on the righthand side of this equation we have 1 WE 4 VD dS D VV dv 494 From Section 49 we recall that V varies as 1r and D as 1r2 for point charges V varies as 1r2 and D as 1r3 for dipoles and so on Hence VD in the first term on the righthand side of eq 494 must vary at least as 1r3 while dS varies as r2 Consequently the first integral in eq 494 must tend to zero as the surface S becomes large Hence eq 494 reduces to 495 WE D VV dv D E dv and since E VV and D eoE 496 148 Electrostatic Fields From this we can define electrostatic energy density wE in Jm as dW 1 1 i D2 wE dv 2 2eo 497 so eq 495 may be written as WE wE dv 498 EXAMPLE 414 Three point charges 1 nC 4 nC and 3 nC are located at 0 0 0 0 0 1 and 1 0 0 respectively Find the energy in the system Solution w w w2 w3 0 Q2V21 G3 V32 a 4TTO Q 1100 000 l00 00l 4ir 10 4 3 36TT 91 7 nJ 1337 nJ Alternatively W 2 2 Qi 2 2 L4TS O1 9 7 nJ 1337 nJ as obtained previously 410 ENERGY DENSITY IN ELECTROSTATIC FIELDS 149 PRACTICE EXERCISE 414 Point charges 2 1 nC Q2 2 nC Q3 3 nC and Q4 4 nC are posi tioned one at a time and in that order at 0 0 0 10 0 0 0 1 and 0 0 1 respectively Calculate the energy in the system after each charge is positioned Answer 0 18 nJ 2918 nJ 6827 nJ EXAMPLE 415 A charge distribution with spherical symmetry has density po 0 r R Pv 0 rR Determine V everywhere and the energy stored in region r R Solution The D field has already been found in Section 46D using Gausss law a Forr RE ar 3e Once E is known V is determined as 3eor C R Since Vr oo oCi 0 b For r E ar 3eo Hence por 6eo From part a Vr R Hence 3e0 d rdr 3eo C 6eo 2en 150 Electrostatic Fields and 6eo Thus from parts a and b V c The energy stored is given by 3e 3R2r2 r W I D E dv eo E 2 dv Forr S R Hence 2ir I o W 2 S e 2 J J J r2 r2 sin 0 2 D5 45eo PRACTICE EXERCISE 415 If V x y xy 2z V find E at 1 2 3 and the electrostatic energy stored in a cube of side 2 m centered at the origin Answer 3ax 2a Vm 02358 nJ SUMMARY 1 The two fundamental laws for electrostatic fields Coulombs and Gausss are pre sented in this chapter Coulombs law of force states that Aireji2 2 Based on Coulombs law we define the electric field intensity E as the force per unit charge that is T Q 4ireJi R point charge only SUMMARY 151 3 For a continuous charge distribution the total charge is given by Q pLdl for line charge Q ps dS for surface charge Q pv dv for volume charge The E field due to a continuous charge distribution is obtained from the formula for point charge by replacing Q with dQ pL dl dQ ps dS or dQ pv dv and integrat ing over the line surface or volume respectively 4 For an infinite line charge and for an infinite sheet of charge F 2e0 5 The electric flux density D is related to the electric field intensity in free space as D eoE The electric flux through a surface S is y I Dds s 6 Gausss law states that the net electric flux penetrating a closed surface is equal to the total charge enclosed that is f Qenc Hence or Pv V D 2 e n c Pvdv first Maxwells equation to be derived When charge distribution is symmetric so that a Gaussian surface where D Dnan is constant can be found Gausss law is useful in determining D that is DndS gene or Dn Gene 152 Electrostatic Fields 7 The total work done or the electric potential energy to move a point charge Q from point A to B in an electric field E is W Q E d l 8 The potential at r due to a point charge Q at r is Vr Q 47rsor r C where C is evaluated at a given reference potential point for example C 0 if Vr oo 0 To determine the potential due to a continuous charge distribution we replace Q in the formula for point charge by dQ pL dl dQ ps dS or dQ pv dv and integrate over the line surface or volume respectively 9 If the charge distribution is not known but the field intensity E is given we find the potential using VEdl W 10 The potential difference VAB the potential at B with reference to A is VAB J dl W VBVA 11 Since an electrostatic field is conservative the net work done along a closed path in a static E field is zero E dl 0 or V X E 0 second Maxwells equation to be derived 12 Given the potential field the corresponding electric field is found using E VV 13 For an electric dipole centered at r with dipole moment p the potential at r is given by Vr P r r 47rcor r3 14 D is tangential to the electric flux lines at every point An equipotential surface or line is one on which V constant At every point the equipotential line is orthogonal to the electric flux line REVIEW QUESTIONS 153 15 The electrostatic energy due to n point charges is WE 2 QkVk 1 For a continuous volume charge distribution DEdv eoEzdv REVIEW QUESTIONS 41 Point charges Q 1 nC and Q2 2 nC are at a distance apart Which of the following statements are incorrect a The force on Ql is repulsive b The force on Q2 is the same in magnitude as that on Qx c As the distance between them decreases the force on Ql increases linearly d The force on Q2 is along the line joining them e A point charge Q3 3 nC located at the midpoint between Q and Q2 experiences no net force 42 Plane z 10 m carries charge 20 nCm2 The electric field intensity at the origin is a 10aVm b 187razVm c 727razVm d 360irazVm 43 Point charges 30 nC 20 nC and 10 nC are located at 102 000 and 15 1 respectively The total flux leaving a cube of side 6 m centered at the origin is a 20 nC b 10 nC c 20 nC d 30 nC e 60 nC 44 The electric flux density on a spherical surface r b is the same for a point charge Q located at the origin and for charge Q uniformly distributed on surface r aa b a Yes b No c Not necessarily 154 Electrostatic Fields 45 The work done by the force F 4ax 3ay 2az N in giving a 1 nC charge a displace f lO 2 7 i 2ay 7az m is ment of a 103 nJ b 60 nJ c 64 nJ d 20 nJ 46 By saying that the electrostatic field is conservative we do not mean that a It is the gradient of a scalar potential b Its circulation is identically zero c Its curl is identically zero d The work done in a closed path inside the field is zero e The potential difference between any two points is zero 47 Suppose a uniform electric field exists in the room in which you are working such that the lines of force are horizontal and at right angles to one wall As you walk toward the wall from which the lines of force emerge into the room are you walking toward a Points of higher potential b Points of lower potential c Points of the same potential equipotential line 48 A charge Q is uniformly distributed throughout a sphere of radius a Taking the potential at infinity as zero the potential at r b a is a b c d Q 0 4irsor2 a Q 4ireor2 Q dr dr Qr 4irena dr dr 49 A potential field is given by V 3x2y yz Which of the following is not true a At point 1 0 1 V and E vanish b x2y 1 is an equipotential line on the xyplane c The equipotential surface V 8 passes through point P2 14 d The electric field at P is 12a 8a az Vm e A unit normal to the equipotential surface V 8 at P is 083a 055aj 007a7 PROBLEMS 410 An electric potential field is produced by point charges 1 juC and 4 2 1 5 and 1 3 1 respectively The energy stored in the field is a 257 mJ b 514 mJ c 1028 mJ d None of the above Answers 41ce 42d 43b 44a 45d 46e 47a 48c 49a 410b 155 located at PROBLEMS 41 Point charges Qx 5 jtC and Q2 4 xC are placed at 3 2 1 and 4 0 6 re spectively Determine the force on Qx 42 Five identical 15C point charges are located at the center and corners of a square defined by 1 x y 1 z 0 a Find the force on the 10C point charge at 0 0 2 b Calculate the electric field intensity at 0 0 2 43 Point charges Qx and Q2 are respectively located at 40 3 and 20 1 If Q2 4 nC find Qx such that a The E at 5 0 6 has no zcomponent b The force on a test charge at 5 0 6 has no jccomponent 44 Charges Q and 3Q are separated by a distance 2 m A third charge is located such that the electrostatic system is in equilibrium Find the location and the value of the third charge in terms of Q 45 Determine the total charge a On line 0 x 5 m if pL 2x2 mCm b On the cylinder p 3 0 z 4 m if ps pz2 nCm2 10 3 c Within the sphere r 4 m if pv Cm 46 Calculate the total charge due to the charge distributions labeled A B C in Fig 423 47 Find E at 5 0 0 due to charge distribution labeled A in Figure 423 48 Due to the charge distribution labeled B in Figure 423 a Find E at point 0 0 3 if ps 5 mCm2 b Find E at point 0 0 3 if ps 5 sin mCm2 49 A circular disk of radius a carries charge ps Cm2 Calculate the potential at 0 0 h P 156 Electrostatic Fields x 2 pv 1 mCm3 ps 5 mCm2 Figure 423 For Problem 46 410 A ring placed along y2 z2 4 x 0 carries a uniform charge of 5 xCm a FindDatP30 0 b If two identical point charges Q are placed at 0 3 0 and 0 3 0 in addition to the ring find the value of Q such that D 0 at P 411 a Show that the electric field at point 0 0 h due to the rectangle described by a x a b y b z 0 carrying uniform charge psCm2 is ab b If a 2 b 5ps 105 find the total charge on the plate and the electric field in tensity at 0 0 10 412 A point charge 100 pC is located at 41 3 while the xaxis carries charge 2 nCm If the plane z 3 also carries charge 5 nCm2 find E at 1 1 1 413 Linex 3 z 1 carries charge 20 nCm while plane x 2 carries charge 4 nCm2 Find the force on a point charge 5 mC located at the origin 414 Point charges are placed at the corners of a square of size 4 m as shown in Figure 424 If Q 15tC find D at 00 6 Q 2 2Q 2 0 2 2Q 2 e PROBLEMS Figure 424 For Problem 414 157 415 State Gausss law Deduce Coulombs law from Gausss law thereby affirming that Gausss law is an alternative statement of Coulombs law and that Coulombs law is im plicit in Maxwells equation V D pv 416 Determine the charge density due to each of the following electric flux densities a D xyax 4x Cm2 b D p sin t ap 2p cos 2z2az Cm2 c D 2 cos 6 ar sin 0 Cm2 417 Let E xyax x2ay find a Electric flux density D b The volume charge density pv 418 Plane x 2y 5 carries charge ps 6 nCm2 Determining E at 1 0 1 419 In free space D 2v2at 4xy az mCm2 Find the total charge stored in the region l x 2 l y 2 1 z 4 420 In a certain region the electric field is given by D 2pz lcos ap pz lsin 0 a0 p2 cos t az Cm2 a Find the charge density b Calculate the total charge enclosed by the volume 0 p 2 0 t x2 0 z 4 c Confirm Gausss law by finding the net flux through the surface of the volume in b 421 The Thomson model of a hydrogen atom is a sphere of positive charge with an electron a point charge at its center The total positive charge equals the electronic charge e Prove 158 Electrostatic Fields that when the electron is at a distance r from the center of the sphere of positive charge it is attracted with a force F where R is the radius of the sphere 422 Three concentric spherical shells r 1 r 2 and r 3 m respectively have charge distributions 2 4 and 5 a Calculate the flux through r 15 m and r 25 m b Find D at r 05 r 25 and r 35 m 423 Given that j Determine D everywhere 424 Let Up nCm3 0 1 P 2 otherwise mCm3 0 1 r 4 r 0 a Find the net flux crossing surface r 2 m and r 6 m b Determine D at r 1 m and r 5 m 425 Find the work done in carrying a 5C charge from Pl 2 4 to R3 5 6 in an elec tric field E ax z 2yzaz Vm 426 Given that the electric field in a certain region is E z 1 sin 0 a z 1 cos a0 p sin az Vm determine the work done in moving a 4nC charge from a Al0 0toB4 00 b S4 0 0 to C4 30 0 c C4 30 0toD4 30 2 d AtoD 427 In an electric field E 20r sin 6 ar lOr cos 6 ae Vm calculate the energy expended in transferring a 10nC charge a From A5 30 0 to B5 90 0 b From A to C 10 30 0 c FromAtoD530 60 d From A to 10 90 60 PROBLEMS 159 428 Let V xy2z calculate the energy expended in transfering a 2C point charge from 1 1 2 to 2 13 429 Determine the electric field due to the following potentials a V x2 2y2 4z2 b V sinx2 y2 z212 c V p2z lsin j d V er sin 6 cos 2t 430 Three point charges gi 1 mC Q2 2mC and Q3 3 mC are respectively located at 0 0 4 2 5 1 and 3 4 6 a Find the potential VP at P1 1 2 b Calculate the potential difference VPQ if Q is 1 2 3 431 In free space V x2yz 3 V Find a Eat 3 4 6 b the charge within the cube 0 xyz 1 432 A spherical charge distribution is given by ra 2 j 0 r a Find V everywhere 433 To verify that E yzax xzay xyaz Vm is truely an electric field show that a V X E 0 b jL E d 0 where L is the edge of the square defined ayOxy2z 1 434 a A total charge Q 60 fiC is split into two equal charges located at 180 intervals around a circular loop of radius 4 m Find the potential at the center of the loop b If Q is split into three equal charges spaced at 120 intervals around the loop find the potential at the center Q c If in the limit pL find the potential at the center 435 For a spherical charge distribution Pv Poa2 r2 0 r a r a a Find E and V for r a b Find E and V for r a c Find the total charge d Show that E is maximum when r 0145a 160 B Electrostatic Fields 436 a Prove that when a particle of constant mass and charge is accelerated from rest in an electric field its final velocity is proportional to the square root of the potential dif ference through which it is accelerated b Find the magnitude of the proportionality constant if the particle is an electron c Through what voltage must an electron be accelerated assuming no change in its mass to require a velocity onetenth that of light At such velocities the mass of a body becomes appreciably larger than its rest mass and cannot be considered constant 437 An electron is projected with an initial velocity uo 107 ms into the uniform field between the parallel plates of Figure 425 It enters the field at the midway between the plates If the electron just misses the upper plate as it emerges from the field a Find the electric field intensity b Calculate its velocity as it emerges from the field Neglect edge effects 438 An electric dipole with p paz C m is placed at x z 0 0 If the potential at 0 1 nm is 9 V find the potential at 1 1 nm 439 Point charges Q and Q are located at 0 d2 0 and 0 d2 0 Show that at point r 6 t where r d Qd sin 6 sin 4 V Find the corresponding E field 440 Determine the work necessary to transfer charges Q mC and Q2 2 mC from infinity to points 2 6 1 and 3 4 0 respectively 441 A point charge Q is placed at the origin Calculate the energy stored in region r a 442 Find the energy stored in the hemispherical region r 2 m 0 6 it where E 2r sin 8 cos j ar r cos 6 cos f ae r sin j a Vm exists 443 If V p2z sin calculate the energy within the region defined by 1 p 4 2 z 2 0 f 7i73 2 cm 10 cm Figure 425 For Problem 437 Chapter 5 ELECTRIC FIELDS IN MATERIAL SPACE The 12 Principles of character 1 Honesty 2 Understanding 3 Compassion 4 Appreciation 5 Patience 6 Discipline 7 Fortitude 8 Perseverance 9 Humor 10 Humility 11 Generosity 12 Respect KATHRYN B JOHNSON J1 INTRODUCTION In the last chapter we considered electrostatic fields in free space or a space that has no materials in it Thus what we have developed so far under electrostatics may be regarded as the vacuum field theory By the same token what we shall develop in this chapter may be regarded as the theory of electric phenomena in material space As will soon be evident most of the formulas derived in Chapter 4 are still applicable though some may require modification Just as electric fields can exist in free space they can exist in material media Materi als are broadly classified in terms of their electrical properties as conductors and noncon ductors Nonconducting materials are usually referred to as insulators or dielectrics A brief discussion of the electrical properties of materials in general will be given to provide a basis for understanding the concepts of conduction electric current and polarization Further discussion will be on some properties of dielectric materials such as susceptibility permittivity linearity isotropy homogeneity dielectric strength and relaxation time The concept of boundary conditions for electric fields existing in two different media will be introduced 2 PROPERTIES OF MATERIALS In a text of this kind a discussion on electrical properties of materials may seem out of place But questions such as why an electron does not leave a conductor surface why a currentcarrying wire remains uncharged why materials behave differently in an electric field and why waves travel with less speed in conductors than in dielectrics are easily an swered by considering the electrical properties of materials A thorough discussion on this subject is usually found in texts on physical electronics or electrical engineering Here a 161 162 Electric Fields in Material Space brief discussion will suffice to help us understand the mechanism by which materials influ ence an electric field In a broad sense materials may be classified in terms of their conductivity a in mhos per meter Um or Siemens per meter Sm as conductors and nonconductors or techni cally as metals and insulators or dielectrics The conductivity of a material usually depends on temperature and frequency A material with high conductivity a 1 is re ferred to as a metal whereas one with low conductivity a sC 1 is referred to as an insu lator A material whose conductivity lies somewhere between those of metals and insula tors is called a semiconductor The values of conductivity of some common materials as shown in Table B 1 in Appendix B From this table it is clear that materials such as copper and aluminum are metals silicon and germanium are semiconductors and glass and rubber are insulators The conductivity of metals generally increases with decrease in temperature At tem peratures near absolute zero T 0K some conductors exhibit infinite conductivity and are called superconductors Lead and aluminum are typical examples of such metals The conductivity of lead at 4K is of the order of 1020 mhosm The interested reader is referred to the literature on superconductivity1 We shall only be concerned with metals and insulators in this text Microscopically the major difference between a metal and an insulator lies in the amount of electrons avail able for conduction of current Dielectric materials have few electrons available for con duction of current in contrast to metals which have an abundance of free electrons Further discussion on the presence of conductors and dielectrics in an electric field will be given in subsequent sections 53 CONVECTION AND CONDUCTION CURRENTS Electric voltage or potential difference and current are two fundamental quantities in electrical engineering We considered potential in the last chapter Before examining how electric field behaves in a conductor or dielectric it is appropriate to consider electric current Electric current is generally caused by the motion of electric charges The current in amperes through a given area is the electric charge passing through the area per unit time That is dQ I 51 dt Thus in a current of one ampere charge is being transferred at a rate of one columb per second The August 1989 issue of the Proceedings of IEEE was devoted to Applications of Superconduc tivity 53 CONVECTION AND CONDUCTION CURRENTS 163 We now introduce the concept of current density J If current A flows through a surface AS the current density is L JnJs or A JnAS 52 assuming that the current density is perpendicular to the surface If the current density is not normal to the surface A J AS Thus the total current flowing through a surface S is 3dS s 53 54 Depending on how is produced there are different kinds of current densities convection current density conduction current density and displacement current density We will con sider convection and conduction current densities here displacement current density will be considered in Chapter 9 What we need to keep in mind is that eq 54 applies to any kind of current density Compared with the general definition of flux in eq 313 eq 54 shows that the current through S is merely the flux of the current density J Convection current as distinct from conduction current does not involve conductors and consequently does not satisfy Ohms law It occurs when current flows through an in sulating medium such as liquid rarefied gas or a vacuum A beam of electrons in a vacuum tube for example is a convection current Consider a filament of Figure 51 If there is a flow of charge of density pv at velocity y from eq 51 the current through the filament is u At 55 I he current densilj m a given point i ihc current through a unit normal area at that point AS A A A Figure 51 Current in a filament 164 Hi Electric Fields in Material Space The vdirected current density Jy is given by A 56 Hence in general J 57 The current is the convection current and J is the convection current density in amperessquare meter Am2 Conduction current requires a conductor A conductor is characterized by a large amount of free electrons that provide conduction current due an impressed electric field When an electric field E is applied the force on an electron with charge e is F eE 58 Since the electron is not in free space it will not be accelerated under the influence of the electric field Rather it suffers constant collision with the atomic lattice and drifts from one atom to another If the electron with mass m is moving in an electric field E with an average drift velocity u according to Newtons law the average change in momentum of the free electron must match the applied force Thus mxx eE T or m 59 where T is the average time interval between collisions This indicates that the drift veloc ity of the electron is directly proportional to the applied field If there are n electrons per unit volume the electronic charge density is given by pv ne 510 Thus the conduction current density is J pvu E aE m or J aE 511 where a ne2rm is the conductivity of the conductor As mentioned earlier the values of a for common materials are provided in Table BI in Appendix B The relationship in eq 511 is known as the point form of Ohms law 54 CONDUCTORS 165 54 CONDUCTORS A conductor has abundance of charge that is free to move Consider an isolated conductor such as shown in Figure 52a When an external electric field Ee is applied the positive free charges are pushed along the same direction as the applied field while the negative free charges move in the opposite direction This charge migration takes place very quickly The free charges do two things First they accumulate on the surface of the con ductor and form an induced surface charge Second the induced charges set up an internal induced field E which cancels the externally applied field Ee The result is illustrated in Figure 52b This leads to an important property of a conductor A perfect conductor cannot contain an electrostatic field within it A conductor is called an equipotential body implying that the potential is the same every where in the conductor This is based on the fact that E W 0 Another way of looking at this is to consider Ohms law J oE To maintain a finite current density J in a perfect conductor a requires that the electric field inside the conductor must vanish In other words E 0 because a in a perfect conductor If some charges are introduced in the interior of such a conductor the charges will move to the surface and redistribute themselves quickly in such a manner that the field inside the conductor vanishes According to Gausss law if E 0 the charge density pv must be zero We conclude again that a perfect conductor cannot contain an electrostatic field within it Under static conditions E 0 pv 0 Vab 0 inside a conductor 512 E E a b Figure 52 a An isolated conductor under the influence of an applied field b a conductor has zero electric field under static conditions 166 Electric Fields in Material Space We now consider a conductor whose ends are maintained at a potential difference V as shown in Figure 53 Note that in this case E 0 inside the conductor as in Figure 52 What is the difference There is no static equilibrium in Figure 53 since the conductor is not isolated but wired to a source of electromotive force which compels the free charges to move and prevents the eventual establishment of electrostatic equilibrium Thus in the case of Figure 53 an electric field must exist inside the conductor to sustain the flow of current As the electrons move they encounter some damping forces called resistance Based on Ohms law in eq 511 we will derive the resistance of the conducting material Suppose the conductor has a uniform cross section S and is of length The direction of the electric field E produced is the same as the direction of the flow of positive charges or current This direction is opposite to the direction of the flow of electrons The electric field applied is uniform and its magnitude is given by V 513 Since the conductor has a uniform cross section 514 Substituting eqs 511 and 513 into eq 514 gives oV oE S t 515 Hence I aS or 516 Figure 53 A conductor of uniform cross section under an applied E field v 54 CONDUCTORS 167 where pc Ia is the resistivity of the material Equation 516 is useful in determining the resistance of any conductor of uniform cross section If the cross section of the conductor is not uniform eq 516 is not applicable However the basic definition of resistance R as the ratio of the potential difference V between the two ends of the conductor to the current through the conductor still applies Therefore applying eqs 460 and 54 gives the re sistance of a conductor of nonuniform cross section that is 517 Note that the negative sign before V fEdl is dropped in eq 517 because E d 0 if 0 Equation 517 will not be utilized until we get to Section 65 Power P in watts is defined as the rate of change of energy W in joules or force times velocity Hence pv dv E u E pvu dv or 518 which is known as Joules law The power density wP in wattsm3 is given by the inte grand in eq 518 that is wP f EJ jE dv For a conductor with uniform cross section dv dS dl so eq 518 becomes 519 P Edl JdS VI S or P I2R which is the more common form of Joules law in electric circuit theory 520 EXAMPLE 51 If J 7 2 cos 6 ar sin 0 ae Am2 calculate the current passing through r a A hemispherical shell of radius 20 cm b A spherical shell of radius 10 cm 168 Electric Fields in Material Space Solution J dS where dS r2 sin 0 dj dd ar in this case fT2 rlK 1 a 2 cos 0 r2 sin 0 JA d0 40 0 r r 0 2 2TT sin 6 dsin 0 00 r02 4TT sin2 02 2 x2 10x 314A b The only difference here is that we have 0 6 w instead of 0 0 7r2 and r 01 Hence 4TT sin2 01 2 0 Alternatively for this case since V J 0 J JS JV J dv 0 PRACTICE EXERCISE 51 For the current density J 0z sin2 ap Am2 find the current through the cylindri cal surface p 2 1 z 5 5 m Answer 754 A EXAMPLE 52 A typical example of convective charge transport is found in the Van de Graaff generator where charge is transported on a moving belt from the base to the dome as shown in Figure 54 If a surface charge density 107 Cm2 is transported at a velocity of 2 ms cal culate the charge collected in 5 s Take the width of the belt as 10 cm Solution If ps surface charge density u speed of the belt and w width of the belt the current on the dome is I psuw The total charge collected in t 5 s is Q It psuwt 1 0 7 X 2 X 01 X 5 100 nC 54 CONDUCTORS 169 charge removal charge placement conducting dome insulating support motor Figure 54 Van de Graaff generator for Example 52 conducting base PRACTICE EXERCISE 52 In a Van de Graaff generator w 01 m u 10 ms and the leakage paths have re sistance 1014 Q If the belt carries charge 05 jtCm2 find the potential difference between the dome and the base Answer 50 MV EXAMPLE 53 A wire of diameter 1 mm and conductivity 5 X 107 Sm has 1029 free electronsm3 when an electric field of 10 mVm is applied Determine a The charge density of free electrons b The current density c The current in the wire d The drift velocity of the electrons Take the electronic charge as e 16 X 10 C Solution In this particular problem convection and conduction currents are the same a pv ne 102916 X 1019 16 X 1010 Cm3 b J oE 5 X 10710 X 103 500 kAm2 c JS 5 X 105 ltd 106 105 0393 A J 5 X 105 d Since pvu u 3125 X 105 ms 170 Electric Fields in Material Space PRACTICE EXERCISE 53 The free charge density in copper is 181 X 10loCm3 For a current density of 8 X 105 Am2 find the electric field intensity and the drift velocity Answer 0138 Vm 442 X 1 T4 ms EXAMPLE 54 A lead a 5 X 106 Sm bar of square cross section has a hole bored along its length of 4 m so that its cross section becomes that of Figure 55 Find the resistance between the square ends Solution Since the cross section of the bar is uniform we may apply eq 516 that is aS where S d2 ntr2 32 TTQ J 9 cm2 Hence R 5 X 1069 TT4 X 10 91A j PRACTICE EXERCISE 54 If the hole in the lead bar of Example 54 is completely filled with copper a 58 X 106 mhosm determine the resistance of the composite bar Answer 8767 xQ 1 cm Figure 55 Cross section of the lead bar of Example 54 3 cm 3 cm 55 POLARIZATION IN DIELECTRICS 171 55 POLARIZATION IN DIELECTRICS In Section 52 we noticed that the main difference between a conductor and a dielectric lies in the availability of free electrons in the atomic outermost shells to conduct current Although the charges in a dielectric are not able to move about freely they are bound by finite forces and we may certainly expect a displacement when an external force is applied To understand the macroscopic effect of an electric field on a dielectric consider an atom of the dielectric as consisting of a negative charge Q electron cloud and a positive charge Q nucleus as in Figure 56a A similar picture can be adopted for a dielectric molecule we can treat the nuclei in molecules as point charges and the electronic structure as a single cloud of negative charge Since we have equal amounts of positive and negative charge the whole atom or molecule is electrically neutral When an electric field E is applied the positive charge is displaced from its equilibrium position in the direction of E by the force F QE while the negative charge is displaced in the opposite direction by the force F QE A dipole results from the displacement of the charges and the dielec tric is said to be polarized In the polarized state the electron cloud is distorted by the applied electric field E This distorted charge distribution is equivalent by the principle of superposition to the original distribution plus a dipole whose moment is p QA 521 where d is the distance vector from Q to Q of the dipole as in Figure 56b If there are N dipoles in a volume Av of the dielectric the total dipole moment due to the electric field is 522 As a measure of intensity of the polarization we define polarization P in coulombsmeter square as the dipole moment per unit volume of the dielectric that is 523 Thus we conclude that the major effect of the electric field E on a dielectric is the cre ation of dipole moments that align themselves in the direction of E This type of dielectric Figure 56 Polarization of a nonpolar atom or molecule 172 Electric Fields in Material Space Figure 57 Polarization of a polar molecule a permanent dipole E 0 b alignment of permanent dipole E 0 a b is said to be nonpolar Examples of such dielectrics are hydrogen oxygen nitrogen and the rare gases Nonpolar dielectric molecules do not possess dipoles until the application of the electric field as we have noticed Other types of molecules such as water sulfur dioxide and hydrochloric acid have builtin permanent dipoles that are randomly oriented as shown in Figure 57a and are said to be polar When an electric field E is applied to a polar molecule the permanent dipole experiences a torque tending to align its dipole moment parallel with E as in Figure 57b Let us now calculate the field due to a polarized dielectric Consider the dielectric ma terial shown in Figure 58 as consisting of dipoles with dipole moment P per unit volume According to eq 480 the potential dV at an exterior point O due to the dipole moment P dv is P as dv 4irsJi2 524 where R2 x x2 y y2 z z2 and R is the distance between the volume element dv at x y z and the field point O x y z We can transform eq 524 into a form that facilitates physical interpretation It is readily shown see Section 77 that the gradient of IR with respect to the primed coordinates is R Rz Thus R2 Oxyz Figure 58 A block of dielectric material with dipole moment p per unit volume 55 POLARIZATION IN DIELECTRICS 173 Applying the vector identity V A V A A V P a R P V P 2 fl 525 Substituting this into eq 524 and integrating over the entire volume v of the dielectric we obtain V v 47reoL R R Applying divergence theorem to the first term leads finally to V P V 4ireJi dS dv 526 where an is the outward unit normal to surface dS of the dielectric Comparing the two terms on the right side of eq 526 with eqs 468 and 469 shows that the two terms denote the potential due to surface and volume charge distributions with densities upon dropping the primes PPs P an P P V V P 527a 527b In other words eq 526 reveals that where polarization occurs an equivalent volume charge density ppv is formed throughout the dielectric while an equivalent surface charge density pps is formed over the surface of the dielectric We refer to pps and ppv as bound or polarization surface and volume charge densities respectively as distinct from free surface and volume charge densities ps and pv Bound charges are those that are not free to move within the dielectric material they are caused by the displacement that occurs on a molecular scale during polarization Free charges are those that are capable of moving over macroscopic distance as electrons in a conductor they are the stuff we control The total positive bound charge on surface S bounding the dielectric is 528a 528b Qb PdS j PpsdS while the charge that remains inside S is Qb I Ppvdv VPJv Thus the total charge of the dielectric material remains zero that is Total charge 4 p dS ppv dv Qb Qb 0 This is expected because the dielectric was electrically neutral before polarization 174 Electric Fields in Material Space We now consider the case in which the dielectric region contains free charge If pv is the free charge volume density the total volume charge density p is given by P Pv Ppv V s oE 529 Hence pv V eoE ppv V 8OE P V D where D P 530 531 We conclude that the net effect of the dielectric on the electric field E is to increase D inside it by amount P In other words due to the application of E to the dielectric material the flux density is greater than it would be in free space It should be noted that the defini tion of D in eq 435 for free space is a special case of that in eq 531 because P 0 in free space 4 We would expect that the polarization P would vary directly as the applied electric field E For some dielectrics this is usually the case and we have P 532 where e known as the electric susceptibility of the material is more or less a measure of how susceptible or sensitive a given dielectric is to electric fields 56 DIELECTRIC CONSTANT AND STRENGTH By substituting eq 532 into eq 531 we obtain D eol x e E eoeE or where D eE e eoer 533 534 535 and er 1 Xe 536 57 LINEAR ISOTROPIC AND HOMOGENEOUS DIELECTRICS 175 In eqs 533 to 536 s is called the permittivity of the dielectric so is the permittiv ity of free space defined in eq 42 as approximately 10936TT Fm and sr is called the dielectric constant or relative permittivity The dielectric constant or relative permittivity e is the ratio of the permittivity of the dielectric to that of free space It should also be noticed that er and e are dimensionless whereas e and so are in faradsmeter The approximate values of the dielectric constants of some common materi als are given in Table B2 in Appendix B The values given in Table B2 are for static or low frequency 1000 Hz fields the values may change at high frequencies Note from the table that er is always greater or equal to unity For free space and nondielectric mate rials such as metals er 1 The theory of dielectrics we have discussed so far assumes ideal dielectrics Practi cally speaking no dielectric is ideal When the electric field in a dielectric is sufficiently large it begins to pull electrons completely out of the molecules and the dielectric becomes conducting Dielectric breakdown is said to have occurred when a dielectric becomes conducting Dielectric breakdown occurs in all kinds of dielectric materials gases liquids or solids and depends on the nature of the material temperature humid ity and the amount of time that the field is applied The minimum value of the electric field at which dielectric breakdown occurs is called the dielectric strength of the dielec tric material The dielectric strength is the maximum electric field that a dielectric can tolerate or withstand without breakdown Table B2 also lists the dielectric strength of some common dielectrics Since our theory of dielectrics does not apply after dielectric breakdown has taken place we shall always assume ideal dielectric and avoid dielectric breakdown 57 LINEAR ISOTROPIC AND HOMOGENEOUS DIELECTRICS Although eqs 524 to 531 are for dielectric materials in general eqs 532 to 534 are only for linear isotropic materials A material is said to be linear if D varies linearly with E and nonlinear otherwise Materials for which s or a does not vary in the region being considered and is therefore the same at all points ie independent of x y z are said to be homogeneous They are said to be inhomogeneous or nonhomogeneous when e is depen dent of the space coordinates The atmosphere is a typical example of an inhomogeneous medium its permittivity varies with altitude Materials for which D and E are in the same direction are said to be isotropic That is isotropic dielectrics are those which have the same properties in all directions For anisotropic or nonisotropic materials D E and P 176 Electric Fields in Material Space are not parallel e or xe has nine components that are collectively referred to as a tensor For example instead of eq 534 we have Dy D yy Ey E 537 for anisotropic materials Crystalline materials and magnetized plasma are anisotropic A dielectric material in which I EK applies is linear if E does noi change with llie applied E tield homogeneous if e does mil change from point lo point and isolropic if i does not change with direction The same idea holds for a conducting material in which J oE applies The material is linear if a does not vary with E homogeneous if a is the same at all points and isotropic if a does not vary with direction For most of the time we will be concerned only with linear isotropic and homoge neous media For such media all formulas derived in Chapter 4 for free space can be applied by merely replacing eo with eoer Thus Coulombs law of eq 44 for example becomes F and eq 496 becomes W dv 538 539 when applied to a dielectric medium EXAMPLE 55 A dielectric cube of side L and center at the origin has a radial polarization given by P av where a is a constant and r xax yay zaz Find all bound charge densities and show explicitly that the total bound charge vanishes Solution For each of the six faces of the cube there is a surface charge pps For the face located at x L2 for example PPS P a The total bound surface charge is ax ahll Q j PpsdS 6 3aL3 L2 L2 JU2 6aL 2 ppsdydz L 57 LINEAR ISOTROPIC AND HOMOGENEOUS DIELECTRICS The bound volume charge density is given by ppv V P a a a 3a and the total bound volume charge is 177 Qv Ppvdv 3a dv 3aL Hence the total charge is Q Qs Gv 3aL3 3aL3 0 PRACTICE EXERCISE 55 A thin rod of cross section A extends along the axis from x 0 to x L The polarization of the rod is along its length and is given by Px ax2 b Calcu late ppv and pps at each end Show explicitly that the total bound charge vanishes in this case Answer 0 2aL b aL1 b proof EXAMPLE 56 The electric field intensity in polystyrene er 255 filling the space between the plates of a parallelplate capacitor is 10 kVm The distance between the plates is 15 mm Calcu late a D b P c The surface charge density of free charge on the plates d The surface density of polarization charge e The potential difference between the plates Solution a D eoeE 10 b P 36TT 155 255 104 2254 nCm2 10 9 36ir 104 137 nCm2 c Ps D an Dn 225 A nCm2 d pps P an Pn 137 nCm2 e V Ed 104 15 X 103 15 V 178 Electric Fields in Material Space PRACTICE EXERCISE 56 A parallelplate capacitor with plate separation of 2 mm has a 1kV voltage applied to its plates If the space between its plates is filled with polystyrene er 255 find E P and pps Answer 500a kVm 6853a jCm26853 Cm2 EXAMPLE 57 A dielectric sphere er 57 of radius 10 cm has a point charge 2 pC placed at its center Calculate a The surface density of polarization charge on the surface of the sphere b The force exerted by the charge on a 4pC point charge placed on the sphere Solution a We apply Coulombs or Gausss law to obtain Q E 4ireo8rr P Pps ir 4irsrr er 1 Q 47 2 X 1312 pCm2 b Using Coulombs law we have 47rerr2 4ir57 100 X 42 X 10 2 4 4ireoerr 1263 arpN 10 9 36TT 57 100 X 10 PRACTICE EXERCISE 57 In a dielectric material Ex 5 Vm and P 33 ay 43 nCm2 Calculate aXe bE c D 57 LINEAR ISOTROPIC AND HOMOGENEOUS DIELECTRICS B 179 Answer a 216 b 5a 167a 667azVm c 1397a 466ay 1863a pCm2 EXAMPLE 58 Find the force with which the plates of a parallelplate capacitor attract each other Also de termine the pressure on the surface of the plate due to the field Solution From eq 426 the electric field intensity on the surface of each plate is where an is a unit normal to the plate and ps is the surface charge density The total force on each plate is oer or F P2 SS Q2 2e 2eS Ps The pressure of forcearea is 2eoer PRACTICE EXERCISE 58 Shown in Figure 59 is a potential measuring device known as an electrometer It is basically a parallelplate capacitor with the guarded plate being suspended from a balance arm so that the force F on it is measurable in terms of weight If S is the area of each plate show that V2 2Fd2 112 Answer Proof 180 Electric Fields in Material Space Figure 59 An electrometer for Practice Exercise 58 58 CONTINUITY EQUATION AND RELAXATION TIME Due to the principle of charge conservation the time rate of decrease of charge within a given volume must be equal to the net outward current flow through the closed surface of the volume Thus current out coming out of the closed surface is out T J dQin dt 540 where Qin is the total charge enclosed by the closed surface Invoking divergence theorem But dQm J dS V J dv dr dt dp dt dt J dt Substituting eqs 541 and 542 into eq 540 gives dv 541 542 J a v dv dt or dt 543 which is called the continuity of current equation It must be kept in mind that the continu ity equation is derived from the principle of conservation of charge and essentially states that there can be no accumulation of charge at any point For steady currents dpvdt 0 and hence V J 0 showing that the total charge leaving a volume is the same as the total charge entering it Kirchhoffs current law follows from this Having discussed the continuity equation and the properties a and e of materials it is appropriate to consider the effect of introducing charge at some interior point of a given 58 CONTINUITY EQUATION AND RELAXATION TIME ffi 181 material conductor or dielectric We make use of eq 543 in conjunction with Ohms law J aE and Gausss law 544 545 Substituting eqs 544 and 545 into eq 543 yields e dt or 546 This is a homogeneous linear ordinary differential equation By separating variables in eq 546 we get and integrating both sides gives dt Pv B at In pv 1 In pvo where In pvo is a constant of integration Thus Pv PvoetlTr where 547 548 549 In eq 548 pm is the initial charge density ie pv at t 0 The equation shows that as a result of introducing charge at some interior point of the material there is a decay of volume charge density pv Associated with the decay is charge movement from the interior point at which it was introduced to the surface of the material The time constant Tr in seconds is known as the relaxation time or rearrangement time Relaxation time is the lime it takes u charge placed in the interior of a material to drop to e l 368 percent ofils initial value 182 Electric Fields in Material Space It is short for good conductors and long for good dielectrics For example for copper a 58 X 107 mhosm er 1 and 10 1 o 36TT 153 X 1019s 58 X 107 550 showing a rapid decay of charge placed inside copper This implies that for good conduc tors the relaxation time is so short that most of the charge will vanish from any interior point and appear at the surface as surface charge On the other hand for fused quartz for instance a 1017 mhosm sr 50 10 9 1 36TT 1017 512 days 551 showing a very large relaxation time Thus for good dielectrics one may consider the in troduced charge to remain wherever placed 59 BOUNDARY CONDITIONS So far we have considered the existence of the electric field in a homogeneous medium If the field exists in a region consisting of two different media the conditions that the field must satisfy at the interface separating the media are called boundary conditions These con ditions are helpful in determining the field on one side of the boundary if the field on the other side is known Obviously the conditions will be dictated by the types of material the media are made of We shall consider the boundary conditions at an interface separating dielectric sr and dielectric er2 conductor and dielectric conductor and free space To determine the boundary conditions we need to use Maxwells equations and Edl DdS Qenc 552 553 Also we need to decompose the electric field intensity E into two orthogonal components E Et En 554 where E and En are respectively the tangential and normal components of E to the inter face of interest A similar decomposition can be done for the electric flux density D 59 BOUNDARY CONDITIONS 183 A DielectricDielectric Boundary Conditions Consider the E field existing in a region consisting of two different dielectrics character ized by E os r l and e2 eoer2 as shown in Figure 510a E1 and E2 in media 1 and 2 respectively can be decomposed as E E ElB E2 E2 E2n We apply eq 552 to the closed path abcda of Figure 510a assuming that the path i very small with respect to the variation of E We obtain 555a 555b is where E Er and En EB As Ah 0 eq 556 becomes Ey E2 u 556 557 Thus the tangential components of E are the same on the two sides of the boundary In other words E undergoes no change on the boundary and it is said to be continuous across the boundary Since D eE D Dn eq 557 can be written as F F 2 ei e2 or D 2 558 that is D undergoes some change across the interface Hence D is said to be discontinu ous across the interface a Figure 510 Dielectricdielectric boundary b 184 Electric Fields in Material Space Similarly we apply eq 553 to the pillbox Gaussian surface of Figure 510b Al lowing Ah 0 gives AQ Ps AS Du AS D2n AS or D2n ps 559 where ps is the free charge density placed deliberately at the boundary It should be borne in mind that eq 559 is based on the assumption that D is directed from region 2 to region 1 and eq 559 must be applied accordingly If no free charges exist at the interface ie charges are not deliberately placed there ps 0 and eq 559 becomes Dm D2n 560 Thus the normal component of D is continuous across the interface that is Dn undergoes no change at the boundary Since D eE eq 560 can be written as EEXn s2E2n 561 showing that the normal component of E is discontinuous at the boundary Equations 557 and 559 or 560 are collectively referred to as boundary conditions they must be satisfied by an electric field at the boundary separating two different dielectrics As mentioned earlier the boundary conditions are usually applied in finding the elec tric field on one side of the boundary given the field on the other side Besides this we can use the boundary conditions to determine the refraction of the electric field across the in terface Consider D1 or E and D2 or E2 making angles 6X and d2 with the normal to the in terface as illustrated in Figure 511 Using eq 557 we have Ex sin 0 Eu E2t E2 sin 62 Figure 511 Refraction of D or E at a dielectricdielectric boundary 59 BOUNDARY CONDITIONS HS 185 or Ei sin di E2 sin 62 Similarly by applying eq 560 or 561 we get eiEi cos di ln D2n s2E2 cos d2 or ExEi COS 6y B2E2 c o s 2 Dividing eq 562 by eq 563 gives tan 61 tan d2 ei e2 Since ei eoen and e2 eoer2 eq 564 becomes 562 tan 91 erl tan 563 564 565 This is the law of refraction of the electric field at a boundary free of charge since ps 0 is assumed at the interface Thus in general an interface between two dielectrics pro duces bending of the flux lines as a result of unequal polarization charges that accumulate on the sides of the interface B ConductorDielectric Boundary Conditions This is the case shown in Figure 512 The conductor is assumed to be perfect ie a oo or pc 0 Although such a conductor is not practically realizable we may regard con ductors such as copper and silver as though they were perfect conductors dielectric e eoet conductor E 0 conductor E 0 a b Figure 512 Conductordielectric boundary 186 US Electric Fields in Material Space To determine the boundary conditions for a conductordielectric interface we follow the same procedure used for dielectricdielectric interface except that we incorporate the fact that E 0 inside the conductor Applying eq 552 to the closed path abcda of Figure 512a gives 0 0Aw 0L EnEtAwEnL0 566 As Ah 0 E 0 567 Similarly by applying eq 553 to the pillbox of Figure 512b and letting Ah 0 we get AQ Dn AS 0 AS because D eE 0 inside the conductor Equation 568 may be written as n AQ D P 568 or n Ps 569 Thus under static conditions the following conclusions can be made about a perfect conductor 1 No electric field may exist within a conductor that is pv 0 E 0 570 2 Since E W 0 there can be no potential difference between any two points in the conductor that is a conductor is an equipotential body 3 The electric field E can be external to the conductor and normal to its surface that is D eoeEt 0 Dn E O G ps 571 An important application of the fact that E 0 inside a conductor is in electrostatic screening or shielding If conductor A kept at zero potential surrounds conductor B as shown in Figure 513 B is said to be electrically screened by A from other electric systems such as conductor C outside A Similarly conductor C outside A is screened by A from B 59 BOUNDARY CONDITIONS Figure 513 Electrostatic screening 187 Thus conductor A acts like a screen or shield and the electrical conditions inside and outside the screen are completely independent of each other C ConductorFree Space Boundary Conditions This is a special case of the conductordielectric conditions and is illustrated in Figure 514 The boundary conditions at the interface between a conductor and free space can be obtained from eq 571 by replacing er by 1 because free space may be regarded as a special dielectric for which er 1 We expect the electric field E to be external to the conductor and normal to its surface Thus the boundary conditions are D saEt 0 Dn eoEn ps 572 It should be noted again that eq 572 implies that E field must approach a conducting surface normally Figure 514 Conductorfree space boundary tree pacc iKinr ih 0i 188 Electric Fields in Material Space EXAMPLE 59 Two extensive homogeneous isotropic dielectrics meet on plane z 0 For z 0 erl 4 and for z 0 sr2 3 A uniform electric field E 5a 2ay 3az kVm exists for z 0 Find a E2 for z 0 b The angles E and E2 make with the interface c The energy densities in Jm3 in both dielectrics d The energy within a cube of side 2 m centered at 3 4 5 Solution Let the problem be as illustrated in Figure 515 a Since az is normal to the boundary plane we obtain the normal components as Eln Ej an Ej az 3 EiB 3a E2n E2 azaz Also Hence E En E E l r E E l B 5a 2ay Figure 515 For Example 59 59 BOUNDARY CONDITIONS i 189 Thus Similarly or E2t Ei 5a 2a D2 Din r2E2n ErlEln ElB J3az 4az 6 3 Thus E2 E2r E2n 5ax 2a 4az kVm b Let i and a2 be the angles E and E2 make with the interface while 0 and 02 are the angles they make with the normal to the interface as shown in Figures 515 that is ai 90 0 a2 90 02 Since Eln 3 and Elt V25 4 tan 0i Elt 1795 0j 609 Hence Alternatively or a 291C an lEj 1 cos0i Similarly Hence cos 0t 04867 0 609 V38 4 2 lr V29 E tan 02 v29 1346 62 534 E2n 4 a2 366 190 Electric Fields in Material Space tan 0 erl Note that is satisfied tan 02 er2 c The energy densities are given by 2 2 672 106 1 r2 597 1 9 3 25 4 16 X 106 2 36x d At the center 3 4 5 of the cube of side 2 m z 5 0 that is the cube is in region 2 with 2 x 4 3 y 5 6 z 4 Hence w2222 w2 dv w2 4 2 43 46 597 X 8juJ 4776 mJ PRACTICE EXERCISE 59 A homogeneous dielectric er 25 fills region 1 x 0 while region 2 x 0 is free space a IfD 12a 10ay 4a nCm2 find D2 and 02 b If E2 12 Vm and 02 60 find and 0j Take 0j and 02 as defined in the previous example Answer a 12a 4ay 16az nCm21975 b 1067 Vm 77 EXAMPLE 510 Region y 0 consists of a perfect conductor while region y 0 is a dielectric medium elr 2 as in Figure 516 If there is a surface charge of 2 nCm2 on the conductor deter mine E and D at a A322 b B4 1 5 Solution a Point A3 2 2 is in the conductor since y 2 0 at A Hence E 0 D b Point fi4 15 is in the dielectric medium since y 1 0 at B Dn p s 2 nCm2 conductor Hence and A t D 2av nCm2 SUMMARY 191 Figure 516 See Example 510 E 2 X 109 X X 109a 36 1131a Vm PRACTICE EXERCISE 510 It is found that E 60ax 20ay 30az mVm at a particular point on the interface between air and a conducting surface Find D and ps at that point Answer 0531a 0177ay 0265az pCm2 0619 pCm2 SUMMARY 1 Materials can be classified roughly as conductors a 1 sr 1 and dielectrics a sC 1 er 1 in terms of their electrical properties a and en where a is the con ductivity and sr is the dielectric constant or relative permittivity 2 Electric current is the flux of electric current density through a surface that is I 3dS 3 The resistance of a conductor of uniform cross section is aS 192 M Electric Fields in Material Space 4 The macroscopic effect of polarization on a given volume of a dielectric material is to paint its surface with a bound charge Qh js pps dS and leave within it an accumu lation of bound charge Qb fvppv dv where pps P an and pp V P 5 In a dielectric medium the D and E fields are related as D sE where e eosr is the permittivity of the medium 6 The electric susceptibility xe er 1 of a dielectric measures the sensitivity of the material to an electric field 7 A dielectric material is linear if D eE holds that is if s is independent of E It is ho mogeneous if e is independent of position It is isotropic if s is a scalar 8 The principle of charge conservation the basis of Kirchhoffs current law is stated in the continuity equation dt 9 The relaxation time Tr elo of a material is the time taken by a charge placed in its interior to decrease by a factor of e 37 percent 10 Boundary conditions must be satisfied by an electric field existing in two different media separated by an interface For a dielectricdielectric interface C1 p D D2n ps or Dln D2r For a dielectricconductor interface E 0 Dn eEn ps because E 0 inside the conductor if ps 0 REVIEW QUESTIONS 51 Which is not an example of convection current a A moving charged belt b Electronic movement in a vacuum tube c An electron beam in a television tube d Electric current flowing in a copper wire 52 When a steady potential difference is applied across the ends of a conducting wire a All electrons move with a constant velocity b All electrons move with a constant acceleration c The random electronic motion will on the average be equivalent to a constant veloc ity of each electron d The random electronic motion will on the average be equivalent to a nonzero con stant acceleration of each electron REVIEW QUESTIONS 193 53 The formula R oS is for thin wires a True b False c Not necessarily 54 Sea water has er 80 Its permittivity is a 81 b 79 c 5162 X 10luFm d 7074 X 1010Fm 55 Both eo and xe are dimensionless a True b False 56 If V D 8 V E and V J jVEina given material the material is said to be a Linear b Homogeneous c Isotropic d Linear and homogeneous e Linear and isotropic f Isotropic and homogeneous 57 The relaxation time of mica a 10 mhosm er 6 is a 5 X 10 1 0s b 106s c 5 hours d 10 hours e 15 hours 58 The uniform fields shown in Figure 517 are near a dielectricdielectric boundary but on opposite sides of it Which configurations are correct Assume that the boundary is charge free and that e2 e 59 Which of the following statements are incorrect a The conductivities of conductors and insulators vary with temperature and fre quency b A conductor is an equipotential body and E is always tangential to the conductor c Nonpolar molecules have no permanent dipoles d In a linear dielectric P varies linearly with E 194 Electric Fields in Material Space a O X b c d e Figure 517 For Review Question 58 0 PROBLEMS 510 The electric conditions charge and potential inside and outside an electric screening are completely independent of one another a True b False Answers 5Id 52c 53c 54d 55b 56d 57e 58e 59b 510a 51 In a certain region J 3r2 cos 6 ar r2 sin d as Am find the current crossing the surface defined by 6 30 0 0 2TT 0 r 2 m 52 Determine the total current in a wire of radius 16 mm if J 500a Am2 P 53 The current density in a cylindrical conductor of radius a is J l0e1plaazAm2 Find the current through the cross section of the conductor PROBLEMS 195 54 The charge 10 4e 3 C is removed from a sphere through a wire Find the current in the wire at t 0 and t 25 s 55 a Let V x2y2z in a region e 2eo defined by 1 x y z 1 Find the charge density pv in the region b If the charge travels at fyay ms determine the current crossing surface 0 x z 05 y 1 56 If the ends of a cylindrical bar of carbon a 3 X 104 of radius 5 mm and length 8 cm are maintained at a potential difference of 9 V find a the resistance of the bar b the current through the bar c the power dissipated in the bar 57 The resistance of round long wire of diameter 3 mm is 404 fikm If a current of 40 A flows through the wire find a The conductivity of the wire and identify the material of the wire b The electric current density in the wire 58 A coil is made of 150 turns of copper wire wound on a cylindrical core If the mean radius of the turns is 65 mm and the diameter of the wire is 04 mm calculate the resistance of the coil 59 A composite conductor 10 m long consists of an inner core of steel of radius 15 cm and an outer sheath of copper whose thickness is 05 cm a Determine the resistance of the conductor b If the total current in the conductor is 60 A what current flows in each metal c Find the resistance of a solid copper conductor of the same length and crosssectional areas as the sheath Take the resistivities of copper and steel as 177 X 108 and 118 X 108 0 m respectively 510 A hollow cylinder of length 2 m has its cross section as shown in Figure 518 If the cylin der is made of carbon a 105 mhosm determine the resistance between the ends of the cylinder Take a 3 cm b 5 cm 511 At a particular temperature and pressure a helium gas contains 5 X 1025 atomsm3 If a 10kVm field applied to the gas causes an average electron cloud shift of 10 8 m find the dielectric constant of helium Figure 518 For Problems 510 and 515 196 Electric Fields in Material Space 512 A dielectric material contains 2 X 1019 polar moleculesm3 each of dipole moment 18 X 1027Cm Assuming that all the dipoles are aligned in the direction of the electric field E 105 ax Vm find P and sr 513 In a slab of dielectric material for which e 248O and V 300z2 V find a D and pv bPandppv 514 For x 0 P 5 sin ay ax where a is a constant Find pps and ppv 515 Consider Figure 518 as a spherical dielectric shell so that 8 eoer for a r b and e eo for 0 r a If a charge Q is placed at the center of the shell find a P for a r b b ppvfora rb c pps at r a and r b 516 Two point charges when located in free space exert a force of 45 uN on each other When the space between them is filled with a dielectric material the force changes to 2 xN Find the dielectric constant of the material and identify the material 517 A conducting sphere of radius 10 cm is centered at the origin and embedded in a dielectric material with e 25eo If the sphere carries a surface charge of 4 nCm2 find E at 3 cm 4 cm 12 cm 518 At the center of a hollow dielectric sphere e eoer is placed a point charge Q If the sphere has inner radius a and outer radius b calculate D E and P 519 A sphere of radius a and dielectric constant er has a uniform charge density of po a At the center of the sphere show that b Find the potential at the surface of the sphere 520 For static timeindependent fields which of the following current densities are possible a J 2x3yax 4x2z 6x2yzaz b J xyax yz lay c J ap z cos 4 az 521 For an anisotropic medium Obtain D for a E 10a 10a Vm b E 10a 203 30az Vm V Dy Dz so 4 1 1 1 4 1 1 1 4 Ex Ey Ez PROBLEMS 8 197 100 2 522 If J Y ap Am find a the rate of increase in the volume charge density b the P t o t a l c u r r e n t p a s s i n g t h r o u g h s u r f a c e d e f i n e d b y p 2 0 z 0 f 2 i r 5 eio 4 523 Given that J ar Am2 at t 01 ms find a the amount of current passing r surface r 2 m b the charge density pv on that surface 524 Determine the relaxation time for each of the following medium a Hard rubber a 1015 Sm e 31eo b Mica ff 1015 Sm e 6eo c Distilled water a 104 Sm e 80eo 525 The excess charge in a certain medium decreases to onethird of its initial value in 20 xs a If the conductivity of the medium is 10 4 Sm what is the dielectric constant of the medium b What is the relaxation time c After 30 is what fraction of the charge will remain 526 Lightning strikes a dielectric sphere of radius 20 mm for which er 25 a 5 X 106 mhosm and deposits uniformly a charge of 10 JLC Determine the initial charge density and the charge density 2 ps later 527 Region 1 z 0 contains a dielectric for which er 25 while region 2 z 0 is char acterized by er 4 Let E 30a 50a 70az Vm and find a D2 b P2 c the angle between Ei and the normal to the surface 528 Given that E 10a 6a 12a Vm in Figure 519 find a Pu b E2 and the angle E2 makes with the yaxis c the energy density in each region 529 Two homogeneous dielectric regions 1 p 4 cm and 2 p 4 cm have dielectric constants 35 and 15 respectively If D2 12ap 6a0 9az nCm2 calculate a Ei and D b P2 and ppv2 c the energy density for each region 530 A conducting sphere of radius a is halfembedded in a liquid dielectric medium of per mittivity s as in Figure 520 The region above the liquid is a gas of permittivity e2 If the total free charge on the sphere is Q determine the electric field intensity everywhere 531 Two parallel sheets of glass er 85 mounted vertically are separated by a uniform air gap between their inner surface The sheets properly sealed are immersed in oil er 30 as shown in Figure 521 A uniform electric field of strength 2000 Vm in the horizontal direction exists in the oil Calculate the magnitude and direction of the electric field in the glass and in the enclosed air gap when a the field is normal to the glass sur faces and b the field in the oil makes an angle of 75 with a normal to the glass surfaces Ignore edge effects 532 a Given that E 15a 8az Vm at a point on a conductor surface what is the surface charge density at that point Assume e e0 b Region y 2 is occupied by a conductor If the surface charge on the conductor is 20 nCm2 find D just outside the conductor 198 11 Electric Fields in Material Space e 3E0 Ei 45B Figure 519 For Problem 528 Figure 520 For Problem 530 glass Figure 521 For Problem 531 oil oil 533 A silvercoated sphere of radius 5 cm carries a total charge of 12 nC uniformly distributed on its surface in free space Calculate a D on the surface of the sphere b D external to the sphere and c the total energy stored in the field Chapter 6 ELECTROSTATIC BOUNDARY VALUE PROBLEMS Our schools had better get on with what is their overwhelmingly most important task teaching their charges to express themselves clearly and with precision in both speech and writing in other words leading them toward mastery of their own language Failing that all their instruction in mathematics and science is a waste of time JOSEPH WEIZENBAUM MIT b1 INTRODUCTION The procedure for determining the electric field E in the preceding chapters has generally been using either Coulombs law or Gausss law when the charge distribution is known or using E W when the potential V is known throughout the region In most practical situations however neither the charge distribution nor the potential distribution is known In this chapter we shall consider practical electrostatic problems where only electro static conditions charge and potential at some boundaries are known and it is desired to find E and V throughout the region Such problems are usually tackled using Poissons1 or Laplaces2 equation or the method of images and they are usually referred to as boundary value problems The concepts of resistance and capacitance will be covered We shall use Laplaces equation in deriving the resistance of an object and the capacitance of a capaci tor Example 65 should be given special attention because we will refer to it often in the remaining part of the text 2 POISSONS AND LAPLACES EQUATIONS Poissons and Laplaces equations are easily derived from Gausss law for a linear mater ial medium V D V eE pv 61 After Simeon Denis Poisson 17811840 a French mathematical physicist 2After Pierre Simon de Laplace 17491829 a French astronomer and mathematician 199 200 Electrostatic BoundaryValue Problems and E VV Substituting eq 62 into eq 61 gives VeVV pv for an inhomogeneous medium For a homogeneous medium eq 63 becomes V2y 62 63 64 This is known as Poissons equation A special case of this equation occurs when pv 0 ie for a chargefree region Equation 64 then becomes V2V 0 65 which is known as Laplaces equation Note that in taking s out of the lefthand side of eq 63 to obtain eq 64 we have assumed that e is constant throughout the region in which V is defined for an inhomogeneous region e is not constant and eq 64 does not follow eq 63 Equation 63 is Poissons equation for an inhomogeneous medium it becomes Laplaces equation for an inhomogeneous medium when pv 0 Recall that the Laplacian operator V2 was derived in Section 38 Thus Laplaces equa tion in Cartesian cylindrical or spherical coordinates respectively is given by 66 67 68 depending on whether the potential is Vx y z Vp 4 z or Vr 6 4 Poissons equation in those coordinate systems may be obtained by simply replacing zero on the righthand side of eqs 66 67 and 68 with pve Laplaces equation is of primary importance in solving electrostatic problems involv ing a set of conductors maintained at different potentials Examples of such problems include capacitors and vacuum tube diodes Laplaces and Poissons equations are not only useful in solving electrostatic field problem they are used in various other field problems 1 r2 d f 2 1 P 3rJ 3 da 1 1 3 V dp 1 r2sin A 32V 8y2 1 P a 36 d2V d2V 230 2 ay sin u 36 0 a2v az2 i r2 0 1 sin2 6 32V 3j 2 0 63 UNIQUENESS THEOREM 201 For example V would be interpreted as magnetic potential in magnetostatics as tempera ture in heat conduction as stress function in fluid flow and as pressure head in seepage 63 UNIQUENESS THEOREM Since there are several methods analytical graphical numerical experimental etc of solving a given problem we may wonder whether solving Laplaces equation in different ways gives different solutions Therefore before we begin to solve Laplaces equation we should answer this question If a solution of Laplaces equation satisfies a given set of boundary conditions is this the only possible solution The answer is yes there is only one solution We say that the solution is unique Thus any solution of Laplaces equation which satisfies the same boundary conditions must be the only solution regardless of the method used This is known as the uniqueness theorem The theorem applies to any solution of Poissons or Laplaces equation in a given region or closed surface The theorem is proved by contradiction We assume that there are two solutions V and V2 of Laplaces equation both of which satisfy the prescribed boundary conditions Thus V 2 0 v v7 V2V2 0 on the boundary We consider their difference vd v2 v which obeys v 2yrf v 2y2 v 2y o Vd 0 on the boundary according to eq 69 From the divergence theorem V A dv I A dS s We let A Vd VVd and use a vector identity V A V VdWd VdV2Vd Wd VVd But V2Vd 0 according to eq 611 so V A VVd VVd Substituting eq 613 into eq 612 gives VVd VVddv j VdWddS 69a 69b 610 611a 611b 612 613 614 From eqs 69 and 611 it is evident that the righthand side of eq 614 vanishes 202 Electrostatic BoundaryValue Problems Hence VVJ2dv 0 Since the integration is always positive or Wrf 0 Vd V2 V constant everywhere in v 615a 615b But eq 615 must be consistent with eq 69b Hence Vd 0 or V V2 everywhere showing that Vx and V2 cannot be different solutions of the same problem This is the uniqueness theorem If a solution lo Laplaces equation can be found liiiit salisties the boundary conditions ihcn the solution is unique Similar steps can be taken to show that the theorem applies to Poissons equation and to prove the theorem for the case where the electric field potential gradient is specified on the boundary Before we begin to solve boundaryvalue problems we should bear in mind the three things that uniquely describe a problem 1 The appropriate differential equation Laplaces or Poissons equation in this chapter 2 The solution region 3 The prescribed boundary conditions A problem does not have a unique solution and cannot be solved completely if any of the three items is missing 64 GENERAL PROCEDURE FOR SOLVING POISSONS OR LAPLACES EQUATION The following general procedure may be taken in solving a given boundaryvalue problem involving Poissons or Laplaces equation 1 Solve Laplaces if pv 0 or Poissons if pv 0 equation using either a direct integration when V is a function of one variable or b separation of variables if V is a function of more than one variable The solution at this point is not unique but expressed in terms of unknown integration constants to be determined 2 Apply the boundary conditions to determine a unique solution for V Imposing the given boundary conditions makes the solution unique 3 Having obtained V find E using E VV and D from D eE 64 GENERAL PROCEDURE FOR SOLVING POISSONS OR LAPLACES EQUATION 203 4 If desired find the charge Q induced on a conductor using Q J ps dS where ps Dn and Dn is the component of D normal to the conductor If necessary the capacitance between two conductors can be found using C QV Solving Laplaces or Poissons equation as in step 1 is not always as complicated as it may seem In some cases the solution may be obtained by mere inspection of the problem Also a solution may be checked by going backward and finding out if it satisfies both Laplaces or Poissons equation and the prescribed boundary conditions EXAMPLE 61 Currentcarrying components in highvoltage power equipment must be cooled to carry away the heat caused by ohmic losses A means of pumping is based on the force transmit ted to the cooling fluid by charges in an electric field The electrohydrodynamic EHD pumping is modeled in Figure 61 The region between the electrodes contains a uniform charge p0 which is generated at the left electrode and collected at the right electrode Cal culate the pressure of the pump if po 25 mCm3 and Vo 22 kV Solution Since p 0 we apply Poissons equation V2V 8 The boundary conditions Vz 0 Vo and Vz d 0 show that V depends only on z there is no p or j dependence Hence d2v dz2 Integrating once gives Integrating again yields dV dz V Az B 2e AreaS Figure 61 An electrohydrodynamic pump for Example 61 204 Electrostatic BoundaryValue Problems where A and B are integration constants to be determined by applying the boundary condi tions When z 0 V Vo Vo 0 0 B B Vo When z d V 0 2e or A 2e d The electric field is given by The net force is F pvE dv p0 dS Edz F PoSVoaz The force per unit area or pressure is p poVo 25 X 1T3 X 22 X 103 550Nm2 PRACTICE EXERCISE 61 In a onedimensional device the charge density is given by pv x 0 and V 0 at x a find V and E If E 0 at Answer a3 A tea 2ae EXAMPLE 62 The xerographic copying machine is an important application of electrostatics The surface of the photoconductor is initially charged uniformly as in Figure 62a When light from the document to be copied is focused on the photoconductor the charges on the lower 64 GENERAL PROCEDURE FOR SOLVING POISSONS OR LAPLACES EQUATION 205 photoconductor light T recombination a I T i t r 1 J b Figure 62 For Example 62 surface combine with those on the upper surface to neutralize each other The image is de veloped by pouring a charged black powder over the surface of the photoconductor The electric field attracts the charged powder which is later transferred to paper and melted to form a permanent image We want to determine the electric field below and above the surface of the photoconductor Solution Consider the modeled version of Figure 62a as in Figure 62b Since pv 0 in this case we apply Laplaces equation Also the potential depends only on x Thus 0 dx2 Integrating twice gives V Ax B Let the potentials above and below be Vx and V2 respectively V1 Axx Bu x a V2 A2x B2 xa 621a 621b 206 Electrostatic BoundaryValue Problems The boundary conditions at the grounded electrodes are V d 0 V2x 0 0 At the surface of the photoconductor Vxx a V2x a Dln D2n ps 622a 622b 623a 623b We use the four conditions in eqs 622 and 623 to determine the four unknown con stants AiA2 B1 andB2 From eqs 621 and 622 0 Ad B B Axd 0 0 B2B2 0 From eqs 621 and 623a Aa B A2a To apply eq 623b recall that D eE e W so that Ps Din D2n e2E2n e e2 ax ax or Ps eiAi e2A2 Solving for Aj and A2 in eqs 624 to 626 we obtain E Aax S s7 d B7 e a s 7 A7ar I s2 d s2 624a 624b 625 626 PRACTICE EXERCISE 62 For the model of Figure 62b if ps 0 and the upper electrode is maintained at Vo while the lower electrode is grounded show that d a a E Voax 2 A 2 a a EXAMPLE 63 64 GENERAL PROCEDURE FOR SOLVING POISSONS OR LAPLACES EQUATION 207 Semiinfinite conducting planes j 0 and f TT6 are separated by an infinitesimal insu lating gap as in Figure 63 If V 0 0 and Vt TT6 100 V calculate V and E in the region between the planes Solution As V depends only on Laplaces equation in cylindrical coordinates becomes Since p 0 is excluded due to the insulating gap we can multiply by p2 to obtain d2V dp 2 0 which is integrated twice to give V Acf B We apply the boundary conditions to determine constants A and B When 4 0 V 0 0 0 BB 0 W h e n 4 fo V Vo Hence gap Figure 63 Potential Vj due to semi infinite conducting planes y 208 B Electrostatic BoundaryValue Problems and Substituting Vo 100 and j0 TT6 gives 600 V and Check 0 V 0 0 V TT6 100 PRACTICE EXERCISE 63 Two conducting plates of size 1 X 5 m are inclined at 45 to each other with a gap of width 4 mm separating them as shown in Figure 64 Determine an approximate value of the charge per plate if the plates are maintained at a potential difference of 50 V Assume that the medium between them has er 15 Answer 222 nC EXAMPLE 64 Two conducting cones 6 TT10 and 6 x6 of infinite extent are separated by an infin itesimal gap at r 0 If V6 TT10 0 and V6 TT6 50 V find V and E between the cones Solution Consider the coaxial cone of Figure 65 where the gap serves as an insulator between the two conducting cones V depends only on 6 so Laplaces equation in spherical coordinates becomes r2sin 6 gap of width 4 mm Figure 64 For Practice Exercise 63 1 m 64 GENERAL PROCEDURE FOR SOLVING POISSONS OR LAPLACES EQUATION 209 Figure 65 Potential V4 due to conducting cones Since r 0 and 0 0 it are excluded we can multiply by r2sin 0 to get Integrating once gives or dV A dV A dd sin 0 Integrating this results in d9 V A F T A A A dd sin 9 J 2 cos 02 sin 92 12 sec2 92 dd tan 02 Jtan 02 tan 92 A In tan 02 B We now apply the boundary conditions to determine the integration constants A and B V9 00 0 0 A In tan 02 B or B A In tan 02 210 M Electrostatic BoundaryValue Problems Hence Also or Thus V A In tan 02 tan 02 V9 62 Vo Vo A In tan 022 tan 02 A In tan 022 V tan 02 tan 02 In tan 022 tan 02 r sin 0 r sin 0 In Taking 0 TT10 02 ir6 and Vo 50 gives tan 022 tan 02 50 In V tan 02 j Ltan7r2oJ In tan TT12 tan TT20 tan 02 and E r sin 0 Check V2V 0 V9 TT10 0 V0 TT6 Vo 64 GENERAL PROCEDURE FOR SOLVING POISSONS OR LAPLACES EQUATION 211 50 V gap For Practice Exercise 64 1 PRACTICE EXERCISE 64 A large conducting cone d 45 is placed on a conducting plane with a tiny gap separating it from the plane as shown in Figure 66 If the cone is connected to a 50V source find V and E at 3 4 2 Answer 2213 V 1136 a Vm a Determine the potential function for the region inside the rectangular trough of infinite length whose cross section is shown in Figure 67 b For Vo 100 V and b 2a find the potential at x a2 y 3a4 Solution a The potential V in this case depends on x and y Laplaces equation becomes v2y dx dr T 0 651 Potential Vx y due to a con ducting rectangular trough Electrostatic BoundaryValue Problems We have to solve this equation subject to the following boundary conditions Vx 0 0 y a 0 652a Vx b 0 y a 0 652b V0 A b y 0 0 652c V0xby a Vo 652d We solve eq 651 by the method of separation of variables that is we seek a product solution of V Let Vx y Xx Yy 653 when X is a function of x only and y is a function of only Substituting eq 653 into eq 651 yields XY YX 0 Dividing through by XY and separating X from Y gives X Y J y 654a Since the lefthand side of this equation is a function of x only and the righthand side is a function of y only for the equality to hold both sides must be equal to a constant X that is r Y 654b The constant X is known as the separation constant From eq 654b we obtain X XX 0 655a and Y Y 0 655b Thus the variables have been separated at this point and we refer to eq 655 as separated equations We can solve for Xx and Yy separately and then substitute our solutions into eq 653 To do this requires that the boundary conditions in eq 652 be separated if possible We separate them as follows V0 y X0Yy 0 X0 0 Vb y XbYy 0 Xb 0 Vx 0 XxY0 0 Y0 0 Vx a X0Ya Vo inseparable 656a 656b 656c 656d To solve for Xc and Yy in eq 655 we impose the boundary conditions in eq 656 We consider possible values of X that will satisfy both the separated equations in eq 655 and the conditions in eq 656 64 GENERAL PROCEDURE FOR SOLVING POISSONS OR LAPLACES EQUATION 213 CASE A If X 0 then eq 655a becomes X 0 or dx2 657 which upon integrating twice yields X Ax B The boundary conditions in eqs 656a and 656b imply that XQt 0 0 0 0 fi or 5 0 and Xx b 00 A b 0 or A 0 because b 0 Hence our solution for X in eq 657 becomes Xx 0 which makes V 0 in eq 653 Thus we regard Xx 0 as a trivial solution and we conclude that A 0 CASE B If X 0 say X or then eq 655a becomes X aX 0 or D2 a2X 0 where D dx that is DX aX showing that we have two possible solutions corresponding to the plus and minus signs For the plus sign eq 658 becomes dX dX aX or a dx dx X 65S a dx or In X ax In A where In i is a constant of integration Thus X Axeax 659a 14 ectrostatic BoundaryValue Problems Similarly for the minus sign solving eq 658 gives X A2eax 659b The total solution consists of what we have in eqs 659a and 659b that is Xx Aeax A2eax 6510 Since cosh ax eax Tajr2 and sinh ax eax eaxl2 or eax cosh ax sinh ax and e ax cosh ax sinh ax eq 6510 can be written as Xx B cosh ax B2 sinh ax 6511 where Bx A A2 and B2 A A2 In view of the given boundary conditions we prefer eq 6511 to eq 6510 as the solution Again eqs 656a and 656b require that Xx 0 0 0 S 1 B2 0 or 5 0 and Xx 6 sinh ab Since a 0 and 0 sinh a cannot be zero This is due to the fact that sinh x 0 if and only if x 0 as shown in Figure 68 Hence B2 0 and Xx 0 This is also a trivial solution and we conclude that X cannot be less than zero CASE C If X 0 say X 32 then eq 655a becomes X 32X 0 cosh x 2 1 sinh x Sketch of cosh x and sinh x showing that sinh x 0 if and only if 0 64 GENERAL PROCEDURE FOR SOLVING POISSONS OR LAPLACES EQUATION 215 that is D1 32X 0 or DX jSX 6512 where V 1 From eqs 658 and 6512 we notice that the difference between Cases 2 and 3 is replacing a byj3 By taking the same procedure as in Case 2 we obtain the solution as Xx t f a eV i l 6513a Since eliix cos 3x j sin fix and eitix cos 3x j sin 3v eq 6513a can be written Xx ga cos 3v i sin fix where g Co C and Co C In view of the given boundary conditions we prefer to use eq 6513b Imposing the conditions in eqs 656a and 656b yields Xx 0 0 0 o 1 0 and Xx b 0 0 0 sin3 Suppose 0 otherwise we get a trivial solution then sin 3b 0 sin nir H 12 34 6513b J 0 6514 Note that unlike sinh v which is zero only when v 0 sin v is zero at an infinite number of points as shown in Figure 69 It should also be noted that n 0 because 3 0 we have already considered the possibility 3 0 in Case 1 where we ended up with a trivial solution Also we do not need to consider n 1 2 3 4 because X j32 1 2 ix 3 jr i isf cv Sketch of sin x showing that sin x 0 at infinite number of points 216 Electrostatic BoundaryValue Problems would remain the same for positive and negative values of n Thus for a given n eq 6513b becomes Xnx gn sin Having found Xx and 6515 6516 we solve eq 655b which is now Y 32Y 0 The solution to this is similar to eq 6511 obtained in Case 2 that is Yy h0 cosh 3y hx sinh j3y The boundary condition in eq 656c implies that Yy 0 00 V l 0 or ho 0 Hence our solution for Yy becomes Yny K sinh 6517 Substituting eqs 6515 and 6517 which are the solutions to the separated equations in eq 655 into the product solution in eq 653 gives Vnx y gnhn sin sinh b b This shows that there are many possible solutions Vb V2 V3 V4 and so on for n 1 2 3 4 and so on By the superposition theorem if V V2 V3 Vn are solutions of Laplaces equa tion the linear combination v c2v2 c3v3 cnvn where cu c2 c 3 cn are constants is also a solution of Laplaces equation Thus the solution to eq 651 is Vx y 2J cn sin sinh i b b 6518 where cn gnhn are the coefficients to be determined from the boundary condition in eq 656d Imposing this condition gives Vx y a Vo 2J cn sin smh i b b 6519 64 GENERAL PROCEDURE FOR SOLVING POISSONS OR LAPLACES EQUATION 217 which is a Fourier series expansion of Vo Multiplying both sides of eq 6519 by sin mKxIb and integrating over 0 x b gives mirx mra mirx mrx VnSin dx cn S l nh sin sin dx Jo b ne b 0 b b By the orthogonality property of the sine or cosine function see Appendix A9 0 m n 6520 sin mx sin nx dx TT2 m n Incorporating this property in eq 6520 means that all terms on the righthand side of eq 6520 will vanish except one term in which m n Thus eq 6520 reduces to b rb mrx mra mrx Vosin dx cn sinh sin r dx or that is o o nwx cos nK b cn sinh mra 1 cos Vob mra b 1 cos mr cn sinh nK b 2 mra 2VO cn smh I cos rnr b nic b J dx n 135 0 n 2 4 6 cn mr sinh 0 mra n odd n even Substituting this into eq 6518 gives the complete solution as Vxy mrx niry sin b sinh b n sinh nK a 6521 6522 Check V2V 0 Vx 0 y 0 Vx b y Vx y 0 Vx y a Vo The solution in eq 6522 should not be a surprise it can be guessed by mere observation of the potential system in Figure 67 From this figure we notice that along x V varies from 218 Electrostatic BoundaryValue Problems 0 at x 0 to 0 at x b and only a sine function can satisfy this requirement Similarly along y V varies from 0 at y 0 to Vo at y a and only a hyperbolic sine function can satisfy this Thus we should expect the solution as in eq 6522 To determine the potential for each point x y in the trough we take the first few terms of the convergent infinite series in eq 6522 Taking four or five terms may be suf ficient b For x a2 and y 3a4 where b 2a we have 2 4 4V 2 n 135 sin 7r4 sinh 3TT8 n sinh rnr2 sin ir4 sinh 3TT8 sin 3TT4 sinh 9TT8 3 sinh 3TT2 T sinh x2 sin 5x4 sinh 15ir4 5 sinh 5TT4 4V 04517 00725 001985 000645 000229 IT 06374Vo It is instructive to consider a special case when A b Ira and Vo 100 V The poten tials at some specific points are calculated using eq 6522 and the result is displayed in Figure 610a The corresponding flux lines and equipotential lines are shown in Figure 610b A simple Matlab program based on eq 6522 is displayed in Figure 611 This selfexplanatory program can be used to calculate Vx y at any point within the trough In Figure 611 Vx bA y 3a4 is typically calculated and found to be 432 volts 10 100 V Equipotential line Flux line 432 540 432 182 250 182 680 954 680 a 10 0 Figure 610 For Example 65 a Vx y calculated at some points b sketch of flux lines and equipotential lines 64 GENERAL PROCEDURE FOR SOLVING POISSONS OR LAPLACES EQUATION 219 SOLUTION OF LAPLACES EQUATION THIS PROGRAM SOLVES THE TWODIMENSIONAL BOUNDARYVALUE PROBLEM DESCRIBED IN FIG 67 a AND b ARE THE DIMENSIONS OF THE TROUGH x AND y ARE THE COORDINATES OF THE POINT OF INTEREST P Vo 1000 a 10 b a x b4 y 3a4 c 4Vopi sum 00 for kl10 n 2k 1 al sinnpixb a2 sinhnpiyb a3 nsinhnpiab sum sum cala2a3 P n sum end diary testout P diary off Figure 611 Matlab program for Example 65 PRACTICE EXERCISE 65 For the problem in Example 65 take Vo 100 V b 2a 2 m find V and E at a xy aa2 b xy 3a2a4 Answer a 4451 V 9925 ay Vm b 165 V 206 ax 7034 ay Vm EXAMPLE 66 In the last example find the potential distribution if Vo is not constant but a Vo 10 sin 3irxb y a Qxb b VQ 2 sin y sin y y a0xb 220 M Electrostatic BoundaryValue Problems Solution a In the last example every step before eq 6519 remains the same that is the solu tion is of the form nirx niry Vx y 2J cn sin sinh tx b b as per eq 6518 But instead of eq 6519 we now have Vy a Vo 10 sin X cn sin sinh b ni b b By equating the coefficients of the sine terms on both sides we obtain For n 3 10 c3 sinh 3ira or 10 sinh 3ira Thus the solution in eq 661 becomes Vxy 10 sin 3TTX sinh sinh b Similarly instead of eq 6519 we have Vo Vy a or 5KX KX 1 2 sin 1 sinh b 10 b Equating the coefficient of the sine terms cn sinh 7TX sinh cn 0 n 15 661 64 GENERAL PROCEDURE FOR SOLVING POISSONS OR LAPLACES EQUATION Forn 1 2 cx sinh or b 221 For n 5 Hence sinhira 1 5ira c5sinh or c5 10 sinh 5ira Vxy irx Try 5irx 5iry 2 sm sinh sin sinh b b b b sinh b 10 sinh 5ira PRACTICE EXERCISE 66 In Example 65 suppose everything remains the same except that Vo is replaced by Vo sin 0 x b y a Find VJC y Answer Vn sin sinh sinh 7ra EXAMPLE 67 Obtain the separated differential equations for potential distribution Vp z in a charge free region Solution This example like Example 65 further illustrates the method of separation of variables Since the region is free of charge we need to solve Laplaces equation in cylindrical coor dinates that is a dv 1 P I d2V d2V We let P dp dp p2 dj Vp 4 z RP ZZ 671 672 222 H Electrostatic BoundaryValue Problems where R P and Z are respectively functions of p j and z Substituting eq 672 into eq 671 gives p dpdp We divide through by RPZ to obtain p 2 dt 2 T 0 pdR 1 d2t pR dp dp P 2P dt dz 1 d2Z Z dz2 673 674 The righthand side of this equation is solely a function of z whereas the lefthand side does not depend on z For the two sides to be equal they must be constant that is JdpdR Jd pR dpdp p 2p dct2 1 d2Z Z dz2 A 2 675 where X2 is a separation constant Equation 675 can be separated into two parts 1 d2Z Zdz 2 A or and Z X2Z 0 Rdp dp Equation 678 can be written as R pdR R dp2 R dp 2 1 d 24 where fx2 is another separation constant Equation 679 is separated as P fo o and p2R pR p2X2 VL2R 0 676 677 678 679 6710 6711 Equations 677 6710 and 6711 are the required separated differential equations Equation 677 has a solution similar to the solution obtained in Case 2 of Example 65 that is Zz cx cosh z c2 sinh Xz 6712 65 RESISTANCE AND CAPACITANCE 223 The solution to eq 6710 is similar to the solution obtained in Case 3 of Example 65 that is P4 c 3 cos fit c4 sin 6713 Equation 6711 is known as the Bessel differential equation and its solution is beyond the scope of this text PRACTICE EXERCISE 67 Repeat Example 67 for Vr 6 f Answer If Vr 0 t Rr F6 0 P 2P 0 R R R F cot 6 F ju2 X2 cosec2 0 F 0 0 65 RESISTANCE AND CAPACITANCE In Section 54 the concept of resistance was covered and we derived eq 516 for finding the resistance of a conductor of uniform cross section If the cross section of the conductor is not uniform eq 516 becomes invalid and the resistance is obtained from eq 517 V jEdl I aEdS 616 The problem of finding the resistance of a conductor of nonuniform cross section can be treated as a boundaryvalue problem Using eq 616 the resistance R or conductance G lR of a given conducting material can be found by following these steps 1 Choose a suitable coordinate system 2 Assume Vo as the potential difference between conductor terminals 3 Solve Laplaces equation V2V to obtain V Then determine E from E from CTE dS 4 Finally obtain R as VJI VV and In essence we assume Vo find and determine R VJI Alternatively it is possible to assume current o find the corresponding potential difference V and determine R from R VIo As will be discussed shortly the capacitance of a capacitor is obtained using a similar technique For a complete solution of Laplaces equation in cylindrical or spherical coordinates see for example D T Paris and F K Hurd Basic Electromagnetic Theory New York McGrawHill 1969 pp 150159 224 U Electrostatic BoundaryValue Problems Generally speaking to have a capacitor we must have two or more conductors car rying equal but opposite charges This implies that all the flux lines leaving one conductor must necessarily terminate at the surface of the other conductor The conductors are some times referred to as the plates of the capacitor The plates may be separated by free space or a dielectric Consider the twoconductor capacitor of Figure 612 The conductors are maintained at a potential difference V given by V V V d 617 where E is the electric field existing between the conductors and conductor 1 is assumed to carry a positive charge Note that the E field is always normal to the conducting surfaces We define the capacitance C of the capacitor as the ratio of the magnitude of the charge on one of the plates to the potential difference between them that is 618 The negative sign before V E d has been dropped because we are interested in the absolute value of V The capacitance C is a physical property of the capacitor and in mea sured in farads F Using eq 618 C can be obtained for any given twoconductor ca pacitance by following either of these methods 1 Assuming Q and determining V in terms of Q involving Gausss law 2 Assuming Vand determining Q in terms of Vinvolving solving Laplaces equation We shall use the former method here and the latter method will be illustrated in Examples 610 and 611 The former method involves taking the following steps 1 Choose a suitable coordinate system 2 Let the two conducting plates carry charges Q and Q Figure 612 A twoconductor ca pacitor 65 RESISTANCE AND CAPACITANCE 225 3 Determine E using Coulombs or Gausss law and find Vfrom V J E d The negative sign may be ignored in this case because we are interested in the absolute value of V 4 Finally obtain C from C QV We will now apply this mathematically attractive procedure to determine the capaci tance of some important twoconductor configurations A ParallelPlate Capacitor Consider the parallelplate capacitor of Figure 613a Suppose that each of the plates has an area S and they are separated by a distance d We assume that plates 1 and 2 respec tively carry charges Q and Q uniformly distributed on them so that Ps Q 619 dielectric e plate area S 1 Figure 613 a Parallelplate capacitor b fringing effect due to a parallelplate capacitor a b 226 Electrostatic BoundaryValue Problems An ideal parallelplate capacitor is one in which the plate separation d is very small com pared with the dimensions of the plate Assuming such an ideal case the fringing field at the edge of the plates as illustrated in Figure 613b can be ignored so that the field between them is considered uniform If the space between the plates is filled with a homo geneous dielectric with permittivity e and we ignore flux fringing at the edges of the plates from eq 427 D psax or 620 ES Hence 621 and thus for a parallelplate capacitor 622 This formula offers a means of measuring the dielectric constant er of a given dielectric By measuring the capacitance C of a parallelplate capacitor with the space between the plates filled with the dielectric and the capacitance Co with air between the plates we find er from c Er co Using eq 496 it can be shown that the energy stored in a capacitor is given by 623 624 To verify this for a parallelplate capacitor we substitute eq 620 into eq 496 and obtain 1 rE i E r 2 J e 2S dv 2E2S2 Q2 d Q2 1 QV 2 eSj 2C 2 as expected 65 RESISTANCE AND CAPACITANCE 227 B Coaxial Capacitor This is essentially a coaxial cable or coaxial cylindrical capacitor Consider length L of two coaxial conductors of inner radius a and outer radius b b a as shown in Figure 614 Let the space between the conductors be filled with a homogeneous dielectric with permit tivity s We assume that conductors 1 and 2 respectively carry Q and Q uniformly dis tributed on them By applying Gausss law to an arbitrary Gaussian cylindrical surface of radius p a p b we obtain Q s j E dS eEp2irpL Hence Neglecting flux fringing at the cylinder ends L 2irspL apdp ap Q b In 2KEL a Thus the capacitance of a coaxial cylinder is given by 625 626 627a 627b 628 C Spherical Capacitor This is the case of two concentric spherical conductors Consider the inner sphere of radius a and outer sphere of radius bb a separated by a dielectric medium with permittivity e as shown in Figure 615 We assume charges Q and Q on the inner and outer spheres dielectric Figure 614 Coaxial capacitor 228 Electrostatic BoundaryValue Problems Figure 615 Spherical capacitor dielectric e respectively By applying Gausss law to an arbitrary Gaussian spherical surface of radius rarb that is Q e E dS sEr4irrz E 4irer2 629 630 The potential difference between the conductors is V E h Q drar 4ire a b Thus the capacitance of the spherical capacitor is 631 632 By letting b t C 47rsa which is the capacitance of a spherical capacitor whose outer plate is infinitely large Such is the case of a spherical conductor at a large distance from other conducting bodiesthe isolated sphere Even an irregularly shaped object of about the same size as the sphere will have nearly the same capacitance This fact is useful in estimating the stray capacitance of an isolated body or piece of equipment Recall from network theory that if two capacitors with capacitance C and C2 are in series ie they have the same charge on them as shown in Figure 616a the total capacitance is C2 or C 633 65 RESISTANCE AND CAPACITANCE 229 Figure 616 Capacitors in a series and b parallel a b If the capacitors arc in parallel ie they have the same voltage across their plates as shown in Figure 616b the total capacitance is C C2 634 Let us reconsider the expressions for finding the resistance R and the capacitance C of an electrical system The expressions were given in eqs 616 and 618 V Q VV fEdl 616 618 The product of these expressions yields 635 which is the relaxation time Tr of the medium separating the conductors It should be re marked that eq 635 is valid only when the medium is homogeneous this is easily in ferred from eqs 616 and 618 Assuming homogeneous media the resistance of various capacitors mentioned earlier can be readily obtained using eq 635 The follow ing examples are provided to illustrate this idea For a parallelplate capacitor Q sS R oS 636 For a cylindrical capacitor c k R b 2KOL In 637 230 Hi Electrostatic BoundaryValue Problems For a spherical capacitor Q 4rre 1 b R 4ira And finally for an isolated spherical conductor C Airsa R 4iroa 638 639 It should be noted that the resistance R in each of eqs 635 to 639 is not the resistance of the capacitor plate but the leakage resistance between the plates therefore a in those equations is the conductivity of the dielectric medium separating the plates A metal bar of conductivity a is bent to form a flat 90 sector of inner radius a outer radius b and thickness t as shown in Figure 617 Show that a the resistance of the bar between the vertical curved surfaces at p a and p b is R oitt and b the resistance between the two horizontal surfaces at z 0 and z t is At R oirb2 a2 Solution a Between the vertical curved ends located at p a and p b the bar has a nonuni form cross section and hence eq 516 does not apply We have to use eq 616 Let a po tential difference Vo be maintained between the curved surfaces at p a and p b so that Figure 617 Metal bar of Exam ple 68 r 65 RESISTANCE AND CAPACITANCE 231 Vp a 0 and Vp b Vo We solve for V in Laplaces equation 2V 0 in cylin drical coordinates Since V Vp 2 d V P dp dp As p 0 is excluded upon multiplying by p and integrating once this becomes oA or dV A dp P Integrating once again yields V Alnp S where A and 5 are constants of integration to be determined from the boundary conditions Vp a 0 0 A In a B or 5 A In a Vp b Vo Vo A In b B A In b A In a A In or A a b l n Hence Thus A In p A In a A I n l n a b a l n a dp J aE dS p fTT2 J dS J p In a dzpdj In In a a oirt as required 232 Electrostatic BoundaryValue Problems b Let Vo be the potential difference between the two horizontal surfaces so that Vz 0 0 and Vz i Vo V Vz so Laplaces equation V2V 0 becomes dz2 o Integrating twice gives V Az B We apply the boundary conditions to determine A and B Vz 0 0 0 0 5 or B 0 Hence Vz t VoVoAt or A V J dS Voa TT p t 2 2 Thus J aE az dS p dj dp a p dcp dp VOGIT b2 a2 At At i I oKb2 a2 Alternatively for this case the cross section of the bar is uniform between the hori zontal surfaces at z 0 and z t and eq 516 holds Hence ab a1 At a2 as required 65 RESISTANCE AND CAPACITANCE 233 n PRACTICE EXERCISE 68 SliIS8 A disc of thickness t has radius b and a central hole of radius a Taking the conduc tivity of the disc as a find the resistance between is I a The hole and the rim of the disc ft V b The two flat sides of the disc A coaxial cable contains an insulating material of conductivity a If the radius of the central wire is a and that of the sheath is b show that the conductance of the cable per unit length is see eq 637 In ba Answer a b 2Kta oirb a 1 J dS 2LoVo In ba The resistance per unit length is and the conductance per unit length is Consider length L of the coaxial cable as shown in Figure 614 Let Vo be the potential dif ference between the inner and outer conductors so that Vp a 0 and Vp b Vo V and E can be found just as in part a of the last example Hence J aE aV a p dS pdf dz a p pmba p dz dj 234 U Electrostatic BoundaryValue Problems PRACTICE EXERCISE 69 A coaxial cable contains an insulating material of conductivity ax in its upper half and another material of conductivity a2 in its lower half similar to the situation in Figure 619b If the radius of the central wire is a and that of the sheath is b show that the leakage resistance of length of the cable is Answer Proof EXAMPLE 610 Conducting spherical shells with radii a 10 cm and b 30 cm are maintained at a po tential difference of 100 V such that Vr b 0 and Vr a 100 V Determine V and E in the region between the shells If sr 25 in the region determine the total charge induced on the shells and the capacitance of the capacitor Solution I Consider the spherical shells shown in Figure 618 V depends only on r and hence Laplaces equation becomes r2 dry dr Since r fc 0 in the region of interest we multiply through by r2 to obtain dr dr Integrating once gives dr Figure 618 Potential Vr due to conducting spherical shells 65 RESISTANCE AND CAPACITANCE 235 or Integrating again gives dV A dr r2 V B r As usual constants A and B are determined from the boundarv conditions When r b V 0 0 B or B b b Hence V A 1 1 b r Also when r a V Vo Vo A 1 1 b a or A 1 1 b a Thus vvn r b 1 1 a b 1 1 r ar eE dS 0 J0 47TEoSrVo L2r r2 sin 5 dd a b a b 236 B Electrostatic BoundaryValue Problems The capacitance is easily determined as Q Vo J a b which is the same as we obtained in eq 632 there in Section 65 we assumed Q and found the corresponding Vo but here we assumed Vo and found the corresponding Q to de termine C Substituting a 01 m b 03 m Vo 100 V yields V 100 10 103 1 5 r 3 Check V2V 0 Vr 03 m 0 Vr 01 m 100 E 100 r2 10 103 ar ar Vm Q 4TT 109 25 100 36r 10 103 4167 nC The positive charge is induced on the inner shell the negative charge is induced on the outer shell Also C Q 4167 X 10 100 4167 pF PRACTICE EXERCISE 610 If Figure 619 represents the cross sections of two spherical capacitors determine their capacitances Let a 1 mm b 3 mm c 2 mm srl 25 and er2 35 Answer a 053 pF b 05 pF Figure 619 For Practice Exer cises 69 610 and 612 a b 65 RESISTANCE AND CAPACITANCE 237 fcXAV In Section 65 it was mentioned that the capacitance C QV of a capacitor can be found by either assuming Q and finding V or by assuming V and finding Q The former approach was used in Section 65 while we have used the latter method in the last example Using the latter method derive eq 622 Solution Assume that the parallel plates in Figure 613 are maintained at a potential difference Vo so that Vx 0 and Vx d Vo This necessitates solving a onedimensional boundary value problem that is we solve Laplaces equation dx Integrating twice gives where A and B are integration constants to be determined from the boundary conditions At x 0 V 0 0 0 B or B 0 and at x d V Vo Vo Ad 0 or Hence Notice that this solution satisfies Laplaces equation and the boundary conditions We have assumed the potential difference between the plates to be Vo Our goal is to find the charge Q on either plate so that we can eventually find the capacitance C QVo The charge on either plate is Q As dS But ps D an eE an where E VV ax t dx On the lower plates an ax so Ps eVn On the upper plates an ax so d and Q d sVo sVoS Ps V and Q T 238 S Electrostatic BoundaryValue Problems As expected Q is equal but opposite on each plate Thus Vn d which is in agreement with eq 622 N 5 PRACTICE EXERCISE 611 H Derive the formula for the capacitance C QVo of a cylindrical capacitor in eq 628 by assuming Vo and finding EXAMPLE 12 Determine the capacitance of each of the capacitors in Figure 620 Take erl 4 er2 6 d 5 mm 51 30 cm2 Solution a Since D and E are normal to the dielectric interface the capacitor in Figure 620a can be treated as consisting of two capacitors Cx and C2 in series as in Figure 616a P p V Op p V Op p C d2 d 2 d The total capacitor C is given by C CXC2 2EoS erlr2 2 C2 d erl er2 1 0 9 3 0 X 1 0 4 4 X 6 36TT 5 X 103 C 2546 pF 10 w2 w2 b Figure 620 For Example 612 6121 65 RESISTANCE AND CAPACITANCE 239 b In this case D and E are parallel to the dielectric interface We may treat the capacitor as consisting of two capacitors Cx and C2 in parallel the same voltage across C and C2 as in Figure 616b eosrl52 eoerlS sosr2S Id Id The total capacitance is 109 30 X 10 36TT 2 5 X 103 C 2653 pF er2 10 6122 Notice that when srl er2 er eqs 6121 and 6122 agree with eq 622 as ex pected PRACTICE EXERCISE 612 Determine the capacitance of 10 m length of the cylindrical capacitors shown in Figure 619 Take a 1 mm b 3 mm c 2 mm erl 25 and er2 35 Answer a 141 nF b 152 nF A cylindrical capacitor has radii a 1 cm and b 25 cm If the space between the plates is filled with an inhomogeneous dielectric with sr 10 pp where p is in centimeters find the capacitance per meter of the capacitor Solution The procedure is the same as that taken in Section 65 except that eq 627a now becomes V Q 2ireQsrpL dp Q 2iTBoL dp 10 p Q r dp Q 2ire0L b 10 p 2irsoL Q 10 b m In 10 p 2irsnL 10 a 240 Electrostatic BoundaryValue Problems Thus the capacitance per meter is L 1 m Q 10 C 4346 pFm 2TT 10 9 36TT 125 A spherical capacitor with a 15 cm 6 4cm has an inhomogeneous dielectric jjjofe lOSor Calculate the capacitance of the capacitor K Answer 113 nF 66 METHOD OF IMAGES The method of images introduced by Lord Kelvin in 1848 is commonly used to determine V E D and ps due to charges in the presence of conductors By this method we avoid solving Poissons or Laplaces equation but rather utilize the fact that a conducting surface is an equipotential Although the method does not apply to all electrostatic problems it can reduce a formidable problem to a simple one luorx stales ilmi ti given charge ctinlijnirmion ahoe an inlinik ciini plane max be replaced bx the charge conliguialioti surface in place of 11 ie Typical examples of point line and volume charge configurations are portrayed in Figure 621a and their corresponding image configurations are in Figure 621b Equipotential surface V 0 Q a Figure 621 Image system a charge configurations above a perfectly conducting plane b image configuration with the conducting plane replaced by equipotential surface 66 METHOD OF IMAGES 241 In applying the image method two conditions must always be satisfied 1 The image charges must be located in the conducting region 2 The image charges must be located such that on the conducting surfaces the po tential is zero or constant The first condition is necessary to satisfy Poissons equation and the second condition ensures that the boundary conditions are satisfied Let us now apply the image theory to two specific problems A A Point Charge Above a Grounded Conducting Plane Consider a point charge Q placed at a distance h from a perfect conducting plane of infinite extent as in Figure 622a The image configuration is in Figure 622b The electric field at point Px y z is given by E E E The distance vectors rt and r2 are given by r x y z 0 0 h x j z h r2 x y z 0 0 h x j z h so eq 641 becomes J2 f xax yay z feaz xax yay z fea 2 y2 z h2f2 x2 y2 hff12 640 641 642 643 644 v o Pxyz a Figure 622 a Point charge and grounded conducting plane b image configuration and field lines Electrostatic BoundaryValue Problems It should be noted that when z 0 E has only the zcomponent confirming that E is normal to the conducting surface The potential at P is easily obtained from eq 641 or 644 using V jEdl Thus vv Q Q v 4ireor1 4irsor2 Q 645 o x2 y2 z hff2 x2 y2 z h2f for z a 0 and V 0 for z 0 Note that Vz 0 0 The surface charge density of the induced charge can also be obtained from eq 644 as ps Dn eJLn Qh 646 h2f2 The total induced charge on the conducting plane is Qi PsdS Qhdxdy By changing variables p2 x2 y2 dx dy p dp d4 Qi n32 647 648 Integrating over j gives 2ir and letting p dp d p2 we obtain 212 649 Q as expected because all flux lines terminating on the conductor would have terminated on the image charge if the conductor were absent B A Line Charge above a Grounded Conducting Plane Consider an infinite charge with density pL Cm located at a distance h from the grounded conducting plane z 0 The same image system of Figure 622b applies to the line charge except that Q is replaced by pL The infinite line charge pL may be assumed to be at 66 METHOD OF IMAGES t 243 x 0 z h and the image pL at x 0 z A so that the two are parallel to the yaxis The electric field at point P is given from eq 421 by 650 651 652 653 654 The distance vectors p t and E p2 are E PL 2irsop given E PL Z 1 27T0p2 by p2 x y z 0 y h x 0 z h so eq 651 becomes E pL xax z haz xax 27TO L X2 Z X2 z Again notice that when z 0 E has only the zcomponent confirming that E is normal to the conducting surface The potential at P is obtained from eq 651 or 654 using V jEdl Thus V V V PL 1 2iT0 PL Pi In ft lnp2 655 2 T T E O P 2 Substituting px pi and p2 p2 in eqs 652 and 653 into eq 655 gives 12 656 for z 0 and V 0 for z 0 Note that Vz 0 0 The surface charge induced on the conducting plane is given by Ps Dn soEz The induced charge per length on the conducting plane is pLh f dx Pi psdx x2 h2 657 658 By letting x h tan a eq 658 becomes Pi pLh r da 659 as expected 244 Electrostatic BoundaryValue Problems EXAMPLE 614 A point charge Q is located at point a 0 b between two semiinfinite conducting planes intersecting at right angles as in Figure 623 Determine the potential at point Px y z and the force on Q Solution The image configuration is shown in Figure 624 Three image charges are necessary to satisfy the conditions in Section 66 From Figure 624a the potential at point Px y z is the superposition of the potentials at P due to the four point charges that is V Q ri r2 4 where r4 xaf From Figure 624b the net force on Q F F F F3 n x a2 y2 z b212 r2 x a2 y 2 z bfyl 2 z bfm 2 z bfm r3 x a2 F2 Q2 Q2 4wso2bf Q2 47reo2a2 32 Q22aax 2baz 4Treo2a2 Ibf r 2 32 a2 b2 2v32 2 I az The electric field due to this system can be determined similarly and the charge induced on the planes can also be found Figure 623 Point charge between two semiinfinite conducting planes o 66 METHOD OF IMAGES 245 o Pxyz a b Figure 624 Determining a the potential at P and b the force on charge Q In general when the method of images is used for a system consisting of a point charge between two semiinfinite conducting planes inclined at an angle in degrees the number of images is given by because the charge and its images all lie on a circle For example when j 180 N 1 as in the case of Figure 622 for 0 90 N 3 as in the case of Figure 623 and for j 60 we expect AT 5 as shown in Figure 625 Q Figure 625 Point charge between two semiinfinite conducting walls inclined at j 60 to each Q Q Q 246 Electrostatic BoundaryValue Problems PRACTICE EXERCISE 614 If the point charge Q 10 nC in Figure 625 is 10 cm away from point O and along the line bisecting t 60 find the magnitude of the force on Q due to the charge induced on the conducting walls A n s w e r 6 0 5 3 i N n iyviasiSESsi 1 Boundaryvalue problems are those in which the potentials at the boundaries of a region are specified and we are to determine the potential field within the region They are solved using Poissons equation if pv 0 or Laplaces equation if pv 0 2 In a nonhomogeneous region Poissons equation is V e VV pv For a homogeneous region e is independent of space variables Poisson becomes s equation V2V In a chargefree region pv 0 Poissons equation becomes Laplaces equation that is v2y o 3 We solve the differential equation resulting from Poissons or Laplaces equation by in tegrating twice if V depends on one variable or by the method of separation of variables if Vis a function of more than one variable We then apply the prescribed boundary con ditions to obtain a unique solution 4 The uniqueness theorem states that if V satisfies Poissons or Laplaces equation and the prescribed boundary condition V is the only possible solution for that given problem This enables us to find the solution to a given problem via any expedient means because we are assured of one and only one solution 5 The problem of finding the resistance R of an object or the capacitance C of a capacitor may be treated as a boundaryvalue problem To determine R we assume a potential difference Vo between the ends of the object solve Laplaces equation find aE dS and obtain R VJI Similarly to determine C we assume a potential difference of Vo between the plates of the capacitor solve Laplaces equation find Q eE dS and obtain C QVo 6 A boundaryvalue problem involving an infinite conducting plane or wedge may be solved using the method of images This basically entails replacing the charge configu ration by itself its image and an equipotential surface in place of the conducting plane Thus the original problem is replaced by an image problem which is solved using techniques covered in Chapters 4 and 5 REVIEW QUESTIONS 247 iREVIEW QUESTIONS 61 Equation V sVV pv may be regarded as Poissons equation for an inhomoge neous medium a T r u e b False 62 In cylindrical coordinates equation dp P dp 10 0 is called l a Maxwells equation b Laplaces equation c Poissons equation d Helmholtzs equation e Lorentzs equation 63 Two potential functions Vi and V2 satisfy Laplaces equation within a closed region and assume the same values on its surface Vx must be equal to V2 a True b False c Not necessarily 64 Which of the following potentials does not satisfy Laplaces equation a V 2x 5 b V 10 xy c V r cos j e V p cos 10 65 Which of the following is not true a 5 cos 3x is a solution to 0x 90x 0 b 10 sin 2x is a solution to jx 4cfx 0 c 4 cosh 3y is a solution to y 9Ry 0 d sinh 2y is a solution to Ry 4Ry 0 e fz 1 where gx sin x and hy sinhy gx hy 248 H Electrostatic BoundaryValue Problems 66 If Vi XlY1 is a product solution of Laplaces equation which of these are not solutions of Laplaces equation a b XyYy 2xy c X x y dX1 Y1 e Xi 2YX 3 67 The capacitance of a capacitor filled by a linear dielectric is independent of the charge on the plates and the potential difference between the plates a True b False i 68 A parallelplate capacitor connected to a battery stores twice as much charge with a given dielectric as it does with air as dielectric the susceptibility of the dielectric is a 0 b I c 2 d 3 v e 4 69 A potential difference Vo is applied to a mercury column in a cylindrical container The mercury is now poured into another cylindrical container of half the radius and the same potential difference Vo applied across the ends As a result of this change of space the re sistance will be increased a 2 times b 4 times c 8 times d 16 times 610 Two conducting plates are inclined at an angle 30 to each other with a point charge between them The number of image charges is a 12 V b 11 6 d 5 e 3 Answers 61a 62c 63a 64c 65b 66de 67a 68b 69d 610b PROBLEMS PROBLEMS 61 In free space V 6xy2z 8 At point P 2 5 find E and pv 62 Two infinitely large conducting plates are located at x 1 and x 4 The space between them is free space with charge distribution nCm3 Find Vatx 2 if Vl 50V I O7T and V4 50 V 63 The region between x 0 and x d is free space and has pv pox dld If Vx 0 0 and Vx d Vo find a V and E b the surface charge densities at x 0 and x d 64 Show that the exact solution of the equation dx2 m 0 x L subject to Vx 0 Vl Vx L V2 x d d o Jo a V x2 y2 2z 10 c V3 pz sin j p 67 Show that the following potentials satisfy Laplaces equation a V e cos 13y sinh 249 65 A certain material occupies the space between two conducting slabs located at y 2 cm When heated the material emits electrons such that pv 501 y2 Cm3 If the slabs are both held at 30 kV find the potential distribution within the slabs Take e 3en 66 Determine which of the following potential field distributions satisfy Laplaces equation 250 Electrostatic BoundaryValue Problems d 2 mm d Vn Vz 0 0 Figure 626 For Problem 611 1 68 Show that E Ex Ey Ez satisfies Laplaces equation 69 Let V A cos nx B sin nxCeny Deny where A B C and are constants Show that V satisfies Laplaces equation 610 The potential field V 2x2yz y3z exists in a dielectric medium having e 2eo a Does V satisfy Laplaces equation b Calculate the total charge within the unit cube 0 xyz 1 m 611 Consider the conducting plates shown in Figure 626 If Vz 0 0 and Vz 2 mm 50 V determine V E and D in the dielectric region er 15 between the plates and ps on the plates 612 The cylindricalcapacitor whose cross section is in Figure 627 has inner and outer radii of 5 mm and 15 mm respectively If Vp 5 mm 100 V and Vp 1 5 mm 0 V calculate V E and D at p 10 mm and ps on each plate Take er 20 613 Concentric cylinders p 2 cm and p 6 cm are maintained at V 60 V and V 2 0 V respectively Calculate V E and D at p 4 cm 614 The region between concentric spherical conducting shells r 05 m and r 1 m is charge free If Vr 05 5 0 V and Vr 1 50 V determine the potential dis tribution and the electric field strength in the region between the shells 615 Find V and E at 3 0 4 due to the two conducting cones of infinite extent shown in Figure 628 Figure 627 Cylindrical capacitor of Problem 612 PROBLEMS 251 V 100 V t Figure 628 Conducting cones of Problem 615 616 The inner and outer electrodes of a diode are coaxial cylinders of radii a 06 m and b 30 mm respectively The inner electrode is maintained at 70 V while the outer elec trode is grounded a Assuming that the length of the electrodes a b and ignoring the effects of space charge calculate the potential at p 15 mm b If an electron is in jected radially through a small hole in the inner electrode with velocity 107 ms find its velocity at p 15mm 617 Another method of finding the capacitance of a capacitor is using energy considerations that is C 2WE vi Using this approach derive eqs 622 628 and 632 618 An electrode with a hyperbolic shape xy 4 is placed above an earthed rightangle corner as in Figure 629 Calculate V and E at point 1 2 0 when the electrode is con nected to a 20V source 619 Solve Laplaces equation for the twodimensional electrostatic systems of Figure 630 and find the potential Vx y 620 Find the potential Vx y due to the twodimensional systems of Figure 631 621 By letting Vp Rp44 be the solution of Laplaces equation in a region where p 0 show that the separated differential equations for R and P are xy 4 Figure 629 For Problem 618 v o vvo a Figure 630 For Problem 619 b c 252 0 Figure 631 For Problem 620 v o vo b PROBLEMS 253 and where X is the separation constant 4 X 0 622 A potential in spherical coordinates is a function of r and 8 but not j Assuming that Vr 6 RrF6 obtain the separated differential equations for R and F in a region for which pv 0 623 Show that the resistance of the bar of Figure 617 between the vertical ends located at 4 0 and p T2 is R lot In bla 624 Show that the resistance of the sector of a spherical shell of conductivity a with cross section shown in Figure 632 where 0 2TT between its base is R 1 1 1 27TCT1 c o s ex i a b 625 A hollow conducting hemisphere of radius a is buried with its flat face lying flush with the earth surface thereby serving as an earthing electrode If the conductivity of earth is a show that the leakage conductance between the electrode and earth is 2iraa 626 The cross section of an electric fuse is shown in Figure 633 If the fuse is made of copper and of thickness 15 mm calculate its resistance 627 In an integrated circuit a capacitor is formed by growing a silicon dioxide layer sr 4 of thickness 1 m over the conducting silicon substrate and covering it with a metal elec trode of area 5 Determine S if a capacitance of 2 nF is desired 628 The parallelplate capacitor of Figure 634 is quarterfilled with mica er 6 Find the capacitance of the capacitor Figure 632 For Problem 624 254 11 Electrostatic BoundaryValue Problems 4 cm 4 cm 3 cm M 1 cmT f 4 cm Figure 633 For Problem 626 629 An airfilled parallel plate capacitor of length L width a and plate separation d has its plates maintained at constant potential difference Vo If a dielectric slab of dielectric constant er is slid between the plates and is withdrawn until only a length x remains between the plates as in Figure 635 show that the force tending to restore the slab to its original position is F eoer 1 a Vj Id 630 A parallelplate capacitor has plate area 200 cm2 and plate separation 3 mm The charge density is 1 xCm2 with air as dielectric Find a The capacitance of the capacitor b The voltage between the plates c The force with which the plates attract each other 631 Two conducting plates are placed at z 2 cm and z 2 cm and are respectively maintained at potentials 0 and 200 V Assuming that the plates are separated by a polypropylene e 225eo Calculate a the potential at the middle of the plates b the surface charge densities at the plates 632 Two conducting parallel plates are separated by a dielectric material with e 56e0 and thickness 064 mm Assume that each plate has an area of 80 cm2 If the potential field dis tribution between the plates is V 3x Ay 2z 6 kV determine a the capaci tance of the capacitor b the potential difference between the plates Figure 634 For Problem 628 10 cm2 2 mm PROBLEMS 255 Figure 635 For Problem 629 633 The space between spherical conducting shells r 5 cm and r 10 cm is filled with a dielectric material for which s 225eo The two shells are maintained at a potential dif ference of 80 V a Find the capacitance of the system b Calculate the charge density on shell r 5 cm 634 Concentric shells r 20 cm and r 30 cm are held at V 0 and V 50 respectively If the space between them is filled with dielectric material e 31e0 a 1012 Sm find a V E and D b the charge densities on the shells c the leakage resistance 635 A spherical capacitor has inner radius a and outer radius d Concentric with the spherical conductors and lying between them is a spherical shell of outer radius c and inner radius b If the regions d r cc r b and b r a are filled with materials with per mittivites eu e2 and e3 respectively determine the capacitance of the system 636 Determine the capacitance of a conducting sphere of radius 5 cm deeply immersed in sea water er 80 637 A conducting sphere of radius 2 cm is surrounded by a concentric conducting sphere of radius 5 cm If the space between the spheres is filled with sodium chloride er 59 calculate the capacitance of the system 638 In an inkjet printer the drops are charged by surrounding the jet of radius 20 fim with a concentric cylinder of radius 600 jm as in Figure 636 Calculate the minimum voltage required to generate a charge 50 fC on the drop if the length of the jet inside the cylinder is 100 xm Take e eo 639 A given length of a cable the capacitance of which is 10 xFkm with a resistance of insu lation of 100 Milkm is charged to a voltage of 100 V How long does it take the voltage to drop to 50 V Liquid reservior r A liquid jet Figure 636 Simplified geometry of an inkjet printer for Problem 638 drop J 256 Electrostatic BoundaryValue Problems Figure 637 For Problem 640 640 The capacitance per unit length of a twowire transmission line shown in Figure 637 is given by C we cosh 2a Determine the conductance per unit length 641 A spherical capacitor has an inner conductor of radius a carrying charge Q and maintained at zero potential If the outer conductor contracts from a radius b to c under internal forces prove that the work performed by the electric field as a result of the contraction is W Qb c 8irebc 642 A parallelplate capacitor has its plates at x 0 d and the space between the plates is filled with an inhomogeneous material with permittivity e e0 1 H I If the plate at V dj x d is maintained at Vo while the plate at x 0 is grounded find a VandE b P c pps at x 0 d 643 A spherical capacitor has inner radius a and outer radius b and filled with an inhomoge neous dielectric with e eokr2 Show that the capacitance of the capacitor is C b a 644 A cylindrical capacitor with inner radius a and outer radius b is filled with an inhomoge neous dielectric having e eokp where A is a constant Calculate the capacitance per unit length of the capacitor 645 If the earth is regarded a spherical capacitor what is its capacitance Assume the radius of the earth to be approximately 6370 km PROBLEMS 257 646 A point charge of 10 nC is located at point P0 0 3 while the conducting plane z 0 is grounded Calculate a V and Eat R635 b The force on the charge due to induced charge on the plane 647 Two point charges of 3 nC and 4 nC are placed respectively at 0 0 1 m and 0 0 2 m while an infinite conducting plane is at z 0 Determine a The total charge induced on the plane b The magnitude of the force of attraction between the charges and the plane 648 Two point charges of 50 nC and 20 nC are located at 3 2 4 and 1 0 5 above the conducting ground plane z 2 Calculate a the surface charge density at 7 2 2 b D at 3 4 8 and c D at 1 1 1 649 A point charge of 10 jttC is located at 1 1 1 and the positive portions of the coordinate planes are occupied by three mutually perpendicular plane conductors maintained at zero potential Find the force on the charge due to the conductors 650 A point charge Q is placed between two earthed intersecting conducting planes that are in clined at 45 to each other Determine the number of image charges and their locations 651 Infinite line x 3 z 4 carries 16 nCm and is located in free space above the conduct ing plane z 0 a Find E at 2 2 3 b Calculate the induced surface charge density on the conducting plane at 5 6 0 652 In free space infinite planes y A and y 8 carry charges 20 nCm2 and 30 nCm2 re spectively If plane y 2 is grounded calculate E at P0 0 0 and Q4 6 2 PART 3 MAGNETOSTATICS Chapter 7 MAGNETOSTATIC FIELDS No honest man can be all things to all people ABRAHAM LINCOLN 71 INTRODUCTION In Chapters 4 to 6 we limited our discussions to static electric fields characterized by E or D We now focus our attention on static magnetic fields which are characterized by H or B There are similarities and dissimilarities between electric and magnetic fields As E and D are related according to D eE for linear material space H and B are related according to B pR Table 71 further shows the analogy between electric and magnetic field quantities Some of the magnetic field quantities will be introduced later in this chapter and others will be presented in the next The analogy is presented here to show that most of the equations we have derived for the electric fields may be readily used to obtain corresponding equations for magnetic fields if the equivalent analo gous quantities are substituted This way it does not appear as if we are learning new concepts A definite link between electric and magnetic fields was established by Oersted1 in 1820 As we have noticed an electrostatic field is produced by static or stationary charges If the charges are moving with constant velocity a static magnetic or magnetostatic field is produced A magnetostatic field is produced by a constant current flow or direct current This current flow may be due to magnetization currents as in permanent magnets electronbeam currents as in vacuum tubes or conduction currents as in currentcarrying wires In this chapter we consider magnetic fields in free space due to direct current Mag netostatic fields in material space are covered in Chapter 8 Our study of magnetostatics is not a dispensable luxury but an indispensable necessity r The development of the motors transformers microphones compasses telephone bell ringers television focusing controls advertising displays magnetically levitated high speed vehicles memory stores magnetic separators and so on involve magnetic phenom ena and play an important role in our everyday life2 Hans Christian Oersted 17771851 a Danish professor of physics after 13 years of frustrating efforts discovered that electricity could produce magnetism 2Various applications of magnetism can be found in J K Watson Applications of Magnetism New York John Wiley Sons 1980 2 6 1 262 Magnetostatic Fields TABLE 71 Analogy between Electric and Magnetic Fields Term Basic laws Force law Source element Field intensity Flux density Relationship between fields Potentials Flux Energy density Poissons equation F f F dQ E D D E v y y wE V2 Electric 222 4ire2 D dS g e n c gE i Vm y Cm 2 sE W f Pidl J Airsr D dS Q CV I E P H F gu H B H A y v Wm V2A Magnetic 4r2 d e n c gu X B Am y Wbm2 vym j o f nidi j 47ri J B d S L Lf i A similar analogy can be found in R S Elliot Electromagnetic theory a simplified representation IEEE Trans Educ vol E24 no 4 Nov 1981 pp 294296 There are two major laws governing magnetostatic fields 1 BiotSavarts law3 and 2 Amperes circuit law4 Like Coulombs law BiotSavarts law is the general law of magnetostatics Just as Gausss law is a special case of Coulombs law Amperes law is a special case of BiotSavarts law and is easily applied in problems involving symmetrical current distribution The two laws of magnetostatics are stated and applied first their derivation is provided later in the chapter 3The experiments and analyses of the effect of a current element were carried out by Ampere and by JeanBaptiste and Felix Savart around 1820 4Andre Marie Ampere 17751836 a French physicist developed Oersteds discovery and intro duced the concept of current element and the force between current elements 72 BIOTSAVARTS LAW 263 72 BIOTSAVARTS LAW BiotSavarts law states that the magnetic field intensity dll produced at a point P as shown in Figure 71 by the differential current clement ill is proportional to the product dl and the sine of the angle a between the clement and the line joining P to the element and is inversely proportional to the square of the distance K between P and the element That is or dH dl sin a R2 kl dl sin a R2 71 72 where k is the constant of proportionality In SI units k l4ir so eq 72 becomes dl sin a dH 4TTRZ 73 From the definition of cross product in eq 121 it is easy to notice that eq 73 is better put in vector form as dH Idl X a Idl XR 74 where R R and aR RR Thus the direction of dL can be determined by the right hand rule with the righthand thumb pointing in the direction of the current the righthand fingers encircling the wire in the direction of dH as shown in Figure 72a Alternatively we can use the righthanded screw rule to determine the direction of dH with the screw placed along the wire and pointed in the direction of current flow the direction of advance of the screw is the direction of dH as in Figure 72b Figure 71 magnetic field dH at P due to current element I dl dH inward 264 a Figure 72 Determining the direction of dH using a the righthand rule or b the righthanded screw rule It is customary to represent the direction of the magnetic field intensity H or current by a small circle with a dot or cross sign depending on whether H or I is out of or into the page as illustrated in Figure 73 Just as we can have different charge configurations see Figure 45 we can have dif ferent current distributions line current surface current and volume current as shown in Figure 74 If we define K as the surface current density in amperesmeter and J as the volume current density in amperesmeter square the source elements are related as 75 Thus in terms of the distributed current sources the BiotSavart law as in eq 74 becomes H H H Id X aR line current KdSXaR surface current 4TTR2 J dv X aR z volume current 4wR2 76 77 78 As an example let us apply eq 76 to determine the field due to a straight current carrying filamentary conductor of finite length AB as in Figure 75 We assume that the conductor is along the zaxis with its upper and lower ends respectively subtending angles H or is out H or is in Figure 73 Conventional representation of H or I a out of the page and b into the page a b 72 BIOTSAVARTS LAW 265 a b c Figure 74 Current distributions a line current b surface current c volume current a2 and a at P the point at which H is to be determined Particular note should be taken of this assumption as the formula to be derived will have to be applied accordingly If we con sider the contribution dH at P due to an element dl at 0 0 z dl Idl X R 4TTR3 But dl dz az and R pap zaz so dl X R P dz i Hence Ipdz H 79 710 711 Figure 75 Field at point P due to a straight filamen tary conductor H into the page 266 w Magnetostatic Fields Letting z p cot a dz p cosec2 a da and eq 711 becomes H 1 f2 p2 cosec2 a da p3 cosec3 a sin a da 4ir I 4irp or I H cos a2 cos 4irp 712 This expression is generally applicable for any straight filamentary conductor of finite length Notice from eq 712 that H is always along the unit vector a ie along concen tric circular paths irrespective of the length of the wire or the point of interest P As a special case when the conductor is semiinfinite with respect to P so that point A is now at 90 0 0 while B is at 0 0 a 90 a2 0 and eq 712 becomes H 4rp 713 Another special case is when the conductor is infinite in length For this case point A is at 0 0 oo while B is at 0 0 a 180 a2 0 so eq 712 reduces to H 2xp 714 To find unit vector a0 in eqs 712 to 714 is not always easy A simple approach is to de termine SJ from e X a 715 where af is a unit vector along the line current and ap is a unit vector along the perpendic ular line from the line current to the field point EXAMPLE 71 The conducting triangular loop in Figure 76a carries a current of 10 A Find H at 0 0 5 due to side i of the loop Solution This example illustrates how eq 712 is applied to any straight thin currentcarrying conductor The key point to keep in mind in applying eq 712 is figuring out a b a2 p and a To find H at 0 0 5 due to side 1 of the loop in Figure 76a consider Figure 72 BIOTSAVARTS LAW 267 a Figure 76 For Example 71 a conducting triangular loop b side 1 of the loop 76b where side 1 is treated as a straight conductor Notice that we join the point of in terest 0 0 5 to the beginning and end of the line current Observe that au a2 and p are assigned in the same manner as in Figure 75 on which eq 712 is based cos a cos 90 0 cos a2 V29 To determine a0 is often the hardest part of applying eq 712 According to eq 715 a BX and ap az so ar az a v Hence Hi irP cos 2 cos 591av mAm 268 Magnetostatic Fields PRACTICE EXERCISE 71 Find H at 0 0 5 due to side 3 of the triangular loop in Figure 76a Answer 3063a 363av mAm EXAMPLE 72 Find H at 3 4 0 due to the current filament shown in Figure 77a Solution Let H Hx Hz where Hx and H are the contributions to the magnetic field intensity at P 3 4 0 due to the portions of the filament along x and z respectively H7 4TTP cos a2 cos At P 3 4 0 p 9 1612 5 90 a2 0 and a is obtained as a unit vector along the circular path through P on plane z 0 as in Figure 77b The direction of a is determined using the righthanded screw rule or the righthand rule From the geometry in Figure 77b 4 3 a0 sin 6 ax cos 6 ay ax ay Alternatively we can determine a from eq 715 At point P a and ap are as illustrated in Figure 77a for Hz Hence 3 a z X ax ayJ ax ay 4 5 a b Figure 77 For Example 72 a current filament along semiinfinite x and zaxes a and ap for Hz only b determining ap for Hz 72 BIOTSAVARTS LAW 269 as obtained before Thus 4TT5 2865ay mAm It should be noted that in this case a0 happens to be the negative of the regular a of cylin drical coordinates Hz could have also been obtained in cylindrical coordinates as z 4TT5 V 4775a mAm Similarly for Hx at P p 4 a2 0 cos a 35 and a az or a ae X ap ax X ay az Hence Thus or 2388 a mAm H Hx Uz 382ax 2865ay 2388a mAm H 4775a0 2388a mAm Notice that although the current filaments appear semiinfinite they occupy the posi tive z and xaxes it is only the filament along the axis that is semiinfinite with respect to point P Thus Hz could have been found by using eq 713 but the equation could not have been used to find Hx because the filament along the xaxis is not semiinfinite with respect to P PRACTICE EXERCISE 72 The positive vaxis semiinfinite line with respect to the origin carries a filamentary current of 2 A in the ay direction Assume it is part of a large circuit Find H at a A2 3 0 b fl3 12 4 Answer a 1458az mAm b 4897a 3673a mAm 270 H Magnetostatic Fields EXAMPLE 73 A circular loop located on x2 y2 9 z 0 carries a direct current of 10 A along a De termine H at 0 0 4 and 0 0 4 Solution Consider the circular loop shown in Figure 78a The magnetic field intensity dH at point P0 0 h contributed by current element d is given by BiotSavarts law AKR3 where d p dj a0 R 0 0 h x y 0 pap haz and d X R 0 pd4 0 p 0 h pi az Hence ph d ap p2 d az dHp i 4xp h dHzaz By symmetry the contributions along ap add up to zero because the radial components produced by pairs of current element 180 apart cancel This may also be shown mathe matically by writing ap in rectangular coordinate systems ie ap cos f ax sin j ay P0 0 h Figure 78 For Example 73 a circular current loop b flux lines due to the current loop 72 BIOTSAVARTS LAW 271 Integrating cos j or sin over 0 j 2TT gives zero thereby showing that Hp 0 Thus H Ip l0 4TTP2 47Tp2 h2f2 or H 2p2 h2f2 a Substituting 0Ap 3h 4 gives H0 0 4 1 0 3 036a7 Am 29 1632 b Notice from rflXR above that if h is replaced by h the zcomponent of dH remains the same while the pcomponent still adds up to zero due to the axial symmetry of the loop Hence H0 0 4 H0 04 036az Am The flux lines due to the circular current loop are sketched in Figure 78b PRACTICE EXERCISE 73 A thin ring of radius 5 cm is placed on plane z 1 cm so that its center is at 001 cm If the ring carries 50 mA along a find H at a 00lcm b 00 10 cm Answer a 400az mAm b 573az mAm EXAMPLE 74 A solenoid of length and radius a consists of N turns of wire carrying current Show that at point P along its axis H cos 62 cos 0az where n N dl and d2 are the angles subtended at P by the end turns as illustrated in Figure 79 Also show that if a at the center of the solenoid H nl 272 Magnetostatic Fields Figure 79 For Example 74 cross section of a solenoid Solution Consider the cross section of the solenoid as shown in Figure 79 Since the solenoid con sists of circular loops we apply the result of Example 73 The contribution to the magnetic field H at P by an element of the solenoid of length dz is la n dz dH7 Idle 2a2 z2m 2a2 z2f2 where dl ndz Nit dz From Figure 79 tan 0 alz that is dz a cosec2 0 dd l sin 6 dd Hence or Thus dHz sin 0 dd Hz sin 0 dO H cos 62 cos di az as required Substituting n Nit gives NI H cos 02 c o s At the center of the solenoid cos V7 12 a2 241 cos 73 AMPERES CIRCUIT LAWMAXWELLS EQUATION 273 and 2 2412 z 2a2 24 If aor02 O 0 180 H PRACTICE EXERCISE 74 If the solenoid of Figure 79 has 2000 turns a length of 75 cm a radius of 5 cm and carries a current of 50 mA along a find H at a 00 0 b 0 0 75 cm c 0050 cm Answer a 6652az him b 6652a2 him c 1317az him 3 AMPERES CIRCUIT LAWMAXWELLS EQUATION Amperes circuit law states that the line integral of the tangential component of H around a dosed path is the same as the net current IK enclosed by the path In other words the circulation of H equals enc that is 716 Amperes law is similar to Gausss law and it is easily applied to determine H when the current distribution is symmetrical It should be noted that eq 716 always holds whether the current distribution is symmetrical or not but we can only use the equation to determine H when symmetrical current distribution exists Amperes law is a special case of BiotSavarts law the former may be derived from the latter By applying Stokes theorem to the lefthand side of eq 716 we obtain e n c V X H W S 717 274 Magnetostatic Fields But 4nc J 718 Comparing the surface integrals in eqs 717 and 718 clearly reveals that V X H J 719 This is the third Maxwells equation to be derived it is essentially Amperes law in differ ential or point form whereas eq 716 is the integral form From eq 719 we should observe that V X H J 0 that is magnetostatic field is not conservative 74 APPLICATIONS OF AMPERES LAW We now apply Amperes circuit law to determine H for some symmetrical current distri butions as we did for Gausss law We will consider an infinite line current an infinite current sheet and an infinitely long coaxial transmission line A Infinite Line Current Consider an infinitely long filamentary current along the zaxis as in Figure 710 To de termine H at an observation point P we allow a closed path pass through P This path on which Amperes law is to be applied is known as an Amperian path analogous to the term Gaussian surface We choose a concentric circle as the Amperian path in view of eq 714 which shows that H is constant provided p is constant Since this path encloses the whole current according to Amperes law j pdf 2irp Amperian path Figure 710 Amperes law applied to an infinite filamentary line current or 74 APPLICATIONS OF AMPERES LAW M 275 720 as expected from eq 714 B Infinite Sheet of Current Consider an infinite current sheet in the z 0 plane If the sheet has a uniform current density K Kyay Am as shown in Figure 711 applying Amperes law to the rectangular closed path Amperian path gives H d enc Kyb 721a To evaluate the integral we first need to have an idea of what H is like To achieve this we regard the infinite sheet as comprising of filaments dH above or below the sheet due to a pair of filamentary currents can be found using eqs 714 and 715 As evident in Figure 711b the resultant dH has only an xcomponent Also H on one side of the sheet is the negative of that on the other side Due to the infinite extent of the sheet the sheet can be re garded as consisting of such filamentary pairs so that the characteristics of H for a pair are the same for the infinite current sheets that is JHoax I Hoax z 0 z 0 721b a Figure 711 Application of Amperes law to an infinite sheet a closed path 12341 b sym metrical pair of current filaments with current along ay 276 Magnetostatic Fields where Ho is yet to be determined Evaluating the line integral of H in eq 721b along the closed path in Figure 711 a gives Udl j H rfl I h h M 0a Hob 0fl Hob 2Hob 1 721c From eqs 721a and 721c we obtain Ho Ky Substituting Ho in eq 721b gives H z0 722 Kyax z0 In general for an infinite sheet of current density K Am 723 where an is a unit normal vector directed from the current sheet to the point of interest C Infinitely Long Coaxial Transmission Line Consider an infinitely long transmission line consisting of two concentric cylinders having their axes along the zaxis The cross section of the line is shown in Figure 712 where the zaxis is out of the page The inner conductor has radius a and carries current while the outer conductor has inner radius b and thickness t and carries return current We want to determine H everywhere assuming that current is uniformly distributed in both conduc tors Since the current distribution is symmetrical we apply Amperes law along the Am Amperian paths Figure 712 Cross section of the 4 yf transmission line the positive direc tion is out of the page perian path for each of the four possible regions 0 ρ a a ρ b b ρ b t and ρ b t For region 0 ρ a we apply Amperes law to path L1 giving H dl Ienc J dS 724 Since the current is uniformly distributed over the cross section J Iπa² az dS ρ dφ dρ az Ienc J dS Iπa² ρ dφ dρ Iπa² πρ² Iρ²a² Hence eq 724 becomes Hφ dl Hφ 2πρ Iρ²a² or Hφ Iρ2πa² 725 For region a ρ b we use path L2 as the Amperian path H dl Ienc I Hφ 2πρ I or Hφ I2πρ 726 since the whole current I is enclosed by L2 Notice that eq 726 is the same as eq 714 and it is independent of a For region b ρ b t we use path L3 getting H dl Hφ 2πφ Ienc 727a where Ienc I J dS 278 Magnetostatic Fields and J in this case is the current density current per unit area of the outer conductor and is along av that is J Thus irb if b2 z 2TT rp 1 p2b2 t2 2bt p dp df Substituting this in eq 727a we have H For region p 6 t we use path L4 getting L4 or Putting eqs 725 to 728 together gives 727b 728 H aw 2rp 0 2bt 0 p a a p Z b pb t p b t 729 The magnitude of H is sketched in Figure 713 Notice from these examples that the ability to take H from under the integral sign is the key to using Amperes law to determine H In other words Amperes law can only be used to find H due to symmetric current distributions for which it is possible to find a closed path over which H is constant in magnitude 74 APPLICATIONS OF AMPERES LAW 279 Figure 713 Plot of H against p U b bt EXAMPLE 75 Planes 2 0 and z 4 carry current K lOa Am and K lOa Am respectively Determine H at a 111 b 0 3 10 Solution Let the parallel current sheets be as in Figure 714 Also let H Ho H4 where Ho and H4 are the contributions due to the current sheets z 0 and z 4 respec tively We make use of eq 723 a At 1 1 1 which is between the plates 0 z 1 4 Ho 12 K X an 12 10ax X a 5av Am H4 l 2 K X a 12 10ax X a 5ay Am Hence H 10ay Am 4 Figure 714 For Example 75 parallel M B t infinite current sheets y 8 8 8 8 z 0 280 Magnetostatic Fields b At 0 3 10 which is above the two sheets z 10 4 0 Ho 12 10a X az 5a Am H4 12 lOaJ X az 5a y Am Hence H 0 Am PRACTICE EXERCISE 75 Plane y 1 carries current K 50az mAm Find H at a 000 b 153 Answer a 25ax mAm b 25a mAm EXAMPLE 76 A toroid whose dimensions are shown in Figure 715 has N turns and carries current De termine H inside and outside the toroid Solution We apply Amperes circuit law to the Amperian path which is a circle of radius p show dotted in Figure 715 Since N wires cut through this path each carrying current the n current enclosed by the Amperian path is NI Hence H d 7enc H 2irp M Figure 715 For Example 76 a toroid with a circular cross section 75 MAGNETIC FLUX DENSITYMAXWELLS EQUATION 281 or H NI 2irp for po a p po a where po is the mean radius of the toroid as shown in Figure 715 An approximate value of His H NI NI approx 2irpo Notice that this is the same as the formula obtained for H for points well inside a very long solenoid 5s a Thus a straight solenoid may be regarded as a special toroidal coil for which po co Outside the toroid the current enclosed by an Amperian path is NI NI 0 and hence H 0 PRACTICE EXERCISE 76 A toroid of circular cross section whose center is at the origin and axis the same as the zaxis has 1000 turns with po 10 cm a 1 cm If the toroid carries a 100mA current find H at a 3 c m 4 cm 0 b 6 cm 9 cm 0 Answer a 0 b 1471 Am 5 MAGNETIC FLUX DENSITYMAXWELLS EQUATION The magnetic flux density B is similar to the electric flux density D As D soE in free space the magnetic flux density B is related to the magnetic field intensity H according to 730 where o is a constant known as the permeability of free space The constant is in henrysmeter Hm and has the value of 4TT X 107 Hm 731 The precise definition of the magnetic field B in terms of the magnetic force will be given in the next chapter 282 Magnetostatic Fields Figure 716 Magnetic flux lines due to a straight Magnetic flux lines wire with current coming out of the page The magnetic flux through a surface S is given by 732 where the magnetic flux f is in webers Wb and the magnetic flux density is in weberssquare meter Wbm2 or teslas The magnetic flux line is the path to which B is tangential at every point in a magnetic field It is the line along which the needle of a magnetic compass will orient itself if placed in the magnetic field For example the magnetic flux lines due to a straight long wire are shown in Figure 716 The flux lines are determined using the same principle followed in Section 410 for the electric flux lines The direction of B is taken as that indicated as north by the needle of the magnetic compass Notice that each flux line is closed and has no beginning or end Though Figure 716 is for a straight currentcarrying conductor it is generally true that magnetic flux lines are closed and do not cross each other regardless of the current distribution In an electrostatic field the flux passing through a closed surface is the same as the charge enclosed that is P D dS Q Thus it is possible to have an isolated electric charge as shown in Figure 717a which also reveals that electric flux lines are not neces sarily closed Unlike electric flux lines magnetic flux lines always close upon themselves as in Figure 717b This is due to the fact that it is not possible to have isolated magnetic closed surface Q closed surface f 0 a b Figure 717 Flux leaving a closed surface due to a isolated electric charge V s D dS Q b magnetic charge Y s B dS 0 76 MAXWELLS EQUATIONS FOR STATIC EM FIELDS 283 N S N S N S N S N S N S N S s B N N N N Figure 718 Successive division of a bar magnet results in pieces with north and south poles showing that magnetic poles cannot be isolated poles or magnetic charges For example if we desire to have an isolated magnetic pole by dividing a magnetic bar successively into two we end up with pieces each having north and south poles as illustrated in Figure 718 We find it impossible to separate the north pole from the south pole An isolated magnetic charge does not exist Thus the total flux through a closed surface in a magnetic field must be zero that is 733 This equation is referred to as the law of conservation of magnetic flux or Gausss law for magnetostatic fields just as D dS Q is Gausss law for electrostatic fields Although the magnetostatic field is not conservative magnetic flux is conserved By applying the divergence theorem to eq 733 we obtain B dS V B dv 0 or V B 0 734 This equation is the fourth Maxwells equation to be derived Equation 733 or 734 shows that magnetostatic fields have no sources or sinks Equation 734 suggests that magnetic field lines are always continuous 6 MAXWELLS EQUATIONS FOR STATIC EM FIELDS Having derived Maxwells four equations for static electromagnetic fields we may take a moment to put them together as in Table 72 From the table we notice that the order in which the equations were derived has been changed for the sake of clarity 284 Magnetostatic Fields TABLE 72 Maxwells Equations for Static EM Fields Differential or Point Form Integral Form Remarks V D pv V B 0 V X E 0 V x H J D dS pv dv Gausss law B dS E d Nonexistence of magnetic monopole Conservativeness of electrostatic field H d J dS Amperes law The choice between differential and integral forms of the equations depends on a given problem It is evident from Table 72 that a vector field is defined completely by specifying its curl and divergence A field can only be electric or magnetic if it satisfies the corresponding Maxwells equations see Problems 726 and 727 It should be noted that Maxwells equations as in Table 72 are only for static EM fields As will be discussed in Chapter 9 the divergence equations will remain the same for timevarying EM fields but the curl equations will have to be modified 77 MAGNETIC SCALAR AND VECTOR POTENTIALS We recall that some electrostatic field problems were simplified by relating the electric po tential V to the electric field intensity E E VV Similarly we can define a potential associated with magnetostatic field B In fact the magnetic potential could be scalar Vm or vector A To define Vm and A involves recalling two important identities see Example 39 and Practice Exercise 39 V X VV 0 V V X A 0 735a 735b which must always hold for any scalar field V and vector field A Just as E VV we define the magnetic scalar potential Vm in amperes as related to H according to H VVm if J 0 736 The condition attached to this equation is important and will be explained Combining eq 736 and eq 719 gives J V X H V X VVm 0 737 77 MAGNETIC SCALAR AND VECTOR POTENTIALS 285 since Vm must satisfy the condition in eq 735a Thus the magnetic scalar potential Vm is only defined in a region where J 0 as in eq 736 We should also note that Vm satisfies Laplaces equation just as V does for electrostatic fields hence V2Vm 0 J 0 738 We know that for a magnetostatic field V B 0 as stated in eq 734 In order to satisfy eqs 734 and 735b simultaneously we can define the vector magnetic potential A in Wbm such that B V X A Just as we defined we can define V dQ 4ireor 739 740 for line current for surface current for volume current 741 742 743 Rather than obtaining eqs 741 to 743 from eq 740 an alternative approach would be to obtain eqs 741 to 743 from eqs 76 to 78 For example we can derive eq 741 from eq 76 in conjunction with eq 739 To do this we write eq 76 as IdV X R R3 744 where R is the distance vector from the line element dV at the source point x1 y z to the field point x y z as shown in Figure 719 and R R that is R r r x xf y yf z zf1 745 Hence v 1 Rj x xax y a y R x xf yf z zT2 286 Magnetostatic Fields R r r Figure 719 Illustration of the source point y z and the field point x y z x y z or R R 746 where the differentiation is with respect to x y and z Substituting this into eq 744 we obtain B M IdV X v 4TT R 747 We apply the vector identity V X F V X F VXF 748 whereis a scalar field and F is a vector field Taking IR and F dV we have Since V operates with respect to x y z while dV is a function of x y z V X dV 0 Hence 749 750 With this equation eq 747 reduces to B V x 4irR Comparing eq 750 with eq 739 shows that verifying eq 741 77 MAGNETIC SCALAR AND VECTOR POTENTIALS 287 By substituting eq 739 into eq 732 and applying Stokess theorem we obtain B dS V X A dS P A d 4 or 751 Thus the magnetic flux through a given area can be found using either eq 732 or 751 Also the magnetic field can be determined using either Vm or A the choice is dictated by the nature of the given problem except that Vm can only be used in a sourcefree region The use of the magnetic vector potential provides a powerful elegant approach to solving EM problems particularly those relating to antennas As we shall notice in Chapter 13 it is more convenient to find B by first finding A in antenna problems EXAMPLE 77 Given the magnetic vector potential A p24 az Wbm calculate the total magnetic flux crossing the surface f ir2 1 p 2 m 0 z 5 m Solution We can solve this problem in two different ways using eq 732 or eq 751 Method 1 B V x A a0 J a0 op 2 dS dp dz a0 Hence TP j B dS 375 Wb 1 p dp dz p z0 5 15 Method 2 We use p I A d fi v2 r3 v4 L where L is the path bounding surface S V 1 f 2 3 and V4 are respectively the evalua tions of A d along the segments of L labeled 1 to 4 in Figure 720 Since A has only a 2component 0 y 3 288 I I Magnetostatic Fields 5 4 3 2 1 L 1 That is i rr 2 Figure 720 For Example 77 375 Wb as obtained previously Note that the direction of the path L must agree with that of dS PRACTICE EXERCISE 77 A current distribution gives rise to the vector magnetic potential A xzyx y2xay 4xyzaz Wbm Calculate a B a t 1 2 5 b The flux through the surface defined b y z 1 0 x 1 1 y 4 Answer a 20ax 40ay 3az Wbm2 b 20 Wb EXAMPLE 78 If plane z 0 carries uniform current K Kyay H 12 Kyax l2Kvax 0 0 This was obtained in Section 74 using Amperes law Obtain this by using the concept of vector magnetic potential 77 MAGNETIC SCALAR AND VECTOR POTENTIALS 289 Solution Consider the current sheet as in Figure 721 From eq 742 lxoKdS dA AKR In this problem K Kyay dS dx dy and for z 0 R R 0 0 z xy0 ixf yf z2112 781 where the primed coordinates are for the source point while the unprimed coordinates are for the field point It is necessary and customary to distinguish between the two points to avoid confusion see Figure 719 Hence dA dx dy ay 4TTX2 yf z2 f z 2 1 2 dB V X dA d Ay ax dz jxoKyz dx dy ax B 4irx2 yf z232 H0Kzax r f dx dy 2n32 782 In the integrand we may change coordinates from Cartesian to cylindrical for convenience so that Hence B 4ir fioKyz 4ir jxoKyzax P d4 dp yo vo 2TT I p 1 too 2 z 2y p0 o 2 f or z 0 By simply replacing z by z in eq 782 and following the same procedure we obtain is H l ax for z 0 290 Magnetostatic Fields Figure 721 For Example 78 infinite current sheet y PRACTICE EXERCISE 78 Repeat Example 78 by using BiotSavarts law to determine H at points 00 h and 00 h 78 DERIVATION OF BIOTSAVARTS LAW AND AMPERES LAW Both BiotSavarts law and Amperes law may be derived using the concept of magnetic vector potential The derivation will involve the use of the vector identities in eq 748 and V X V X A VV A V2A 752 Since BiotSavarts law as given in eq 74 is basically on line current we begin our derivation with eqs 739 and 741 that is 4TTR 4TT R 753 where R is as denned in eq 745 If the vector identity in eq 748 is applied by letting F dl and IR eq 753 becomes 754 Since V operates with respect to x y z and dl is a function of xr y z V X dl 0 Also x xf y yf z z2 R 12 755 78 DERIVATION OF BIOTSAVARTS LAW AND AMPERES LAW 291 jl x xax y yay z zaz a R x xf y yY 32 756 where a is a unit vector from the source point to the field point Thus eq 754 upon dropping the prime in d becomes 4TT L R2 which is BiotSavarts law Using the identity in eq 752 with eq 739 we obtain V X B VV A V2A It can be shown that for a static magnetic field VA 0 so that upon replacing B with xoH and using eq 719 eq 758 becomes V2A AIOV X H or V2A i 757 758 759 760 which is called the vector Poissons equation It is similar to Poissons equation V2V pvle in electrostatics In Cartesian coordinates eq 760 may be decomposed into three scalar equations V2AX V2Ay 761 which may be regarded as the scalar Poissons equations It can also be shown that Amperes circuit law is consistent with our definition of the magnetic vector potential From Stokess theorem and eq 739 H d V X H dS V X V X AdS From eqs 752 759 and 760 VxVxAV2A 762 2 I agnetostatic Fields Substituting this into eq 762 yields H d J dS I which is Amperes circuit law SUMMARY 1 The basic laws BiotSavarts and Amperes that govern magnetostatic fields are dis cussed BiotSavarts law which is similar to Coulombs law states that the magnetic field intensity dH at r due to current element d at r is dR Id X R in Am where R r r and R R For surface or volume current distribution we replace d with K dS or J dv respectively that is Id Jdv 2 Amperes circuit law which is similar to Gausss law states that the circulation of H around a closed path is equal to the current enclosed by the path that is or V X H J IeiK J d S third Maxwells equation to be derived When current distribution is symmetric so that an Amperian path on which H is constant can be found Amperes law is useful in determining H that is Im or H enc 3 The magnetic flux through a surface S is given by f BdS inWb where B is the magnetic flux density in Wbm2 In free space where fio 4ir X 10 7 Hm permeability of free space 4 Since an isolated or free magnetic monopole does not exist the net magnetic flux through a closed surface is zero f t B dS 0 REVIEW QUESTIONS 293 or V B 0 fourth Maxwells equation to be derived 5 At this point all four Maxwells equations for static EM fields have been derived namely V D Pv VB 0 V X E 0 V X H J 6 The magnetic scalar potential Vm is defined as H W m if J 0 and the magnetic vector potential A as B V X A where V A 0 With the definition of A the magnetic flux through a surface S can be found from V A d L where L is the closed path defining surface S see Figure 320 Rather than using BiotSavarts law the magnetic field due to a current distribution may be found using A a powerful approach that is particularly useful in antenna theory For a current element d at r the magnetic vector potential at r is A AKR R r r l 7 Elements of similarity between electric and magnetic fields exist Some of these are listed in Table 71 Corresponding to Poissons equation V2V pvle for example is V2A nJ 71 One of the following is not a source of magnetostatic fields a A dc current in a wire b A permanent magnet c An accelerated charge d An electric field linearly changing with time e A charged disk rotating at uniform speed 294 Magnetostatic Fields 72 Identify the configuration in Figure 722 that is not a correct representation of and H 73 Consider points A B C D and on a circle of radius 2 as shown in Figure 723 The items in the right list are the values of a at different points on the circle Match these items with the points in the list on the left a b c d e A B C D 77 h 1 ii iii iv v vi vii viii ax a a a v a ay V2 ax a a x ay ax ay 74 The zaxis carries filamentary current of IOTT A along az Which of these is incorrect a H aAm at 050 b H a Am at 5 TT4 0 c H 08ax 06a at34 0 d H a 0 a t 5 3ir2 0 75 Plane y 0 carries a uniform current of 30az niAm At 1 10 2 the magnetic field intensity is a 5xmJm b 15 a d b c H O e Figure 722 For Review Question 72 I REVIEW QUESTIONS Ji 295 Figure 723 For Review Question 73 c 4775axAm d 1885avnAm e None of the above 76 For the currents and closed paths of Figure 724 calculate the value of jL H d 77 Which of these statements is not characteristic of a static magnetic field a It is solenoidal b It is conservative c It has no sinks or sources d Magnetic flux lines are always closed e The total number of flux lines entering a given region is equal to the total number of flux lines leaving the region Si 30 A Figure 724 For Review Question 76 30 A c d 296 Magnetostatic Fields Figure 725 For Review Question 710 Volume 78 Two identical coaxial circular coils carry the same current but in opposite direc tions The magnitude of the magnetic field B at a point on the axis midway between the coils is a Zero b The same as that produced by one coil c Twice that produced by one coil d Half that produced by one coil 79 One of these equations is not Maxwells equation for a static electromagnetic field in a linear homogeneous medium a V B 0 b V X D 0 c 0 B d nJ d D dS Q e V2A nJ 710 Two bar magnets with their north poles have strength Qml 20 A m and Qm2 10 A m magnetic charges are placed inside a volume as shown in Figure 725 The magnetic flux leaving the volume is a 200 Wb b 30 Wb c 10 Wb d OWb e lOWb Answers 71c 72c 73 aii bvi ci dv eiii 74d 75a 76 a 10 A b 20 A c 0 d 10 A 77b 78a 79e 710d PROBLEMS 71 a State BiotSavarts law b The y and zaxes respectively carry filamentary currents 10 A along ay and 20 A along az Find H at 3 4 5 PROBLEMS 297 Figure 726 For Problem 73 72 A conducting filament carries current from point A0 0 a to point 50 0 b Show that at point Px y 0 H Vx 73 Consider AB in Figure 726 as part of an electric circuit Find H at the origin due to AB 74 Repeat Problem 73 for the conductor AB in Figure 727 75 Line x 0 y 0 0 z 10m carries current 2 A along az Calculate H at points a 5 0 0 b 5 5 0 c 5 15 0 d 5 1 5 0 76 a Find H at 0 0 5 due to side 2 of the triangular loop in Figure 76a b Find H at 0 0 5 due to the entire loop 77 An infinitely long conductor is bent into an L shape as shown in Figure 728 If a direct current of 5 A flows in the current find the magnetic field intensity at a 2 2 0 b0 2 0 and c 00 2 Figure 727 For Problem 74 4A 298 Magnetostatic Fields Figure 728 Current filament for Problem 77 5 A 5A 78 Find H at the center C of an equilateral triangular loop of side 4 m carrying 5 A of current as in Figure 729 79 A rectangular loop carrying 10 A of current is placed on z 0 plane as shown in Figure 730 Evaluate H at a 2 2 0 b 4 2 0 c 4 8 0 d 0 0 2 710 A square conducting loop of side 2a lies in the z 0 plane and carries a current in the counterclockwise direction Show that at the center of the loop H wa 711 a A filamentary loop carrying current is bent to assume the shape of a regular polygon of n sides Show that at the center of the polygon nl ir H 2irr sin n where r is the radius of the circle circumscribed by the polygon b Apply this to cases when n 3 and n 4 and see if your results agree with those for the triangular loop of Problem 78 and the square loop of Problem 710 respectively Figure 729 Equilateral triangular loop for Problem 78 PROBLEMS 299 Figure 730 Rectangular loop of Problem 79 c As n becomes large show that the result of part a becomes that of the circular loop of Example 73 712 For the filamentary loop shown in Figure 731 find the magnetic field strength at O 713 Two identical current loops have their centers at 0 0 0 and 0 0 4 and their axes the same as the zaxis so that the Helmholtz coil is formed If each loop has radius 2 m and carries current 5 A in a calculate H at a 000 b 002 714 A 3cmlong solenoid carries a current of 400 mA If the solenoid is to produce a mag netic flux density of 5 mWbm how many turns of wire are needed 715 A solenoid of radius 4 mm and length 2 cm has 150 turnsm and carries current 500 mA Find a H at the center b H at the ends of the solenoid 716 Plane x 10 carries current 100 mAm along az while line x 1 y 2 carries fila mentary current 20TT mA along ar Determine H at 4 3 2 717 a State Amperes circuit law b A hollow conducting cylinder has inner radius a and outer radius b and carries current along the positive zdirection Find H everywhere 10 A 100 cm 10 A Figure 731 Filamentary loop of Problem 712 not drawn to scale 300 Magnetostatic Fields 718 a An infinitely long solid conductor of radius a is placed along the zaxis If the con ductor carries current in the z direction show that H 2at 2ira within the conductor Find the corresponding current density b If 3 A and a 2 cm in part a find H at 0 1 cm 0 and 0 4 cm 0 719 If H yax xay Am on plane z 0 a determine the current density and b verify Amperes law by taking the circulation of H around the edge of the rectangle Z 0 0 x 3 1 y 4 720 In a certain conducting region H yzx2 y2ax y2xzay 4x2y2az Am a Determine J at 5 2 3 b Find the current passing through x 10 yz 2 c Show that V B 0 721 An infinitely long filamentary wire carries a current of 2 A in the zdirection Calculate a B a t 3 4 7 b The flux through the square loop described by 2 p 6 0 z 4 90 722 The electric motor shown in Figure 732 has field 106 H sin 2j a Am Calculate the flux per pole passing through the air gap if the axial length of the pole is 20 cm 723 Consider the twowire transmission line whose cross section is illustrated in Figure 733 Each wire is of radius 2 cm and the wires are separated 10 cm The wire centered at 0 0 Figure 732 Electric motor pole of Problem 722 armature PROBLEMS 301 4 cmKJ 10 cm Figure 733 Twowire line of Problem 723 carries current 5 A while the other centered at 10 cm 0 carries the return current Find Hat a 5 cm 0 b 10 cm 5 cm 724 Determine the magnetic flux through a rectangular loop a X b due to an infinitely long conductor carrying current as shown in Figure 734 The loop and the straight conductors are separated by distance d 725 A brass ring with triangular cross section encircles a very long straight wire concentrically as in Figure 735 If the wire carries a current show that the total number of magnetic flux lines in the ring is r b a In a b 2wb L b Calculate V if a 30 cm b 10 cm h 5 cm and 10 A 726 Consider the following arbitrary fields Find out which of them can possibly represent electrostatic or magnetostatic field in free space a A y cos axax y exaz b B ap c C r2 sin 6 a0 Figure 734 For Problem 724 I 302 Magnetostatic Fields hv Brass ring Figure 735 Cross section of a brass ring enclosing a long straight wire for Problem 725 727 Reconsider the previous problem for the following fields a D y2zax 2x yzay JC lz2az z 1 sin0 b E cos 4 aD H P P c F 2 cos 6 ar sin d ae 728 For a current distribution in free space A 2x2y yzax xy2 xz3ay 6xyz 2jc2y2 az Wbm a Calculate B b Find the magnetic flux through a loop described by x 1 0 y z 2 c Show that V A 0 and V B 0 729 The magnetic vector potential of a current distribution in free space is given by A 15p sin j az Wbm Find H at 3 ir4 10 Calculate the flux through p 5 0 0 w2 0 z 10 730 A conductor of radius a carries a uniform current with J Joaz Show that the magnetic vector potential for p a is A iaJop2az 731 An infinitely long conductor of radius a is placed such that its axis is along the zaxis The vector magnetic potential due to a direct current Io flowing along a in the conductor is given by A fxox2 y2 az Wbm Find the corresponding H Confirm your result using Amperes law PROBLEMS 303 732 The magnetic vector potential of two parallel infinite straight current filaments in free space carrying equal current in opposite direction is JX1 d p A In a 2TT p where d is the separation distance between the filaments with one filament placed along the zaxis Find the corresponding magnetic flux density B 733 Find the current density J to in free space A az Wbm P 734 Prove that the magnetic scalar potential at 0 0 z due to a circular loop of radius a shown in Figure 78a is V m 1 735 A coaxial transmission line is constructed such that the radius of the inner conductor is a and the outer conductor has radii 3a and 4a Find the vector magnetic potential within the outer conductor Assume Az 0 for p 3a 736 The zaxis carries a filamentary current 12 A along az Calculate Vm at 4 30 2 if Vm Oat10 60 7 737 Plane z 2 carries a current of 50ay Am If Vm 0 at the origin find Vm at a 2 0 5 b 10 3 1 738 Prove in cylindrical coordinates that a V X VV 0 b V V X A 0 739 IfR r r and R show that R R fl3 where V and V are del operators with respect to x y z and x y z respectively Chapter 8 MAGNETIC FORCES MATERIALS AND DEVICES Do all the good you can By all the means you can In all the ways you can In all the places you can At all the times you can To all the people you can As long as ever you can JOHN WESLEY 81 INTRODUCTION Having considered the basic laws and techniques commonly used in calculating magnetic field B due to currentcarrying elements we are prepared to study the force a magnetic field exerts on charged particles current elements and loops Such a study is important to problems on electrical devices such as ammeters voltmeters galvanometers cyclotrons plasmas motors and magnetohydrodynamic generators The precise definition of the mag netic field deliberately sidestepped in the previous chapter will be given here The con cepts of magnetic moments and dipole will also be considered Furthermore we will consider magnetic fields in material media as opposed to the magnetic fields in vacuum or free space examined in the previous chapter The results of the preceding chapter need only some modification to account for the presence of materi als in a magnetic field Further discussions will cover inductors inductances magnetic energy and magnetic circuits 82 FORCES DUE TO MAGNETIC FIELDS There are at least three ways in which force due to magnetic fields can be experienced The force can be a due to a moving charged particle in a B field b on a current element in an external B field or c between two current elements 304 82 FORCES DUE TO MAGNETIC FIELDS 305 A Force on a Charged Particle According to our discussion in Chapter 4 the electric force Fe on a stationary or moving electric charge Q in an electric field is given by Coulombs experimental law and is related to the electric field intensity E as Fe QE 81 This shows that if Q is positive Fe and E have the same direction A magnetic field can exert force only on a moving charge From experiments it is found that the magnetic force Fm experienced by a charge Q moving with a velocity u in a magnetic field B is Fm Qn X B 82 This clearly shows that Fm is perpendicular to both u and B From eqs 81 and 82 a comparison between the electric force e and the magnetic force Fm can be made Fe is independent of the velocity of the charge and can perform work on the charge and change its kinetic energy Unlike Fe Fm depends on the charge ve locity and is normal to it Fm cannot perform work because it is at right angles to the direc tion of motion of the charge Fm d 0 it does not cause an increase in kinetic energy of the charge The magnitude of Fm is generally small compared to Fe except at high ve locities For a moving charge Q in the presence of both electric and magnetic fields the total force on the charge is given by F F F or F gE u X B 83 This is known as the Lorentz force equation1 It relates mechanical force to electrical force If the mass of the charged particle moving in E and B fields is m by Newtons second law of motion du m 84 The solution to this equation is important in determining the motion of charged particles in E and B fields We should bear in mind that in such fields energy transfer can be only by means of the electric field A summary on the force exerted on a charged particle is given in Table 81 Since eq 82 is closely parallel to eq 81 which defines the electric field some authors and instructors prefer to begin their discussions on magnetostatics from eq 82 just as discussions on electrostatics usually begin with Coulombs force law After Hendrik Lorentz 18531928 who first applied the equation to electric field motion 306 W Magnetic Forces Materials and Devices TABLE Force on a Charged Particle State of Particle E Field B Field Combined E and B Fields Stationary Moving Qu X B QE 2E u X B B Force on a Current Element To determine the force on a current element dl of a currentcarrying conductor due to the magnetic field B we modify eq 82 using the fact that for convection current see eq 57 J Pu From eq 75 we recall the relationship between current elements Idl KdS idv Combining eqs 85 and 86 yields I dl pvu dv dQu Alternatively dl dl dQ dQ u dt dt 85 86 Hence Idl dQu 87 This shows that an elemental charge dQ moving with velocity u thereby producing con vection current element dQ u is equivalent to a conduction current element dl Thus the force on a current element dl in a magnetic field B is found from eq 82 by merely re placing Qu by dl that is d Idl X B 88 If the current is through a closed path L or circuit the force on the circuit is given by 89 F b Idl X B i In using eq 88 or 89 we should keep in mind that the magnetic field produced by the current element dl does not exert force on the element itself just as a point charge does not exert force on itself The B field that exerts force on dl must be due to another element In other words the B field in eq 88 or 89 is external to the current element dl If instead of the line current element dl we have surface current elements K dS 82 FORCES DUE TO MAGNETIC FIELDS 307 or a volume current element J dv we simply make use of eq 86 so that eq 88 becomes dF KdS XB or dF J dv X B while eq 89 becomes F KdSXB or F J d v X B 88a 89a From eq 88 The magnetic field B is defined as the force per unit current element Alternatively B may be defined from eq 82 as the vector which satisfies FJq u X B just as we defined electric field E as the force per unit charge FJq Both of these defini tions of B show that B describes the force properties of a magnetic field C Force between Two Current Elements Let us now consider the force between two elements dx and I2 d2 According to BiotSavarts law both current elements produce magnetic fields So we may find the force dd on element dl due to the field dB2 produced by element I2 d2 as shown in Figure 81 From eq 88 But from BiotSavarts law Hence ddF 7 dx X dB2 xo2 d2 X aRii 810 811 812 Figure 81 Force between two current loops 308 Magnetic Forces Materials and Devices This equation is essentially the law of force between two current elements and is analogous to Coulombs law which expresses the force between two stationary charges From eq 812 we obtain the total force F on current loop 1 due to current loop 2 shown in Figure 81 as F 4TT X dl2 X 813 L JL2 Although this equation appears complicated we should remember that it is based on eq 810 It is eq 89 or 810 that is of fundamental importance The force F2 on loop 2 due to the magnetic field Bx from loop 1 is obtained from eq 813 by interchanging subscripts 1 and 2 It can be shown that F2 F thus F and F2 obey Newtons third law that action and reaction are equal and opposite It is worth while to mention that eq 813 was experimentally established by Oersted and Ampere Biot and Savart Amperes colleagues actually based their law on it EXAMPLE 81 A charged particle of mass 2 kg and charge 3 C starts at point 1 2 0 with velocity 4ax 3az ms in an electric field 123 lOa Vm At time t 1 s determine a The acceleration of the particle b Its velocity c Its kinetic energy d Its position Solution a This is an initialvalue problem because initial values are given According to Newtons second law of motion F ma QE where a is the acceleration of the particle Hence QE 3 a 12a 10ay 18a 15ayms2 du d a ux uy uz 18ax 15a b Equating components gives dux dt 18KX 18r A 15 v 15 B dt y 811 812 82 FORCES DUE TO MAGNETIC FIELDS 309 dt 0 M 7 C 813 where A B and C are integration constants But at t 0 u Aax 3az Hence uxt 0 44 0 A or A 4 uyt 0 00 0 B or B 0 uzt O 3 H 3 C Substituting the values of A B and C into eqs 811 to 813 gives ur wx MV Mj 18f 4 15f 3 Hence ut 1 s 22a 15a 3az ms c Kinetic energy KE m ju2 2222 152 32 718J d u xyz 18r 4 15 3 Equating components yields ux 18 4 x 9r2 4f 814 dt y uz 3 z dt At t 0 JC j z 1 2 0 hence xt 0 1 1 0 A yf 0 2 2 0 B zf 0 0 0 0 C or C 0 Substituting the values of Ab Bu and C into eqs 814 to 816 we obtain x y z 9r2 4 1 752 2 30 or A 1 or 5 2 815 816 817 Hence at t 1 j z 14 55 3 By eliminating tm eq 817 the motion of the particle may be described in terms of y and z 310 Magnetic Forces Materials and Devices PRACTICE EXERCISE 81 A charged particle of mass 1 kg and charge 2 C starts at the origin with zero initial velocity in a region where E 3az Vm Find a The force on the particle b The time it takes to reach point P0 0 12 m c Its velocity and acceleration at P d Its KE at P Answer a 6az N b 2 s c 12az ms 6az ms2 d 72 J EXAMPLE 82 A charged particle of mass 2 kg and 1 C starts at the origin with velocity 3av ms and travels in a region of uniform magnetic field B lOa Wbm At t 4 s calculate a The velocity and acceleration of the particle b The magnetic force on it c Its KE and location d Find the particles trajectory by eliminating t e Show that its KE remains constant Solution du a F m Qu X B dt du Q a u X B dt m Hence uxax uyy uzaz ux uy uz 0 0 10 By equating components we get dux di duz dt 5ur 5 0 Ujiy 821 822 823 82 FORCES DUE TO MAGNETIC FIELDS 311 We can eliminate ux or uy in eqs 821 and 822 by taking second derivatives of one equation and making use of the other Thus d2ux duy dt2 5 25 or d ux d7 25ux 0 which is a linear differential equation with solution see Case 3 of Example 65 ux d cos 5 C2 sin 5 824 From eqs 821 and 824 5M 5C sin 5f 5C2 cos 5t 825 dt or uy d sin 5 C2 cos 5 We now determine constants Co Cu and C2 using the initial conditions At t 0 u 3ar Hence ux 0 0 Cj 1 C2 0 C 0 uy 3 3 d 0 C2 1 C2 3 uz 0 0 Co Substituting the values of Co C and C2 into eqs 823 to 825 gives u ux uy uz 3 sin 5 3 cos 5t 0 826 Hence and b or Uf 4 3 sin 20 3 cos 20 0 2739ax 1224ay ms du a 15 cos 5f 15 sin 5t 0 if af 4 6101a 13703avms2 F ma 122ax 274avN F gu X B 1X27398 1224av X 10a 122a 274a N 312 Magnetic Forces Materials and Devices c KE l2m u2 122 27392 12242 9 J ux 3 sin 5f x cos 5 bx dy 3 uy 3 cos 5f y sin 5t b2 at 5 dz dt 827 828 829 where bu b2 and b3 are integration constants At t 0 x y z 0 0 0 and hence xt 0 0 0 1 06 yt 0 0 0 0 b2 62 0 8210 z 0 0 0 3 Substituting the values of bt b2 and b3 into eqs 827 to 829 we obtain x y z 06 06 cos 5 06 sin 5f 0 At t 4 s x y z 03552 05478 0 d From eq 8210 we eliminate t by noting that x 062 y2 062 cos2 5t sin2 5 z 0 or x 062 y2 062 z 0 which is a circle on plane z 0 centered at 06 0 0 and of radius 06 m Thus the parti cle gyrates in an orbit about a magnetic field line e KE m u2 2 9 cos2 5t 9 sin2 5t 9 J which is the same as the KE at t 0 and t 4 s Thus the uniform magnetic field has no effect on the KE of the particle Note that the angular velocity cu QBIm and the radius of the orbit r uju where MO is the initial speed An interesting application of the idea in this example is found in a common method of focusing a beam of electrons The method employs a uniform mag netic field directed parallel to the desired beam as shown in Figure 82 Each electron emerging from the electron gun follows a helical path and is back on the axis at the same focal point with other electrons If the screen of a cathode ray tube were at this point a single spot would appear on the screen 82 FORCES D U E TO MAGNETIC FIELDS W 313 focal point Figure 82 Magnetic focusing of a beam of electrons a helical paths of electrons b end view of paths a b PRACTICE EXERCISE 82 A proton of mass m is projected into a uniform field B Boaz with an initial veloc ity aax 3ar a Find the differential equations that the position vector r xax yay zaz must satisfy b Show that a solution to these equations is a x sin oit 0 a y cos ut where w eBJm and e is the charge on the proton c Show that this solution de scribes a circular helix in space Answer a a cos ut a sin cat j3 b and c Proof at at at EXAMPLE 83 A charged particle moves with a uniform velocity 4ax ms in a region where E 20 ay Vm and B Boaz Wbm2 Determine Bo such that the velocity of the particle remains constant Solution If the particle moves with a constant velocity it implies that its acceleration is zero In other words the particle experiences no net force Hence 0 2 20av 4ax X Boa or 20av ABoay Thus Bo 5 This example illustrates an important principle employed in a velocity filter shown in Figure 83 In this application E B and u are mutually perpendicular so that Qu X B is 314 Magnetic Forces Materials and Devices charged u particles Aperture Particles with constant velocity F mQuXB Figure 83 A velocity filter for charged particles directed opposite to QE regardless of the sign of the charge When the magnitudes of the two vectors are equal QuB QE or This is the required critical speed to balance out the two parts of the Lorentz force Parti cles with this speed are undeflected by the fields they are filtered through the aperture Particles with other speeds are deflected down or up depending on whether their speeds are greater or less than this critical speed PRACTICE EXERCISE 83 Uniform E and B fields are oriented at right angles to each other An electron moves with a speed of 8 X 106 ms at right angles to both fields and passes undeflected through the field a If the magnitude of B is 05 mWbm2 find the value of E b Will this filter work for positive and negative charges and any value of mass Answer a 4 kVm b Yes EXAMPLE 84 A rectangular loop carrying current I2 is placed parallel to an infinitely long filamentary wire carrying current Ix as shown in Figure 84a Show that the force experienced by the loop is given by 2x 1 iPo 1 po 82 FORCES DUE TO MAGNETIC FIELDS 315 a 2H w Figure 84 For Example 84 a rectangular loop inside the field produced by an infinitely long wire b forces acting on the loop and wire b Solution Let the force on the loop be F 4 I dh X B F F7 where F b F2 F3 and F 4 are respectively the forces exerted on sides of the loop labeled 1 2 3 and 4 in Figure 84b Due to the infinitely long wire a 2TTPO Hence F I2 d2 X Bl I2 dz az X 2irpo ao attractive Fj is attractive because it is directed toward the long wire that is F is along ap due to the fact that loop side 1 and the long wire carry currents along the same direction Similarly F 3 I2 d2 X B I2 i7 X zb 2TTPO a 2TTPO a F2 72 I ip ap X Acl Po Cl In 2TT P O repulsive parallel 316 Magnetic Forces Materials and Devices dpap x VJih Po a 2TT In az parallel The total force Fe on the loop is the sum of Fl5 F2 F3 and F4 that is 1 1 F 2w po a which is an attractive force trying to draw the loop toward the wire The force Fw on the wire by Newtons third law is F see Figure 84b PRACTICE EXERCISE 84 In Example 84 find the force experienced by the infinitely long wire if lx 10 A I2 5 A po 20 cm a 10 cm b 30 cm Answer Sa tN 83 MAGNETIC TORQUE AND MOMENT Now that we have considered the force on a current loop in a magnetic field we can deter mine the torque on it The concept of a current loop experiencing a torque in a magnetic field is of paramount importance in understanding the behavior of orbiting charged parti cles dc motors and generators If the loop is placed parallel to a magnetic field it expe riences a force that tends to rotate it The torque T or mechanical moincnl of force on ihe loop is the cclor product of the force F and iho momem arm r That is T r X F 814 and its units are Newtonmeters N m Let us apply this to a rectangular loop of length and width w placed in a uniform magnetic field B as shown in Figure 85a From this figure we notice that d is parallel to B along sides 12 and 34 of the loop and no force is exerted on those sides Thus F d X B I d X B 0 dz az X B dz a z X B 0 e 83 MAGNETIC TORQUE AND MOMENT 317 it 2wf 1 3 4 f l 1 B I fl i I axis of rotation a b Figure 85 Rectangular planar loop in a uniform magnetic field or F Fo Fo 0 815 where F0 IB because B is uniform Thus no force is exerted on the loop as a whole However Fo and Fo act at different points on the loop thereby creating a couple If the normal to the plane of the loop makes an angle a with B as shown in the crosssectional view of Figure 85b the torque on the loop is T FO wsina or T Bliw sin a But w S the area of the loop Hence T BIS sin a We define the quantity m ISa 816 817 818 as the magnetic dipole moment in Am2 of the loop In eq 818 an is a unit normal vector to the plane of the loop and its direction is determined by the righthand rule fingers in the direction of current and thumb along an The magnetic dipolc moment is the product of current and area of the loop its di rection is normal to the loop Introducing eq 818 in eq 817 we obtain T m X B 819 318 Magnetic Forces Materials and Devices This expression is generally applicable in determining the torque on a planar loop of any arbitrary shape although it was obtained using a rectangular loop The only limitation is that the magnetic field must be uniform It should be noted that the torque is in the direc tion of the axis of rotation the zaxis in the case of Figure 85a It is directed such as to reduce a so that m and B are in the same direction In an equilibrium position when m and B are in the same direction the loop is perpendicular to the magnetic field and the torque will be zero as well as the sum of the forces on the loop 84 A MAGNETIC DIPOLE A bar magnet or a small filamentary current loop is usually referred to as a magnetic dipole The reason for this and what we mean by small will soon be evident Let us de termine the magnetic field B at an observation point Pr 8 4 due to a circular loop carry ing current as in Figure 86 The magnetic vector potential at P is 820 It can be shown that at far field r a so that the loop appears small at the observation point A has only 0component and it is given by 821a or A jxjwa sin 4TIT2 om X 47rr2 ar 821b Pr 6 0 Figure 86 Magnetic field at P due to a current loop 84 A MAGNETIC DIPOLE 319 where m Iira2az the magnetic moment of the loop and a X ar sin d a0 We deter mine the magnetic flux density B from B V X A as B 2 cos 6 ar sin 6 822 It is interesting to compare eqs 821 and 822 with similar expressions in eqs 480 and 482 for electrical potential V and electric field intensity E due to an elec tric dipole This comparison is done in Table 82 in which we notice the striking similari TABLE 82 Comparison between Electric and Magnetic Monopoles and Dipoles Electric V Monopoie point charge Qcasd E Qd 2 cos B ar sin1 0 Dipole two point charge Magnetic Does not exist Qm Monopoie point charge A sin 0i 0 4irr2 B Me 2 cos Hr sin 9ae 47TC3 Dipole small current loop or bar magnet 320 Magnetic Forces Materials and Devices Figure 87 The B lines due to magnetic dipoles a a small current loop with m IS b a bar magnet with m Qm a b ties between B as far field due to a small current loop and E at far field due to an electric dipole It is therefore reasonable to regard a small current loop as a magnetic dipole The B lines due to a magnetic dipole are similar to the E lines due to an electric dipole Figure 87a illustrates the B lines around the magnetic dipole m IS A short permanent magnetic bar shown in Figure 87b may also be regarded as a magnetic dipole Observe that the B lines due to the bar are similar to those due to a small current loop in Figure 87a Consider the bar magnet of Figure 88 If Qm is an isolated magnetic charge pole strength and is the length of the bar the bar has a dipole moment Qm Notice that Qm does exist however it does not exist without an associated Qm See Table 82 When the bar is in a uniform magnetic field B it experiences a torque TmXB2JXB 823 where points in the direction southtonorth The torque tends to align the bar with the external magnetic field The force acting on the magnetic charge is given by F QmB 824 Since both a small current loop and a bar magnet produce magnetic dipoles they are equiv alent if they produce the same torque in a given B field that is when T QJB ISB Hence QJ IS showing that they must have the same dipole moment 825 826 n s r l F B Figure 88 A bar magnet in an external magnetic field 84 A MAGNETIC DIPOLE 321 EXAMPLE 85 Determine the magnetic moment of an electric circuit formed by the triangular loop of Figure 89 Solution From Problem 118c the equation of a plane is given by Ax By Cz D 0 where D A2 B2 C2 Since points 2 0 0 0 2 0 and 0 0 2 lie on the plane these points must satisfy the equation of the plane and the constants A B C and D can be determined Doing this gives x y z 2 as the plane on which the loop lies Thus we can use m ISan where S loop area X base X height 2 V22 V2sin 60 4 sin 60 If we define the plane surface by a function fxyz x y z 2 0 V a ay az V 3 We choose the plus sign in view of the direction of the current in the loop using the right hand rule m is directed as in Figure 89 Hence a ay a m 5 4 sin 60 r V 3 10ax ay a A m2 Figure 89 Triangular loop of Example 85 322 B Magnetic Forces Materials and Devices PRACTICE EXERCISE 85 A rectangular coil of area 10 cm2 carrying current of 50 A lies on plane 2x 6y 3z 7 such that the magnetic moment of the coil is directed away from the origin Calculate its magnetic moment Answer 1429a 4286a 2143az X 102 A m2 EXAMPLE 86 A small current loop L with magnetic moment 53 Am is located at the origin while another small loop current L2 with magnetic moment 3ay A m2 is located at 4 3 10 Determine the torque on L2 Solution The torque T2 on the loop L2 is due to the field Bj produced by loop L Hence T2 m2 X B Since m for loop Lx is along az we find Bj using eq 822 B 4irr 2 cos 9 ar sin 8 ag Using eq 223 we transform m2 from Cartesian to spherical coordinates m2 3av 3 sin 6 sin 4 ar cos 6 sin 0 ae cos t a At 4 3 10 r V4 2 32 102 5V5 2 V5 Hence p 5 1 1 tan 6 sin0 1 z 10 2 V COS P y 3 3 4 tan j sin 0 c o s j B 4 x X 1 0 7 X 5 4 1 j I j ar H j ae 47T625V5 VV5 V5 107 4ar a 625 m2 3 5V5 5V5 5 85 MAGNETIZATION IN MATERIALS 323 and T 1 0 4V5aA X 4ar 1 3 a r 6 625 5 V5 4293 X 10 6a r 3878ae 24a0 0258ar 1665a lO3a0nN m PRACTICE EXERCISE 86 If the coil of Practice Exercise 85 is surrounded by a uniform field 06ax 043 05a Wbm2 a Find the torque on the coil b Show that the torque on the coil is maximum if placed on plane 2x 8 4z V84 Calculate the value of the maximum torque Answer a 003a 002av 002a N m b 004387 N m 85 MAGNETIZATION IN MATERIALS Our discussion here will parallel that on polarization of materials in an electric field We shall assume that our atomic model is that of an electron orbiting about a positive nucleus We know that a given material is composed of atoms Each atom may be regarded as consisting of electrons orbiting about a central positive nucleus the electrons also rotate or spin about their own axes Thus an internal magnetic field is produced by electrons or biting around the nucleus as in Figure 810a or electrons spinning as in Figure 810b Both of these electronic motions produce internal magnetic fields B that are similar to the magnetic field produced by a current loop of Figure 811 The equivalent current loop has a magnetic moment of m IbSan where S is the area of the loop and Ib is the bound current bound to the atom Without an external B field applied to the material the sum of ms is zero due to random orientation as in Figure 812a When an external B field is applied the magnetic Figure 810 a Electron orbiting around the nucleus b electron spin nucleus 23 electron J electron a 324 Magnetic Forces Materials and Devices Figure 811 Circular current loop equivalent to electronic motion of Figure 810 moments of the electrons more or less align themselves with B so that the net magnetic moment is not zero as illustrated in Figure 812b The magnetization M in amperesmeter is the magnetic dipole moment per unit volume If there are N atoms in a given volume Av and the kth atom has a magnetic moment m M lim k 0 Av 827 A medium for which M is not zero everywhere is said to be magnetized For a differential volume dv the magnetic moment is dm M dv From eq 821b the vector magnetic potential due to dm is dX X AKR1 dv According to eq 746 R X Rdv B 0 M 0 Hgurc 8 2 Magnetic dipole mo ment in a volume Av a before B is applied b after B is applied a b 85 MAGNETIZATION IN MATERIALS 325 Hence Using eq 748 gives A I M X V dv 4TT R 828 VXMVX Substituting this into eq 828 yields 4TT Jy R 4TT JV fl Applying the vector identity V X F dv J F X r f S to the second integral we obtain 4TT JV J 4TT JS R Ho f ibdv JXO 829 4ir v R 4TT S R Comparing eq 829 with eqs 742 and 743 upon dropping the primes gives h V X M and 830 831 where Jb is the bound volume current density or magnetization volume current density in amperes per meter square Kb is the bound surface current density in amperes per meter and an is a unit vector normal to the surface Equation 829 shows that the potential of a magnetic body is due to a volume current density Jb throughout the body and a surface current Kb on the surface of the body The vector M is analogous to the polarization P in dielectrics and is sometimes called the magnetic polarization density of the medium In another sense M is analogous to H and they both have the same units In this respect as J V X H so is Jb V X M Also Jb and Kb for a magnetized body are similar to ppv and pps for a polarized body As is evident in eqs 829 to 831 Jh and Kh can be derived from M therefore ib and Kb are not commonly used 326 HI Magnetic Forces Materials and Devices In free space M 0 and we have V X H it or V X B 832 where Jy is the free current volume density In a material medium M i 0 and as a result B changes so that xiJJJ V X H V X M or B M 833 The relationship in eq 833 holds for all materials whether they are linear or not The concepts of linearity isotropy and homogeneity introduced in Section 57 for dielectric media equally apply here for magnetic media For linear materials M in Am depends linearly on H such that 834 where m is a dimensionless quantity ratio of M to H called magnetic susceptibility of the medium It is more or less a measure of how susceptible or sensitive the material is to a magnetic field Substituting eq 834 into eq 833 yields B xol or where 835 836 837 The quantity x ioxr is called the permeability of the material and is measured in henrysmeter the henry is the unit of inductance and will be defined a little later The di mensionless quantity xr is the ratio of the permeability of a given material to that of free space and is known as the relative permeability of the material It should be borne in mind that the relationships in eqs 834 to 837 hold only for linear and isotropic materials If the materials are anisotropic eg crystals eq 833 still holds but eqs 834 to 837 do not apply In this case fi has nine terms similar to e in eq 537 and consequently the fields B H and M are no longer parallel B ixr A 1 ioxrH Xm E Mo 86 CLASSIFICATION OF MAGNETIC MATERIALS 327 86 CLASSIFICATION OF MAGNETIC MATERIALS In general we may use the magnetic susceptibility m or the relative permeability ir to classify materials in terms of their magnetic property or behavior A material is said to be nonmagnetic if ym 0 or jxr 1 it is magnetic otherwise Free space air and materials with Xm 0 or fir 1 are regar3eTasfT61imagnetic Roughly speaking magnetic materials may be grouped into three major classes dia magnetic paramagnetic and ferromagnetic This rough classification is depicted in Figure 813 A material is said to be diamagnetic if it has xr S 1 ie very small nega tive Xm It is paramagnetic if pr S 1 ie very small positive xm If Mr 1 ie verY large positive xm the material is ferromagnetic Table B3 in Appendix B presents the values fir for some materials From the table it is apparent that for most practical purposes we may assume that ir 1 for diamagnetic and paramagnetic materials Thus we may regard diamagnetic and paramagnetic materials as linear and nonmagnetic Ferromagnetic materials are always nonlinear and magnetic except when their temperatures are above curie temperature to be explained later The reason for this will become evident as we more closely examine each of these three types of magnetic materials Diamagnetism occurs in materials where the magnetic fields due to electronic motions of orbiting and spinning completely cancel each other Thus the permanent or intrinsic magnetic moment of each atom is zero and the materials are weakly affected by a magnetic field For most diamagnetic materials eg bismuth lead copper silicon diamond sodium chloride xm is of the order of 1T5 In certain types of materials called super conductors at temperatures near absolute zero perfect diamagnetism occurs xm 1 or jjir 0 and B 0 Thus superconductors cannot contain magnetic fields2 Except for superconductors diamagnetic materials are seldom used in practice Although the diamag netic effect is overshadowed by other stronger effects in some materials all materials exhibit diamagnetism Materials whose atoms have nonzero permanent magnetic moment may be paramag netic or ferromagnetic Paramagnetism occurs in materials where the magnetic fields pro Magnetic Materials Linear Diamagnetics Xm0 M r s 10 Paramagnetics Xm 0 fir a Nonlinear Ferromagnetics Xm 0 nr a Figure 813 Classification of magnetic materials 2An excellent treatment of superconductors is found in M A Plonus Applied Electromagnetics New York McGrawHill 1978 pp 375388 Also the August 1989 issue of the Proceedings of IEEE is devoted to superconductivity 328 Magnetic Forces Materials and Devices duced by orbital and spinning electrons do not cancel completely Unlike diamagnetism paramagnetism is temperature dependent For most paramagnetic materials eg air plat inum tungsten potassium m is ofthe order 105 to 103 and is temperature depen dent Such materials find application in masers Ferromagnetism occurs in materials whose atoms have relatively large permanent magnetic moment They are called ferromagnetic materials because the best known member is iron Other members are cobalt nickel and their alloys Ferromagnetic materi als are very useful in practice As distinct from diamagnetic and paramagnetic materials ferromagnetic materials have the following properties 1 They are capable of being magnetized very strongly by a magnetic field 2 They retain a considerable amount of their magnetization when removed from the field 3 They lose their ferromagnetic properties and become linear paramagnetic materials when the temperature is raised above a certain temperature known as the curie tem perature Thus if a permanent magnet is heated above its curie temperature 770C for iron it loses its magnetization completely 4 They are nonlinear that is the constitutive relation B xoirH does not hold for ferromagnetic materials because xr depends on B and cannot be represented by a single value Thus the values of xr cited in Table B3 for ferromagnetics are only typical For example for nickel xr 50 under some conditions and 600 under other conditions As mentioned in Section 59 for conductors ferromagnetic materials such as iron and steel are used for screening or shielding to protect sensitive electrical devices from dis turbances from strong magnetic fields A typical example of an iron shield is shown in Figure 814a where the compass is protected Without the iron shield the compass gives an erroneous reading due to the effect of the external magnetic field as in Figure 814b For perfect screening it is required that the shield have infinite permeability Even though B juoH M holds for all materials including ferromagnetics the relationship between B and H depends on previous magnetization of a ferromagnetic Iron shield N s I b Figure 814 Magnetic screening a iron shield protecting a small compass b compass gives erroneous reading without the shield 86 CLASSIFICATION OF MAGNETIC MATERIALS 329 materialits magnetic history Instead of having a linear relationship between B and H ie B fiH it is only possible to represent the relationship by a magnetization curve or BH curve A typical BH curve is shown in Figure 815 First of all note the nonlinear relation ship between B and H Second at any point on the curve fi is given by the ratio BH and not by dBdH the slope of the curve If we assume that the ferromagnetic material whose BH curvein Figure 815 is ini tially unmagnetized as H increases due to increase in current from O to maximum applied field intensity HmdX curve OP is produced This curve is referred to as the virgin or initial magnetization curve After reaching saturation at P if H is decreased B does not follow the initial curve but lags behind H This phenomenon of B lagging behind H is called hysteresis which means to lag in Greek If H is reduced to zero B is not reduced to zero but to Bn which is referred to as the permanent flux density The value of Br depends on max the maximum applied field in tensity The existence of Br is the cause of having permanent magnets If H increases neg atively by reversing the direction of current B becomes zero when H becomes Hc which is known as the coercive field intensity Materials for which Hc is small are said to be mag netically hard The value of Hc also depends on Hmm Further increase in H in the negative direction to reach Q and a reverse in its direction to reach P gives a closed curve called a hysteresis loop The shape of hysteresis loops varies from one material to another Some ferrites for example have an almost rectangu lar hysteresis loop and are used in digital computers as magnetic information storage devices The area of a hysteresis loop gives the energy loss hysteresis loss per unit volume during one cycle of the periodic magnetization of the ferromagnetic material This energy loss is in the form of heat It is therefore desirable that materials used in electric generators motors and transformers should have tall but narrow hysteresis loops so that hysteresis losses are minimal Initial magnetization curve Figure 815 Typical magnetization BH curve 330 Magnetic Forces Materials and Devices EXAMPLE 87 Region 0 z 2 m is occupied by an infinite slab of permeable material xr 25 If B 0yx 5xay mWbm2 within the slab determine a J b ih c M d Kb on z 0 Solution a By definition J V X H V X B 1 4ir X 1025 V dx dy dB a 106 5 1010 X 4775azkAmz b h XmJ Mr DJ 154775az 103 7163a7kAm2 c M XmH B 1510yax 5xay 10 Air X 10725 3 4775vax 2387xav kAm d Kb M X an Since z 0 is the lower side of the slab occupying 0 z 2 an az Hence Kb 4775jax 2387xav X a 2387xax 4775jaTkAm PRACTICE EXERCISE 87 In a certain region i 46x0 find a Xm b H c M B We mWbm2 Answer a 36 b mOea Am c 6228eyaz Am 87 MAGNETIC BOUNDARY CONDITIONS We define magnetic boundary conditions as the conditions that H or B field must satisfy at the boundary between two different media Our derivations here are similar to those in Section 59 We make use of Gausss law for magnetic fields B dS 0 838 87 MAGNETIC BOUNDARY CONDITIONS 331 and Amperes circuit law H d I 339 Consider the boundary between two magnetic media 1 and 2 characterized respec tively by ix and x2 as in Figure 816 Applying eq 838 to the pillbox Gaussian surface of Figure 816a and allowing Ah 0 we obtain ln AS B2n AS 840 Thus or 841 since B H Equation 841 shows that the normal component of B is continuous at the boundary It also shows that the normal component of H is discontinuous at the boundary H undergoes some change at the interface Similarly we apply eq 839 to the closed path abcda of Figure 816b where surface current K on the boundary is assumed normal to the path We obtain Ah Z I H2n Ah As Ah 0 eq 842 leads to H Aw H M H Ak Hit H2 K 842 843 f H a b Figure 816 Boundary conditions between two magnetic media a for B b for H 332 Magnetic Forces Materials and Devices This shows that the tangential component of H is also discontinuous Equation 843 may be written in terms of B as K 844 In the general case eq 843 becomes H H2 X anl2 K 845 where anl2 is a unit vector normal to the interface and is directed from medium 1 to medium 2 If the boundary is free of current or the media are not conductors for K is free current density K 0 and eq 843 becomes I H l r H2 or 846 Thus the tangential component of H is continuous while that of B is discontinuous at the boundary If the fields make an angle 6 with the normal to the interface eq 841 results in cos 0 Bln B2n B2 cos while eq 846 produces Mi B2 sin 0 Hu H2t sin 62 847 848 Dividing eq 848 by eq 847 gives r tan I tan02 849 which is similar to eq 565 the law of refraction for magnetic flux lines at a boundary with no surface current EXAMPLE 88 Given that H 2a x 6ay 4az Am in region y x 2 0 where 50 cal culate a M and B b H2 and B2 in region y x 2 0 where ju2 2ito Solution Since j x 2 0 i s a plane y x 2 o r y x 2 i s region 1 in Figure 817 A point in this region may be used to confirm this For example the origin 0 0 is in this 87 MAGNETIC BOUNDARY CONDITIONS 333 Figure 817 For Example 88 region since 0 0 2 0 If we let the surface of the plane be described by jx y y x 2 a unit vector normal to the plane is given by a M ay a V2 M 1H 5 lX264 24av 16a7 Am b HlB H aBaB 264 B JHJ AionnH 4TT X 10752 64 1257a 377ay 2513aiWbm2 1101 110 Vl J V2 But Hence Hln Hlf Ult H Hln 2 6 4 44 0 2ar 2av 4a7 Using the boundary conditions we have H2 H u 4az or H2n HIB 4a 4ay 10a 10a 334 f i Magnetic Forces Materials and Devices Thus H 2 H2n H2 8 a x 12av 4a Am and B 2 fi2H2 jxojxr2n2 4TT X 10 728 12 4 2011a 3016ay 1005a PRACTICE EXERCISE 88 Region 1 described by 3x Ay 10 is free space whereas region 2 described by 3x Ay 10 is a magnetic material for which J OJU0 Assuming that the boundary between the material and free space is current free find B7 if B 01a 04av 02a Wbm2 Answer 1052a 1264a 2az Wbm2 EXAMPLE 89 The xyplane serves as the interface between two different media Medium 1 z 0 is filled with a material whose Mr 6 and medium 2 z 0 is filled with a material whose Hr 4 If the interface carries current 1Mo av mAm and B2 5a 8a mWbm2 find HiandB Solution In the previous example K 0 so eq 846 was appropriate In this example however K 0 and we must resort to eq 845 in addition to eq 841 Consider the problem as illustrated in Figure 818 Let B Bx By Bz in mWbm2 But 2 nr24 Bin B2n 5a K y B7 8 mAm Figure 818 For Example 89 881 882 87 MAGNETIC BOUNDARY CONDITIONS 335 and H B 1 Bxax Byay Bzaz mAm 883 Having found the normal components we can find the tangential components using H H2 X anl2 K or H X anl2 H2 X aBl2 K 884 Substituting eqs 882 and 883 into eq 884 gives B A 5vay Bzaz X az 5a 8az X a av 6 4x M Equating components yields By 0 From eqs 881 and 885 5 4 o r 6 T 4 B 15a 8azmWbm2 H 025ax 133a mAm Ml Mo and 885 H2 125a 2az mAm Mo Note that Hlx is 1O mAm less than H2x due to the current sheet and also that Bin B2n PRACTICE EXERCISE 89 A unit normal vector from region 2 ft 2MO to region 1 ft Mo is a2i 6ax 2a 3az7 If H 0ax ay 12az Am and H2 4az Am determine a H b The surface current density K on the interface c The angles Bj and B2 make with the normal to the interface 5ay Answer a 5833 b 486a 864a 395a Am c7627 7762 336 Magnetic Forces Materials and Devices 88 INDUCTORS AND INDUCTANCES A circuit or closed conducting path carrying current produces a magnetic field B which causes a flux J B dS to pass through each turn of the circuit as shown in Figure 819 If the circuit has N identical turns we define the flux linkage X as X NY 850 Also if the medium surrounding the circuit is linear the flux linkage X is proportional to the current producing it that is or X 851 where Lisa constant of proportionality called the inductance of the circuit The inductance L is a property of the physical arrangement of the circuit A circuit or part of a circuit that has inductance is called an inductor From eqs 850 and 851 we may define induc tance L of an inductor as the ratio of the magnetic flux linkage X to the current through the inductor that is X 852 The unit of inductance is the henry H which is the same as webersampere Since the henry is a fairly large unit inductances are usually expressed in millihenrys mH The inductance denned by eq 852 is commonly referred to as selfinductance since the linkages are produced by the inductor itself Like capacitances we may regard induc tance as a measure of how much magnetic energy is stored in an inductor The magnetic energy in joules stored in an inductor is expressed in circuit theory as Wm L 853 Figure 819 Magnetic field B produced by a circuit 88 INDUCTORS AND INDUCTANCES 337 or L 854 Thus the selfinductance of a circuit may be defined or calculated from energy considera tions If instead of having a single circuit we have two circuits carrying current I and I2 as shown in Figure 820 a magnetic interaction exists between the circuits Four component fluxes n f2 V 21 and f22 a r e produced The flux f 2 for example is the flux passing through circuit 1 due to current I2 in circuit 2 If B2 in the field due to I2 and S is the area of circuit 1 then 855 We define the mutual inductance Mn as the ratio of the flux linkage X12 Nfu on circuit 1 to current I2 that is x12 12 T I i 856 Similarly the mutual inductance M2 is defined as the flux linkages of circuit 2 per unit current that is M7I h 857a It can be shown by using energy concepts that if the medium surrounding the circuits is linear ie in the absence of ferromagnetic material M12 M2 857b The mutual inductance Mi2 or M2l is expressed in henrys and should not be confused with the magnetization vector M expressed in amperesmeter Figure 820 Magnetic interaction between two circuits 338 B Magnetic Forces Materials and Devices We define the selfinductance of circuits 1 and 2 respectively as L x w 858 and U x22 859 where V 2The total energy in the magnetic field is the sum of the energies due to Lh L2 andMI2 orM21 that is Wm W2 W12 2 L2I2 2 860 The positive sign is taken if currents and I2 flow such that the magnetic fields of the two circuits strengthen each other If the currents flow such that their magnetic fields oppose each other the negative sign is taken As mentioned earlier an inductor is a conductor arranged in a shape appropriate to store magnetic energy Typical examples of inductors are toroids solenoids coaxial trans mission lines and parallelwire transmission lines The inductance of each of these induc tors can be determined by following a procedure similar to that taken in determining the capacitance of a capacitor For a given inductor we find the selfinductance L by taking these steps 1 Choose a suitable coordinate system 2 Let the inductor carry current 3 Determine B from BiotSavarts law or from Amperes law if symmetry exists and calculate f from V B dS X NY 4 Finally find L from L The mutual inductance between two circuits may be calculated by taking a similar proce dure In an inductor such as a coaxial or a parallelwire transmission line the inductance produced by the flux internal to the conductor is called the internal inductance Lm while that produced by the flux external to it is called external inductance Lext The total induc tance L is Mn el Just as it was shown that for capacitors e RC a 861 635 it can be shown that LexlC 89 MAGNETIC ENERGY 339 862 Thus Lext may be calculated using eq 862 if C is known A collection of formulas for some fundamental circuit elements is presented in Table 83 All formulas can be derived by taking the steps outlined above3 89 MAGNETIC ENERGY Just as the potential energy in an electrostatic field was derived as eE dv WE D E dv 496 we would like to derive a similar expression for the energy in a magnetostatic field A simple approach is using the magnetic energy in the field of an inductor From eq 853 853 The energy is stored in the magnetic field B of the inductor We would like to express eq 853 in terms of B or H Consider a differential volume in a magnetic field as shown in Figure 821 Let the volume be covered with conducting sheets at the top and bottom surfaces with current A conducting sheets Figure 821 A differential volume in a magnetic field Additional formulas can be found in standard electrical handbooks or in H Knoepfel Pulsed High Magnetic Fields Amsterdam NorthHolland 1970 pp 312324 o TABLE 83 A Collection of Formulas for Inductance of Common Elements 1 Wire L L 877 2 3 Hollow L 2TI a Parallel L IT cylinder a wires f J l n 3 4 Coaxial conductor L In 7T a 5 Circular loop ii 1 In 2 2TT d 27rpo po d 6 Solenoid L 7 Torus of circular cross section L voN2Po Vp2 o a2 21 8 Sheet L to 2 89 MAGNETIC ENERGY 341 We assume that the whole region is filled with such differential volumes From eq 852 each volume has an inductance AL AT tiHAxAz A A where A H Ay Substituting eq 863 into eq 853 we have AWm AL A2 iiH2 Ax Ay Az or l 2 m j The magnetostatic energy density wm in Jm3 is defined as AWm 1 wm lim Av0 Av Hence 1 1 B2 2 2 2 Thus the energy in a magnetostatic field in a linear medium is Wm wmdv or which is similar to eq 496 for an electrostatic field 863 864 865 866 EXAMPLE 810 Calculate the selfinductance per unit length of an infinitely long solenoid Solution We recall from Example 74 that for an infinitely long solenoid the magnetic flux inside the solenoid per unit length is B nH uln 342 Magnetic Forces Materials and Devices where n N number of turns per unit length If S is the crosssectional area of the so lenoid the total flux through the cross section is Y BS Since this flux is only for a unit length of the solenoid the linkage per unit length is X nV im2IS and thus the inductance per unit length is L X V xn2S Hm PRACTICE EXERCISE 810 A very long solenoid with 2 X 2 cm cross section has an iron core pr 1000 and 4000 turnsmeter If it carries a current of 500 mA find a Its selfinductance per meter b The energy per meter stored in its field Answer a 8042 Hm b 1005 Jm EXAMPLE 811 Determine the selfinductance of a coaxial cable of inner radius a and outer radius b Solution The selfinductance of the inductor can be found in two different ways by taking the four steps given in Section 88 or by using eqs 854 and 866 Method 1 Consider the cross section of the cable as shown in Figure 822 We recall from eq 729 that by applying Amperes circuit law we obtained for region 1 0 p a and for region 2 a p b B 2waj a 89 MAGNETIC ENERGY 343 zaxis a b Figure 822 Cross section of the coaxial cable a for region 1 0 p a b for region 2 a p b for Example 811 We first find the internal inductance Lin by considering the flux linkages due to the inner conductor From Figure 822a the flux leaving a differential shell of thickness dp is dYi 5 dp dz dp dz 2ira The flux linkage is dxPl multiplied by the ratio of the area within the path enclosing the flux to the total area that is because is uniformly distributed over the cross section for dc excitation Thus the total flux linkages within the differential flux element are flip dp dz lira2 a2 For length of the cable X r dp dz 8TT 8TT The internal inductance per unit length given by m i Hm 8111 8112 is independent of the radius of the conductor or wire Thus eqs 8111 and 8112 are also applicable to finding the inductance of any infinitely long straight conductor of finite radius 344 Magnetic Forces Materials and Devices We now determine the external inductance Lext by considering the flux linkages between the inner and the outer conductor as in Figure 822b For a differential shell of thickness dp df2 B2 dp dz 2Kp dp dz In this case the total current is enclosed within the path enclosing the flux Hence X pa Jz0 al dp dz i b In 2irp 2K a 2 b L e x t I 2 a Thus or the inductance per length is Hm Method 2 It is easier to use eqs 854 and 866 to determine L that is 2W where or L 1 f B1 I H dv dv 2 Hence 2 B 2T2 2 p dp dj dz 4TT a r dz dct p 3 dp 0 J0 J0 8TT P 6P dt dz 89 MAGNETIC ENERGY 345 and as obtained previously PRACTICE EXERCISE 811 Calculate the selfinductance of the coaxial cable of Example 811 if the inner con ductor is made of an inhomogeneous material having a 2aJ p aot uj b 1 b Answer 1 In In 8TT IT I a 1 a EXAMPLE 812 Determine the inductance per unit length of a twowire transmission line with separation distance d Each wire has radius a as shown in Figure 637 Solution We use the two methods of the last example Method 1 We determine Lin just as we did in the last example Thus for region 0pawe obtain alt as in the last example For region a p d a the flux linkages between the wires are cda T rt j X2 The flux linkages produced by wire 1 are dp dz In 2TT a X alt alt d a 1 In 8TT 2TT a By symmetry the same amount of flux produced by current in wire 2 Hence the total linkages are X 2XiX2 If d 5s a the selfinductance per unit length is Hm 346 11 Magnetic Forces Materials and Devices Method 2 From the last example Lm 8 Now B2dv 1 ii22 Lext I2 2 l dz p dp dj dz da 2ir a Since the two wires are symmetrical L 2 Lin Lext as obtained previously PRACTICE EXERCISE 812 Two 10 copper wires 2588 mm in diameter are placed parallel in air with a sepa ration distance d between them If the inductance of each wire is 12 jiiHm calculate a Lin and Lext per meter for each wire b The separation distance d Answer a 005115 juHm b 4079 cm EXAMPLE 813 Two coaxial circular wires of radii a and bb a are separated by distance hh a b as shown in Figure 823 Find the mutual inductance between the wires Solution Let current flow in wire 1 At an arbitrary point P on wire 2 the magnetic vector poten tial due to wire 1 is given by eq 821a namely A Ah2 b2f2 Ifi b b 810 MAGNETIC CIRCUITS 347 Figure 823 Two coaxial circular wires for Example 813 Hence and u 212 d2 4r x7ra b 2h3 lbl PRACTICE EXERCISE 813 Find the mutual inductance of two coplanar concentric circular loops of radii 2 m and 3 m Answer 2632 H 810 MAGNETIC CIRCUITS The concept of magnetic circuits is based on solving some magnetic field problems using circuit approach Magnetic devices such as toroids transformers motors generators and relays may be considered as magnetic circuits The analysis of such circuits is made simple if an analogy between magnetic circuits and electric circuits is exploited Once this is done we can directly apply concepts in electric circuits to solve their analogous magnetic circuits The analogy between magnetic and electric circuits is summarized in Table 84 and portrayed in Figure 824 The reader is advised to pause and study Table 84 and Figure 824 First we notice from the table that two terms are new We define the magnetomotive force mmf 9 in ampereturns as 867 348 Magnetic Forces Materials and Devices TABLE 84 Analogy between Electric and Magnetic Circuits Electric Conductivity a Field intensity E Current J dS Current density J oE Electromotive force emf V Resistance R Conductance G R V Ohms law R or V E IR Kirchoffs laws E 0 J V 2 RI 0 Magnetic Permeability j Field intensity H Magnetic flux V B dS Flux density B fiH Magnetomotive force mmf 9 Reluctance 2ft Permeance 9 gft Ohms law gft or 9 Hi Nl Kirchhoffs laws The source of mmf in magnetic circuits is usually a coil carrying current as in Figure 824 We also define reluctance 2ft in ampereturnsweber as 868 where and S are respectively the mean length and the crosssectional area of the mag netic core The reciprocal of reluctance is permeance 3 The basic relationship for circuit elements is Ohms law V IR 869 Based on this Kirchhoffs current and voltage laws can be applied to nodes and loops of a given magnetic circuit just as in an electric circuit The rules of adding voltages and for Figure 824 Analogy between a an electric circuit and b a magnetic circuit a b 811 FORCE ON MAGNETIC MATERIALS 349 combining series and parallel resistances also hold for mmfs and reluctances Thus for n magnetic circuit elements in series and 3 3 For n magnetic circuit elements in parallel and 870 871 872 873 Some differences between electric and magnetic circuits should be pointed out Unlike an electric circuit where current flows magnetic flux does not flow Also conductivity a is independent of current density J in an electric circuit whereas permeability JX varies with flux density B in a magnetic circuit This is because ferromagnetic nonlinear materials are normally used in most practical magnetic devices These differences notwithstanding the magnetic circuit concept serves as an approximate analysis of practical magnetic devices 811 FORCE ON MAGNETIC MATERIALS It is of practical interest to determine the force that a magnetic field exerts on a piece of magnetic material in the field This is useful in electromechanical systems such as electro magnets relays rotating machines and magnetic levitation Consider for example an electromagnet made of iron of constant relative permeability as shown in Figure 825 The coil has N turns and carries a current If we ignore fringing the magnetic field in the air gap is the same as that in iron Bln B2n To find the force between the two pieces of iron we calculate the change in the total energy that would result were the two pieces of the magnetic circuit separated by a differential displacement d The work required to effect dlA Figure 825 An electromagnet 2F 350 Magnetic Forces Materials and Devices the displacement is equal to the change in stored energy in the air gap assuming constant current that is Fdl dWm 2 1 B 2 MoSdl 874 where S is the crosssectional area of the gap the factor 2 accounts for the two air gaps and the negative sign indicates that the force acts to reduce the air gap or that the force is at tractive Thus F 2 B2S 875 Note that the force is exerted on the lower piece and not on the currentcarrying upper piece giving rise to the field The tractive force across a single gap can be obtained from eq 875 as F B2S 876 Notice the similarity between eq 876 and that derived in Example 58 for electrostatic case Equation 876 can be used to calculate the forces in many types of devices includ ing relays rotating machines and magnetic levitation The tractive pressure in Nm2 in a magnetized surface is P F S B1 877 which is the same as the energy density wm in the air gap EXAMPLE 814 The toroidal core of Figure 826a has po 10 cm and a circular cross section with a 1 cm If the core is made of steel x 1000 io and has a coil with 200 turns calcu late the amount of current that will produce a flux of 05 mWb in the core a b Figure 826 a Toroidal core of Example 814 b its equivalent elec tric circuit analog 811 FORCE ON MAGNETIC MATERIALS 351 Solution This problem can be solved in two different ways using the magnetic field approach direct or using the electric circuit analog indirect Method 1 Since p0 is large compared with a from Example 76 UNI floHrNI Hence B BS HolirNI tea 2TTP0 or 8TT 210 X 10205 X 103 4TT X 1071000200l X 10 3979 A Method 2 The toroidal core in Figure 826a is analogous to the electric circuit of Figure 826b From the circuit and Table 84 or 3979 A as obtained previously PRACTICE EXERCISE 814 A conductor of radius a is bent into a circular loop of mean radius po see Figure 826a If p0 10 cm and 2a 1 cm calculate the internal inductance of the loop Answer 3142 nH EXAMPLE 815 In the magnetic circuit of Figure 827 calculate the current in the coil that will produce a magnetic flux density of 15 Wbm in the air gap assuming that fi 50xo and that all branches have the same crosssectional area of 10 cm2 352 Magnetic Forces Materials and Devices Figure 827 Magnetic circuit of Exam ple 815 10 cm Solution The magnetic circuit of Figure 827 is analogous to the electric circuit of Figure 828 In Figure 827 Sft 2ft2 2l3 and 3la are the reluctances in paths 143 123 35 and 16 and 56 air gap respectively Thus 3i3i7 3 X 10s 30 X 10z 4TT X 10 75010 X 20TT 9 X 10 3 4TT X 1075010 X 1 X 102 09 X 10s 20TT 5 X 108 Ja 4r x 107l10 X 104 20ir We combine 91 and 9l2 as resistors in parallel Hence 2ft2ft7 21 15 X 108 The total reluctance is 20x 74 X 108 91 a b Figure 828 Electric circuit analog of the magnetic circuit in Figure 827 811 FORCE ON MAGNETIC MATERIALS 353 The But mmf is rr BaS Hence i a 1 44 52ft N 16 T A NI YJ 15 X 10 X 104 X 74 X 108 400 X 20TT PRACTICE EXERCISE 815 The toroid of Figure 826a has a coil of 1000 turns wound on its core If p 0 10 cm and a 1 cm what current is required to establish a magnetic flux of 05 mWb a If the core is nonmagnetic b If the core has xr 500 Answer a 7958 A b 1592 A EXAMPLE 816 A Ushaped electromagnet shown in Figure 829 is designed to lift a 400kg mass which includes the mass of the keeper The iron yoke jxr 3000 has a cross section of 40 cm2 and mean length of 50 cm and the air gaps are each 01 mm long Neglecting the reluc tance of the keeper calculate the number of turns in the coil when the excitation current is 1 A Solution The tractive force across the two air gaps must balance the weight Hence F 2B2 aS mg NI Figure 829 Ushaped electromagnet for Example 816 iron yoke keeper weisht 354 Magnetic Forces Materials and Devices or But Since mgno 400 X 98 X 4TT X 10 i S 40 X Bn 111 Wbm2 cm ZJL ta 2 X 01 10 3 10b Air X 10 7 X 40 X 10 4 50 X 102 48TT 4TT X 10 X 3000 X 40 X 10 3 NI NI 6 5 11 5 X 10 48TT Po 11 L 7V 162 11 X 111 X 01 X 10 6 X 4TT X 107 X 1 PRACTICE EXERCISE 816 Find the force across the air gap of the magnetic circuit of Example 815 Answer 8952 N SUMMARY 1 The Lorentz force equation F gE u X B m du dt relates the force acting on a particle with charge Q in the presence of EM fields It ex presses the fundamental law relating EM to mechanics 2 Based on the Lorentz force law the force experienced by a current element Idl in a magnetic field B is dV Idl X B From this the magnetic field B is defined as the force per unit current element SUMMARY 355 3 The torque on a current loop with magnetic moment m in a uniform magnetic field B is T m X B ISan X B 4 A magnetic dipole is a bar magnet or a small filamental current loop it is so called due to the fact that its B field lines are similar to the E field lines of an electric dipole 5 When a material is subjected to a magnetic field it becomes magnetized The magne tization M is the magnetic dipole moment per unit volume of the material For linear material where m is the magnetic susceptibility of the material 6 In terms of their magnetic properties materials are either linear diamagnetic or para magnetic or nonlinear ferromagnetic For linear materials B xH xol M where x permeability and xr ilxo relative permeability of the material For nonlinear material B fiH H that is JX does not have a fixed value the relationship between B and H is usually represented by a magnetization curve 7 The boundary conditions that H or B must satisfy at the interface between two differ ent media are H H2 X anl2 K or Hu H2 if K 0 where anl2 is a unit vector directed from medium 1 to medium 2 8 Energy in a magnetostatic field is given by Wm B Udv For an inductor carrying current m V 2 Thus the inductance L can be found using L B Hdv 9 The inductance L of an inductor can also be determined from its basic definition the ratio of the magnetic flux linkage to the current through the inductor that is X NY I I 356 II Magnetic Forces Materials and Devices Thus by assuming current we determine B and P B dS and finally find L NYI 10 A magnetic circuit can be analyzed in the same way as an electric circuit We simply keep in mind the similarity between NI 4 H TO and V IR that is Thus we can apply Ohms and Kirchhoffs laws to magnetic circuits just as we apply them to electric circuits 11 The magnetic pressure or force per unit surface area on a piece of magnetic material is F 1 S 2 B2 2xo where B is the magnetic field at the surface of the material 81 Which of the following statements are not true about electric force Fe and magnetic force Fm on a charged particle a E and Fc are parallel to each other whereas B and Fm are perpendicular to each other b Both Fe and Fm depend on the velocity of the charged particle c Both Fe and m can perform work d Both Fc and Fm are produced when a charged particle moves at a constant velocity e Fm is generally small in magnitude compared to Fe f Fe is an accelerating force whereas Fm is a purely deflecting force 82 Two thin parallel wires carry currents along the same direction The force experienced by one due to the other is a Parallel to the lines b Perpendicular to the lines and attractive c Perpendicular to the lines and repulsive d Zero 83 The force on differential length d at point P in the conducting circular loop in Figure 830 is a Outward along OP b Inward along OP o o o o B o REVIEW QUESTIONS i 357 Q Figure 830 For Review Questions 83 and 84 O O O c In the direction of the magnetic field d Tangential to the loop at P 84 The resultant force on the circular loop in Figure 830 has the magnitude of a 2wpJB b irpllB c 2PJB d Zero 85 What is the unit of magnetic charge a Amperemeter square b Coulomb c Ampere d Amperemeter 86 Which of these materials requires the least value of magnetic field strength to magne tize it a Nickel b Silver c Tungsten d Sodium chloride 87 Identify the statement that is not true of ferromagnetic materials a They have a large m b They have a fixed value of fir c Energy loss is proportional to the area of the hysteresis loop d They lose their nonlinearity property above the curie temperature 358 B Magnetic Forces Materials and Devices 88 Which of these formulas is wrong a Bu B2n b B2 Vfi2 B c Hu d a2i X Ht H2 K where an2i is a unit vector normal to the interface and di rected from region 2 to region 1 89 Each of the following pairs consists of an electric circuit term and the corresponding mag netic circuit term Which pairs are not corresponding a V and S5 b GandSP c e and n d IR and tf9l e 2 0 and 2 f 0 810 A multilayer coil of 2000 turns of fine wire is 20 mm long and has a thickness 5 mm of winding If the coil carries a current of 5 mA the mmf generated is a lOAt b 500 At c 2000 At d None of the above Answers 81 bc 82b 83a 84d 85d 86a 87b 88c 89cd 810a PROBLEMS 81 An electron with velocity u 3ar 12aY 4az X 105ms experiences no net J force at a point in a magnetic field B Wax 20av 30a mWbm2 Find E at that point 82 A charged particle of mass 1 kg and charge 2 C starts at the origin with velocity 10az ms in a magnetic field B 1 a Wbm2 Find the location and the kinetic energy of the parti cle at t 2 s 83 A particle with mass 1 kg and charge 2 C starts from rest at point 2 3 4 in a region where E 4a v Vm and B 5ar Wbm2 Calculate a The location of the particle at t I s b Its velocity and KE at that location 84 A 2mC charge starts at point 0 1 2 with a velocity of 5ax ms in a magnetic field B 6av Wbm Determine the position and velocity of the particle after 10 s assuming that the mass of the charge is 1 gram Describe the motion of the charge PROBtEMS i 359 Figure 831 For Problem 85 85 By injecting an electron beam normally to the plane edge of a uniform field Boaz elec trons can be dispersed according to their velocity as in Figure 831 a Show that the electrons would be ejected out of the field in paths parallel to the input beam as shown b Derive an expression for the exit distance d above entry point 86 Given that B 6xa 9yay 3zaz Wbm2 find the total force experienced by the rec tangular loop on z 0 plane shown in Figure 832 87 A current element of length 2 cm is located at the origin in free space and carries current 12 mA along ax A filamentary current of 15az A is located along x 3 y 4 Find the force on the current filament 88 Three infinite lines L b L2 and L3 defined by x 0 y 0 x 0 y 4 x 3 y 4 respectively carry filamentary currents 100 A 200 A and 300 A along az Find the force per unit length on a L2 due to L b L due to L2 c L3 due to Lj d L3 due to Lx and L2 State whether each force is repulsive or attractive Figure 832 For Problem 86 5A 1 2 3 360 Magnetic es Materials and Devices Figure 833 For Problem 89 89 A conductor 2 m long carrying 3A is placed parallel to the zaxis at distance p0 10 cm as shown in Figure 833 If the field in the region is cos 43 ap Wbm2 how much work is required to rotate the conductor one revolution about the zaxis 810 A conducting triangular loop carrying a current of 2 A is located close to an infinitely long straight conductor with a current of 5 A as shown in Figure 834 Calculate a the force on side 1 of the triangular loop and b the total force on the loop 811 A threephase transmission line consists of three conductors that are supported at points A B and C to form an equilateral triangle as shown in Figure 835 At one instant con ductors A and B both carry a current of 75 A while conductor C carries a return current of 150 A Find the force per meter on conductor C at that instant 812 An infinitely long tube of inner radius a and outer radius b is made of a conducting magnetic material The tube carries a total current and is placed along the zaxis If it is exposed to a constant magnetic field Boap determine the force per unit length acting on the tube 5 A Figure 834 For Problem 810 2m 4 m 2m B 7 5 A PROBLEMS Figure 835 For Problem 811 361 813 An infinitely long conductor is buried but insulated from an iron mass fi 2000o as shown in Figure 836 Using image theory estimate the magnetic flux density at point P 814 A galvanometer has a rectangular coil of side 10 by 30 mm pivoted about the center of the shorter side It is mounted in radial magnetic field so that a constant magnetic field of 04 Wbm always acts across the plane of the coil If the coil has 1000 turns and carries current 2 mA find the torque exerted on it 05a mWbm2at10 0 0 FindB 815 A small magnet placed at the origin produces B at a 030 b 3 4 0 c 1 1 D 816 A block of iron 5000uo is placed in a uniform magnetic field with 15 Wbm If iron consists of 85 X 1028 atomsm3 calculate a the magnetization M b the average magnetic current y 30 mm o p 0 A 20 mm iron 20 mm x Figure 836 For Problem 813 362 Magnetic Forces Materials and Devices 817 In a certain material for which n 65x0 H 10ax 25av 40az Am find a The magnetic susceptibility xm of the material b The magnetic flux density B c The magnetization M d The magnetic energy density 818 In a ferromagnetic material 45to B 4a mWbm2 calculate a Xm b H c M d Jb 819 The magnetic field intensity is H 1200 Am in a material when B 2 Wbm2 When H is reduced to 400 Am B 14 Wbm2 Calculate the change in the magnetization M 820 An infinitely long cylindrical conductor of radius a and permeability xoxr is placed along the zaxis If the conductor carries a uniformly distributed current along a7 find M and Jb for 0 p a 821 If M yx xay in a cube of size a find Jb Assume ko is a constant 822 a For the boundary between two magnetic media such as is shown in Figure 816 show that the boundary conditions on the magnetization vector are Mu M2t K and m l n Xml Xml Xm Xml b If the boundary is not current free show that instead of eq 849 we obtain tan 0 HI tan 62 L B sin d7 823 If Mi 2fio for region 1 0 j it and p2 5o for region 2 IT tj 2ir and B2 10ap 15a0 20az mWbm2 Calculate a B b the energy densities in the two media 824 The interface 2x y 8 between two media carries no current If medium 1 2x y 8 is nonmagnetic with H 4aA 3av az Am Find a the mag netic energy density in medium 1 b M2 and B2 in medium 2 2x 8 with H 10io c the angles H and H2 make with the normal to the interface 825 The interface 4x 5z 0 between two magnetic media carries current 35av Am If H 25ax 30av 45a Am in region 4x 5z 0 where firl 5 calculate H2 in region 4x 5z 0 where fir2 10 T PROBLEMS 363 826 The plane z 0 separates air z 0 y O from iron z 0 2000 Given that H 10ax 15av 3a Am in air find B in iron and the angle it makes with the interface 827 Region 0 s 2 m is filled with an infinite slab of magnetic material fi 25juo If the surfaces of the slab at z 0 and z 2 respectively carry surface currents 30a Am and 40av Am as in Figure 837 calculate H and B for a z 0 b 0 z 2 c z 2 828 In a certain region for which x m 19 H 5x2yzaA 10xy2zav 15xyzV Am How much energy is stored in 0 x 1 0 v 2 1 z 2 1 829 The magnetization curve for an iron alloy is approximately given by B H H2n Wbm2 Find a ir when H 210 Am b the energy stored per unit volume in the alloy as H increases from 0 to 210 Am 830 a If the cross section of the toroid of Figure 715 is a square of side a show that the self inductance of the toroid is L In 2po a b If the toroid has a circular cross section as in Figure 715 show that L where p o a 831 When two parallel identical wires are separated by 3 m the inductance per unit length is 25 juHm Calculate the diameter of each wire Mo Figure 837 For Problem 827 40 ax Am z 0 30 ax Am 364 Magnetic Forces Materials and Devices 832 A solenoid with length 10 cm and radius 1 cm has 450 turns Calculate its inductance 833 The core of a toroid is 12 cm and is made of material with xr 200 If the mean radius of the toroid is 50 cm calculate the number of turns needed to obtain an inductance of 25 H 834 Show that the mutual inductance between the rectangular loop and the infinite line current of Figure 84 is r a p0 Calculate M12 when a b po 1 m 835 Prove that the mutual inductance between the closed wound coaxial solenoids of length and 2 i X turns Ny and N2 and radii r and r2 with rx r2 is 836 A cobalt ring jxr 600 has a mean radius of 30 cm If a coil wound on the ring carries 12 A calculate the number of turns required to establish an average magnetic flux density of 15 Wbm in the ring 837 Refer to Figure 827 If the current in the coil is 05 A find the mmf and the magnetic field intensity in the air gap Assume that i 500no and that all branches have the same crosssectional area of 10 cm2 838 The magnetic circuit of Figure 838 has current 10 A in the coil of 2000 turns Assume that all branches have the same cross section of 2 cm2 and that the material of the core is iron with nr 1500 Calculate R 9 and V for a The core b The air gap 12 cm 06 cm Figure 838 For Problem 838 02 A 1 500 turns PROBLEMS Figure 839 For Problem 839 365 L 42 cm 839 Consider the magnetic circuit in Figure 839 Assuming that the core 1000xohasa uniform cross section of 4 cm2 determine the flux density in the air gap 840 An electromagnetic relay is modeled as shown in Figure 840 What force is on the arma ture moving part of the relay if the flux in the air gap is 2 mWb The area of the gap is 03 cm2 and its length 15 mm 841 A toroid with air gap shown in Figure 841 has a square cross section A long conductor carrying current 72 is inserted in the air gap If 7 200 mA N 750 p0 10 cm a 5 mm and ia 1 mm calculate a The force across the gap when 72 0 and the relative permeability of the toroid is 300 b The force on the conductor when 72 2 m A and the permeability of the toroid is in finite Neglect fringing in the gap in both cases 842 A section of an electromagnet with a plate below it carrying a load is shown in Figure 842 The electromagnet has a contact area of 200 cm2 per pole with the middle pole having a winding of 1000 turns with 7 3 A Calculate the maximum mass that can be lifted Assume that the reluctance of the electromagnet and the plate is negligible 843 Figure 843 shows the cross section of an electromechanical system in which the plunger moves freely between two nonmagnetic sleeves Assuming that all legs have the same crosssectional area S show that F 2 2 2 a a 2xx Figure 840 For Problem 840 366 B Magnetic Forces Materials and Devices Figure 841 For Problem 841 Figure 842 For Problem 842 1 mm Figure 843 For Problem 843 u J nmagnetic sleeve PART 4 WAVES AND APPLICATIONS Chapter y MAXWELLS EQUATIONS Do you want to be a hero Dont be the kind of person who watches others do great things or doesnt know whats happening Go out and make things happen The people who get things done have a burning desire to make things happen get ahead serve more people become the best they can possibly be and help improve the world around them GLENN VAN EKEREN 91 INTRODUCTION In Part II Chapters 4 to 6 of this text we mainly concentrated our efforts on electrostatic fields denoted by Ex y z Part III Chapters 7 and 8 was devoted to magnetostatic fields represented by HJC y z We have therefore restricted our discussions to static or time invariant EM fields Henceforth we shall examine situations where electric and magnetic fields are dynamic or time varying It should be mentioned first that in static EM fields electric and magnetic fields are independent of each other whereas in dynamic EM fields the two fields are interdependent In other words a timevarying electric field necessarily involves a corresponding timevarying magnetic field Second timevarying EM fields represented by Ex y z t and Hx y z t are of more practical value than static EM fields However familiarity with static fields provides a good background for understand ing dynamic fields Third recall that electrostatic fields are usually produced by static elec tric charges whereas magnetostatic fields are due to motion of electric charges with uniform velocity direct current or static magnetic charges magnetic poles timevarying fields or waves are usually due to accelerated charges or timevarying currents such as shown in Figure 91 Any pulsating current will produce radiation timevarying fields It is worth noting that pulsating current of the type shown in Figure 91b is the cause of ra diated emission in digital logic boards In summary charges electrostatic fields steady currenis magnclosiatic fields timevarying currenis electromagnetic fields or wavesj Our aim in this chapter is to lay a firm foundation for our subsequent studies This will involve introducing two major concepts 1 electromotive force based on Faradays ex periments and 2 displacement current which resulted from Maxwells hypothesis As a result of these concepts Maxwells equations as presented in Section 76 and the boundary 369 370 Maxwells Equations a b 0 Figure 91 Various types of timevarying current a sinusoidal b rectangular c triangular conditions for static EM fields will be modified to account for the time variation of the fields It should be stressed that Maxwells equations summarize the laws of electromag netism and shall be the basis of our discussions in the remaining part of the text For this reason Section 95 should be regarded as the heart of this text 92 FARADAYS LAW After Oersteds experimental discovery upon which BiotSavart and Ampere based their laws that a steady current produces a magnetic field it seemed logical to find out if mag netism would produce electricity In 1831 about 11 years after Oersteds discovery Michael Faraday in London and Joseph Henry in New York discovered that a timevarying magnetic field would produce an electric current According to Faradays experiments a static magnetic field produces no current flow but a timevarying field produces an induced voltage called electromotive force or simply emf in a closed circuit which causes a flow of current Faraday discovered that the induced emf iM in volts in any closed circuit is equal to the time rale of change of the magnetic flux linkage by the circuit This is called Faradays law and it can be expressed as dt dt emf 91 where N is the number of turns in the circuit and V is the flux through each turn The neg ative sign shows that the induced voltage acts in such a way as to oppose the flux produc For details on the experiments of Michael Faraday 17911867 and Joseph Henry 17971878 see W F Magie A Source Book in Physics Cambridge MA Harvard Univ Press 1963 pp 472519 92 FARADAYS LAW 371 battery Figure 92 A circuit showing emfproducing field and electrostatic field E ing it This is known as Lenzs law2 and it emphasizes the fact that the direction of current flow in the circuit is such that the induced magnetic field produced by the induced current will oppose the original magnetic field Recall that we described an electric field as one in which electric charges experience force The electric fields considered so far are caused by electric charges in such fields the flux lines begin and end on the charges However there are other kinds of electric fields not directly caused by electric charges These are emfproduced fields Sources of emf include electric generators batteries thermocouples fuel cells and photovoltaic cells which all convert nonelectrical energy into electrical energy Consider the electric circuit of Figure 92 where the battery is a source of emf The electrochemical action of the battery results in an emfproduced field Ey Due to the accu mulation of charge at the battery terminals an electrostatic field Ee VV also exists The total electric field at any point is E Ey Ee 92 Note that Ey is zero outside the battery Ey and Ee have opposite directions in the battery and the direction of Ee inside the battery is opposite to that outside it If we integrate eq 92 over the closed circuit E d f Ey d 0 Efdl through battery 93a where Ee d 0 because Ee is conservative The emf of the battery is the line integral of the emfproduced field that is d 93b since Eyand Ee are equal but opposite within the battery see Figure 92 It may also be re garded as the potential difference VP VN between the batterys opencircuit terminals It is important to note that 1 An electrostatic field Ee cannot maintain a steady current in a closed circuit since LEedl 0 2 An emfproduced field Eyis nonconservative 3 Except in electrostatics voltage and potential difference are usually not equivalent 2After Heinrich Friedrich Emil Lenz 18041865 a Russian professor of physics 372 B Maxwells Equations 93 TRANSFORMER AND MOTIONAL EMFs Having considered the connection between emf and electric field we may examine how Faradays law links electric and magnetic fields For a circuit with a single turn N 1 eq 91 becomes V 94 In terms of E and B eq 94 can be written as yemf f E d B h dt 4 95 where P has been replaced by Js B dS and S is the surface area of the circuit bounded by the closed path L It is clear from eq 95 that in a timevarying situation both electric and magnetic fields are present and are interrelated Note that d and JS in eq 95 are in ac cordance with the righthand rule as well as Stokess theorem This should be observed in Figure 93 The variation of flux with time as in eq 91 or eq 95 may be caused in three ways 1 By having a stationary loop in a timevarying B field 2 By having a timevarying loop area in a static B field 3 By having a timevarying loop area in a timevarying B field Each of these will be considered separately A Stationary Loop in TimeVarying B Fit transformer emf This is the case portrayed in Figure 93 where a stationary conducting loop is in a time varying magnetic B field Equation 95 becomes 96 Increasing Bt Figure 93 Induced emf due to a stationary loop in a time varying B field ed B 93 TRANSFORMER AND MOTIONAL EMFS 373 This emf induced by the timevarying current producing the timevarying B field in a sta tionary loop is often referred to as transformer emf in power analysis since it is due to transformer action By applying Stokess theorem to the middle term in eq 96 we obtain V X E dS I dS For the two integrals to be equal their integrands must be equal that is 97 dt 98 This is one of the Maxwells equations for timevarying fields It shows that the time varying E field is not conservative V X E 0 This does not imply that the principles of energy conservation are violated The work done in taking a charge about a closed path in a timevarying electric field for example is due to the energy from the timevarying mag netic field Observe that Figure 93 obeys Lenzs law the induced current flows such as to produce a magnetic field that opposes Bf B Moving Loop in Static B Field Motional emf When a conducting loop is moving in a static B field an emf is induced in the loop We recall from eq 82 that the force on a charge moving with uniform velocity u in a mag netic field B is Fm Qu X B We define the motional electric field Em as 82 Em u X B 99 If we consider a conducting loop moving with uniform velocity u as consisting of a large number of free electrons the emf induced in the loop is 910 This type of emf is called motional emf or fluxcutting emf because it is due to motional action It is the kind of emf found in electrical machines such as motors generators and al ternators Figure 94 illustrates a twopole dc machine with one armature coil and a two bar commutator Although the analysis of the dc machine is beyond the scope of this text we can see that voltage is generated as the coil rotates within the magnetic field Another example of motional emf is illustrated in Figure 95 where a rod is moving between a pair 374 11 Maxwells Equations Figure 94 A directcurrent machine of rails In this example B and u are perpendicular so eq 99 in conjunction with eq 82 becomes or and eq 910 becomes Fm U X B Fm KB Vem By applying Stokess theorem to eq 910 V X E J d S V X u X B dS s s or V X Em V X u X B 911 912 913 914 Notice that unlike eq 96 there is no need for a negative sign in eq 910 because Lenzs law is already accounted for Bin Figure 95 Induced emf due to a moving loop in a static B field 93 TRANSFORMER AND MOTIONAL EMFS 375 To apply eq 910 is not always easy some care must be exercised The following points should be noted 1 The integral in eq 910 is zero along the portion of the loop where u 0 Thus d is taken along the portion of the loop that is cutting the field along the rod in Figure 95 where u has nonzero value 2 The direction of the induced current is the same as that of Em or u X B The limits of the integral in eq 910 are selected in the opposite direction to the induced current thereby satisfying Lenzs law In eq 913 for example the integration over L is along av whereas induced current flows in the rod along ay C Moving Loop in TimeVarying Field This is the general case in which a moving conducting loop is in a timevarying magnetic field Both transformer emf and motional emf are present Combining eqs 96 and 910 gives the total emf as 915 f V f 9J E d f flB ft u X B d or from eqs 98 and 914 V X E V X u X B dt 916 Note that eq 915 is equivalent to eq 94 so Vemf can be found using either eq 915 or 94 In fact eq 94 can always be applied in place of eqs 96 910 and 915 EXAMPLE 91 A conducting bar can slide freely over two conducting rails as shown in Figure 96 Calcu late the induced voltage in the bar a If the bar is stationed at y 8 cm and B 4 cos 106f az mWbm2 b If the bar slides at a velocity u 20aj ms and B 4az mWbm2 c If the bar slides at a velocity u 20ay ms and B 4 cos 106r y az mWbm2 Figure 96 For Example 91 0 6 cm B 0 0 376 Maxwells Equations Solution a In this case we have transformer emf given by dB dt dS 008 006 y0 sin Wtdxdy 4103008006 sin l06t 192 sin 106V The polarity of the induced voltage according to Lenzs law is such that point P on the bar is at lower potential than Q when B is increasing b This is the case of motional emf Vemf x B d uy X Baz dxax uB 204103006 48 mV c Both transformer emf and motional emf are present in this case This problem can be solved in two ways Method 1 Using eq 915 Kmf I dS U X B d r006 ry 911 x0 0 0 20ay X 410 3 cos106f yaj dxax 006 240 cos106f 80103006 cos106r y 240 00810 y 240 cos 106f 4810j cos106f y 240 cos106f y 240 cos 106 912 because the motional emf is negligible compared with the transformer emf Using trigono metric identity A B A B cos A cos B 2 sin sin Veirf 480 sin MO 6 sin V 913 93 TRANSFORMER AND MOTIONAL EMFS Method 2 Alternatively we can apply eq 94 namely Vemt where dt 377 914 BdS 006 4 cos106r y dx dy y0 JJt 4006 sin106f y 024 sin106r y 024 sin 10f mWb But Hence u y ut 20 V 024 sin106r 200 024 sin 106f mWb yemf 024106 20 cos106r 20f 024106 cos 106f mV dt 240 cos106f y 240 cos 106f V 915 which is the same result in 912 Notice that in eq 911 the dependence of y on time is taken care of in u X B d and we should not be bothered by it in dBdt Why Because the loop is assumed stationary when computing the transformer emf This is a subtle point one must keep in mind in applying eq 911 For the same reason the second method is always easier PRACTICE EXERCISE 91 Consider the loop of Figure 95 If B 05az Wbm2 R 20 0 10 cm and the rod is moving with a constant velocity of 8ax ms find a The induced emf in the rod b The current through the resistor c The motional force on the rod d The power dissipated by the resistor Answer a 04 V b 20 mA c a x mN d 8 mW 378 Maxwells Equations EXAMPLE 92 The loop shown in Figure 97 is inside a uniform magnetic field B 50 ax mWbm2 If side DC of the loop cuts the flux lines at the frequency of 50 Hz and the loop lies in the jzplane at time t 0 find a The induced emf at t 1 ms b The induced current at t 3 ms Solution a Since the B field is time invariant the induced emf is motional that is yemf u x B d where d dDC dzaz u dt dt p AD 4 cm a 2TT IOOTT As u and d are in cylindrical coordinates we transform B into cylindrical coordinates using eq 29 B BQax Bo cos j ap sin t a0 where Bo 005 Hence u X B 0 pco 0 Bo cos 4 Bo sin 4 0 puBo cos az Figure 97 For Example 92 polarity is for increasing emf 93 TRANSFORMER AND MOTIONAL EMFS 379 and uXBdl puBo cos f dz 004100TT005 COS t dz 02ir cos 0 dz r 003 V e m f 02TT COS 4 dz 6TT COS f m V 40 To determine j recall that co dt 0 cof C o dt where Co is an integration constant At t 0 0 TT2 because the loop is in the yzplane at that time Co TT2 Hence CO TT and mf 6TT cosf cor 6TT sinlOOirf mV At f 1 ms yemf 6TT sinOlTr 5825 mV b The current induced is Vem R 607rsin100xrmA At t 3 ms i 60TT sin037r mA 01525 A PRACTICE EXERCISE 92 Rework Example 92 with everything the same except that the B field is changed to a B 50av mWbm2that is the magnetic field is oriented along the ydirection b B 002ir ax Wbm2that is the magnetic field is time varying Answer a 1793 mV 01108 A b 205 jtV 4192 mA EXAMPLE 93 The magnetic circuit of Figure 98 has a uniform cross section of 10 3 m2 If the circuit is energized by a current ixi 3 sin IOOTT A in the coil of N 200 turns find the emf induced in the coil of N2 100 turns Assume that JX 500 xo 380 11 Maxwells Equations A ho r l lpo 10 cm LJN2 Dl Solution The flux in the circuit is w Figure 98 Magnetic circuit of Example 93 o 2irpo According to Faradays law the emf induced in the second coil is V2 N2 r dt 2Kp0 dt 100 200 500 4TT X 107 103 300TT COS IOOTT 2x 10 X 102 6TTCOS 100irV PRACTICE EXERCISE 93 A magnetic core of uniform cross section 4 cm2 is connected to a 120V 60Hz generator as shown in Figure 99 Calculate the induced emf V2 in the second ary coil Aaswer 72 V Vtfc T i N2 300 V2 Figure 99 For Practice Exercise 93 94 DISPLACEMENT CURRENT 381 94 DISPLACEMENT CURRENT In the previous section we have essentially reconsidered Maxwells curl equation for elec trostatic fields and modified it for timevarying situations to satisfy Faradays law We shall now reconsider Maxwells curl equation for magnetic fields Amperes circuit law for timevarying conditions For static EM fields we recall that V x H J 917 But the divergence of the curl of any vector field is identically zero see Example 310 Hence V V X H 0 V J The continuity of current in eq 543 however requires that 918 919 Thus eqs 918 and 919 are obviously incompatible for timevarying conditions We must modify eq 917 to agree with eq 919 To do this we add a term to eq 917 so that it becomes V X H J Jd 920 where id is to be determined and defined Again the divergence of the curl of any vector is zero Hence In order for eq 921 to agree with eq 919 921 922a or h 320 results V X H dD dt in 1 dX dt 922b 923 This is Maxwells equation based on Amperes circuit law for a timevarying field The term Jd dDdt is known as displacement current density and J is the conduction current 382 Maxwells Equations a Figure 910 Two surfaces of integration showing the need for Jd in Amperes circuit law density J aE3 The insertion of Jd into eq 917 was one of the major contributions of Maxwell Without the term Jd electromagnetic wave propagation radio or TV waves for example would be impossible At low frequencies Jd is usually neglected compared with J However at radio frequencies the two terms are comparable At the time of Maxwell highfrequency sources were not available and eq 923 could not be verified experimen tally It was years later that Hertz succeeded in generating and detecting radio waves thereby verifying eq 923 This is one of the rare situations where mathematical argu ment paved the way for experimental investigation Based on the displacement current density we define the displacement current as ld jddS dt dS 924 We must bear in mind that displacement current is a result of timevarying electric field A typical example of such current is the current through a capacitor when an alternating voltage source is applied to its plates This example shown in Figure 910 serves to illus trate the need for the displacement current Applying an unmodified form of Amperes circuit law to a closed path L shown in Figure 910a gives 925 H d J dS enc where is the current through the conductor and Sx is the flat surface bounded by L If we use the balloonshaped surface S2 that passes between the capacitor plates as in Figure 910b 926 H d J dS Ieac 0 4 because no conduction current J 0 flows through S2 This is contradictory in view of the fact that the same closed path L is used To resolve the conflict we need to include the Recall that we also have J pvii as the convection current density 94 DISPLACEMENT CURRENT 383 displacement current in Amperes circuit law The total current density is J Jd In eq 925 id 0 so that the equation remains valid In eq 926 J 0 so that H d I id dS I D dS I dt dt 927 So we obtain the same current for either surface though it is conduction current in S and displacement current in S2 EXAMPLE 94 A parallelplate capacitor with plate area of 5 cm2 and plate separation of 3 mm has a voltage 50 sin 103r V applied to its plates Calculate the displacement current assuming e 2eo Solution D eE s d dD dt e dv d dt Hence which is the same as the conduction current given by 2 s dt dt 5 dt 36TT 3 X 103 dt d dt 103 X 50 cos 10t dV dt 1474 cos 103nA PRACTICE EXERCISE 94 In free space E 20 cos at 50xj ay Vm Calculate a h b H c w Answer a 20aso sin wt 50J ay Am2 b 04 wso cosuit 50x az Am c15 X 1010rads 384 Maxwells Equations 95 MAXWELLS EQUATIONS IN FINAL FORMS James Clerk Maxwell 18311879 is regarded as the founder of electromagnetic theory in its present form Maxwells celebrated work led to the discovery of electromagnetic waves4 Through his theoretical efforts over about 5 years when he was between 35 and 40 Maxwell published the first unified theory of electricity and magnetism The theory comprised all previously known results both experimental and theoretical on electricity and magnetism It further introduced displacement current and predicted the existence of electromagnetic waves Maxwells equations were not fully accepted by many scientists until they were later confirmed by Heinrich Rudolf Hertz 18571894 a German physics professor Hertz was successful in generating and detecting radio waves The laws of electromagnetism that Maxwell put together in the form of four equations were presented in Table 72 in Section 76 for static conditions The more generalized forms of these equations are those for timevarying conditions shown in Table 91 We notice from the table that the divergence equations remain the same while the curl equa tions have been modified The integral form of Maxwells equations depicts the underlying physical laws whereas the differential form is used more frequently in solving problems For a field to be qualified as an electromagnetic field it must satisfy all four Maxwells equations The importance of Maxwells equations cannot be overemphasized because they summarize all known laws of electromagnetism We shall often refer to them in the remaining part of this text Since this section is meant to be a compendium of our discussion in this text it is worthwhile to mention other equations that go hand in hand with Maxwells equations The Lorentz force equation u X B 928 TABLE 91 Generalized Forms of Maxwells Equations Differential V D pv V B O V X E V X H J Form 3B dt 3D at 1 D s 9 B I P H L Integral dS j p rfS 0 dt Form rfv BrfS Remarks Gausss law Nonexistence of isolated magnetic charge Faradays law dS Amperes circuit law This is also referred to as Gausss law for magnetic fields 4The work of James Clerk Maxwell 18311879 a Scottish physicist can be found in his book A Treatise on Electricity and Magnetism New York Dover vols 1 and 2 1954 95 MAXWELLS EQUATIONS IN FINAL FORMS M 385 is associated with Maxwells equations Also the equation of continuity V J 929 dt is implicit in Maxwells equations The concepts of linearity isotropy and homogeneity of a material medium still apply for timevarying fields in a linear homogeneous and isotropic medium characterized by a e and fi the constitutive relations D eE eoE P B ixH noH M J CTE pvu hold for timevarying fields Consequently the boundary conditions Eu E2t or Ej E2 X anl2 0 u H2t K or H H2 X anl2 K Din D2n p or D D2 anl2 p Bm B2n 0 or B2 B aBl2 0 930a 930b 930c 931a 931b 931c 931d remain valid for timevarying fields However for a perfect conductor a in a time varying field and hence E 0 H 0 J 0 BB 0 E 0 932 933 For a perfect dielectric a 0 eq 931 holds except that K 0 Though eqs 928 to 933 are not Maxwells equations they are associated with them To complete this summary section we present a structure linking the various poten tials and vector fields of the electric and magnetic fields in Figure 911 This electromag netic flow diagram helps with the visualization of the basic relationships between field quantities It also shows that it is usually possible to find alternative formulations for a given problem in a relatively simple manner It should be noted that in Figures 910b and c we introduce pm as the free magnetic density similar to pv which is of course zero Ae as the magnetic current density analogous to J Using terms from stress analysis the principal relationships are typified as a compatibility equations V B pm 0 934 386 Maxwells Eolations vv a VxVx 0 Vx Vx V b c Figure 911 Electromagnetic flow diagram showing the relationship between the potentials and vector fields a electrostatic system b magnetostatic system c electromagnetic system Adapted with permission from IEE Publishing Dept and b constitutive equations and c equilibrium equations and B uH D eE V D Pv dt 96 TIMEVARYING POTENTIALS 387 96 TIMEVARYING POTENTIALS For static EM fields we obtained the electric scalar potential as pvdv and the magnetic vector potential as V A AireR fiJ dv 4wR 940 941 We would like to examine what happens to these potentials when the fields are time varying Recall that A was defined from the fact that V B 0 which still holds for time varying fields Hence the relation B V X A 942 holds for timevarying situations Combining Faradays law in eq 98 with eq 942 gives V X E V X A 943a or V X E dt 943b Since the curl of the gradient of a scalar field is identically zero see Practice Exercise 310 the solution to eq 943b is dt or dt 944 945 From eqs 942 and 945 we can determine the vector fields B and E provided that the potentials A and V are known However we still need to find some expressions for A and V similar to those in eqs 940 and 941 that are suitable for timevarying fields From Table 91 or eq 938 we know that V D pv is valid for timevarying condi tions By taking the divergence of eq 945 and making use of eqs 937 and 938 we obtain VE V 2 V V A e dt 388 Maxwells Equations or VV V A dt e Taking the curl of eq 942 and incorporating eqs 923 and 945 results in VX V X A uj en VV dt V dt dV 946 where D sE and B fiH have been assumed By applying the vector identity V X V X A VV A V2A to eq 947 V2A VV A f dt d2A rdt2 947 948 949 A vector field is uniquely defined when its curl and divergence are specified The curl of A has been specified by eq 942 for reasons that will be obvious shortly we may choose the divergence of A as V A J dV dt 950 This choice relates A and V and it is called the Lorentz condition for potentials We had this in mind when we chose V A 0 for magnetostatic fields in eq 759 By imposing the Lorentz condition of eq 950 eqs 946 and 949 respectively become 951 and 2 V2A JUS d2V dt2 a2 A dt2 Pv e y xj 952 which are wave equations to be discussed in the next chapter The reason for choosing the Lorentz condition becomes obvious as we examine eqs 951 and 952 It uncouples eqs 946 and 949 and also produces a symmetry between eqs 951 and 952 It can be shown that the Lorentz condition can be obtained from the continuity equation there fore our choice of eq 950 is not arbitrary Notice that eqs 64 and 760 are special static cases of eqs 951 and 952 respectively In other words potentials V and A satisfy Poissons equations for timevarying conditions Just as eqs 940 and 941 are 97 TIMEHARMONIC FIELDS 389 the solutions or the integral forms of eqs 64 and 760 it can be shown that the solu tions5 to eqs 951 and 952 are V P dv AKSR and A AKR 953 954 The term pv or J means that the time t in pvx y z t or Jx y z t is replaced by the retarded time t given by 955 where R r r is the distance between the source point r and the observation point r and 1 u 956 xe is the velocity of wave propagation In free space u c 3 X 1 0 ms is the speed of light in a vacuum Potentials V and A in eqs 953 and 954 are respectively called the retarded electric scalar potential and the retarded magnetic vector potential Given pv and J V and A can be determined using eqs 953 and 954 from V and A E and B can be determined using eqs 945 and 942 respectively 97 TIMEHARMONIC FIELDS So far our time dependence of EM fields has been arbitrary To be specific we shall assume that the fields are time harmonic A timeharmonic field is one thai varies periodically or sinusoidally wiih time Not only is sinusoidal analysis of practical value it can be extended to most waveforms by Fourier transform techniques Sinusoids are easily expressed in phasors which are more convenient to work with Before applying phasors to EM fields it is worthwhile to have a brief review of the concept of phasor Aphasor z is a complex number that can be written as z x jy r 957 1983 pp 291292 390 Maxwells Equations or r r e r cos j j sin 958 where j V 1 x is the real part of z y is the imaginary part of z r is the magnitude of z given by r and cj is the phase of z given by tan1 l 959 960 Here x y z r and 0 should not be mistaken as the coordinate variables although they look similar different letters could have been used but it is hard to find better ones The phasor z can be represented in rectangular form as z x jy or in polar form as z r r e The two forms of representing z are related in eqs 957 to 960 and illustrated in Figure 912 Addition and subtraction of phasors are better performed in rec tangular form multiplication and division are better done in polar form Given complex numbers z x jy r z x jy r the following basic properties should be noted Addition Subtraction Multiplication Division x2 x2 and z2 x2 jy2 r2 j2 y2 961a y2 961b 961c 961d lm co rads Figure 912 Representation of a phasor z x jy r t Re 97 TIMEHARMONIC FIELDS H 391 Square Root Complex Conjugate Z Vr Z x jy rj re Other properties of complex numbers can be found in Appendix A2 To introduce the time element we let 961e 961f 962 where 6 may be a function of time or space coordinates or a constant The real Re and imaginary Im parts of rejeeJo are respectively given by and Re rejt r cos ut 0 Im rei4 r sin art 0 963 964a 964b Thus a sinusoidal current 70 7O coswt 0 for example equals the real part of IoejeeM The current 70 h sinco 0 which is the imaginary part of Ioeee01t can also be represented as the real part of Ioejeejuej90 because sin a cosa 90 However in performing our mathematical operations we must be consistent in our use of either the real part or the imaginary part of a quantity but not both at the same time The complex term Ioeje which results from dropping the time factor ejo in 70 is called the phasor current denoted by 7 that is s ioJ 70 0 965 where the subscript s denotes the phasor form of 70 Thus 70 70 coscof 0 the in stantaneous form can be expressed as Re 966 In general a phasor could be scalar or vector If a vector A y z t is a timeharmonic field the phasor form of A is Asx y z the two quantities are related as A Re XseJo For example if A Ao cos ut j3x ay we can write A as A Re Aoej0x a u Comparing this with eq 967 indicates that the phasor form of A is 967 968 s AQ 969 392 M Maxwells Equations Notice from eq 967 that 970 Re AM showing that taking the time derivative of the instantaneous quantity is equivalent to mul tiplying its phasor form byyco That is 3A Similarly 971 972 Note that the real part is chosen in eq 967 as in circuit analysis the imaginary part could equally have been chosen Also notice the basic difference between the instanta neous form AJC y z t and its phasor form Asx y z the former is time dependent and real whereas the latter is time invariant and generally complex It is easier to work with A and obtain A from As whenever necessary using eq 967 We shall now apply the phasor concept to timevarying EM fields The fields quanti ties Ex y z t Dx y z t Hx y z t Bx y z t Jx y z t and pvx y z i and their derivatives can be expressed in phasor form using eqs 967 and 971 In phasor form Maxwells equations for timeharmonic EM fields in a linear isotropic and homogeneous medium are presented in Table 92 From Table 92 note that the time factor eJa disappears because it is associated with every term and therefore factors out resulting in time independent equations Herein lies the justification for using phasors the time factor can be suppressed in our analysis of timeharmonic fields and inserted when necessary Also note that in Table 92 the time factor e01 has been assumed It is equally possible to have assumed the time factor eja in which case we would need to replace every y in Table 92 with j TABLE 92 TimeHarmonic Maxwells Equations Assuming Time Factor e Point Form Integral Form Dv dS I pvs dv B5 dS 0 V D v V Bv 0 V X E s joiB k E s d ju I Bs dS V X H Js juDs Hsdl Js joiDs dS 97 TIMEHARMONIC FIELDS 393 EXAMPLE 95 Evaluate the complex numbers 73 4 a z 1 112 Solution a This can be solved in two ways working with z in rectangular form or polar form Method 1 working in rectangular form Let Z3Z4 where 3 j z4 3 j4 the complex conjugate of 3 j4 3 4 To find the complex conjugate of a complex number simply replace every with j z5 1 76 and Hence z3z4 j4 4 and 1 j63 4 3 4 2714 4 3 24 2 7 7 I 4 Multiplying and dividing z by 27 j4 rationalization we have 4 j327 yi4 150 J25 Zl 27 yl427 j14 272 142 01622 0027 01644 946 Method 2 working in polar form z3j 190 z4 3 j4 5 53130 5 5313 394 Maxwells Equations Hence as obtained before b Let where and Hence z5 I j6 V37 9946 zb 2 jf V5 26562 5 53130 1 905 53130 and V37 99465 53130 1 90 9946 01644 946 V37 01622 70027 12 Zs Z7lj V245 4 78 4V5634O V2 45 V2 4V56340 4V5 01581 71084 45 634 z2 V01581 108472 03976 7542 PRACTICE EXERCISE 95 Evaluate these complex numbers b 6 W 5 3 ejn Answer a 024 j032 b 2903 J8707 97 TIMEHARMONIC FIELDS 395 EXAMPLE 96 Given that A 10 cos 108 10 60 az and Bs 20 a 10 ej2B ay express A in phasor form and B in instantaneous form Solution where u 10 Hence A Re10eM10AH If A Re 0eJbU lw az e Re A O A 0ej 90 e23a j B a 10e23ay j20a v 2 2 3 B Re Be0 Re 20ejw7r2ax lOw2TJ3a 2TT 20 cos art 7r2av 10 cos I wf lav 20 sin ot ax 10 cos r jav PRACTICE EXERCISE 96 If P 2 sin Qt x TT4 av and Qs ejax a sin Try determine the phasor form of P and the instantaneous form of Qv Answer 2eju Jx4av sin x y coswf jca ar EXAMPLE 97 The electric field and magnetic field in free space are given by E cos l06f 3z a Vm P H cos l06f 3z a0 Am Express these in phasor form and determine the constants Ho and 3 such that the fields satisfy Maxwells equations 396 Maxwells Equations Solution The instantaneous forms of E and H are written as E Re EseJal H Re HseJ where co 106 and phasors Es and Hs are given by 50 H E ea H eza p p For free space pv 0 a 0 e eo and ft fio so Maxwells equations become VB ioVH 0 VHa 0 dE dt V X H S j iii V X E fio 971 972 973 974 975 976 Substituting eq 972 into eqs 973 and 974 it is readily verified that two Maxwells equations are satisfied that is Now V X Hs V X P V P Substituting eqs 972 and 977 into eq 975 we have JHOI3 mz 50 M 977i or o3 50 aeo Similarly substituting eq 972 into 976 gives P or 978 979 97 TIMEHARMONIC FIELDS 397 Multiplying eq 978 with eq 979 yields or Mo 50 Ho 50V sJno 7 0 1 3 2 6 Dividing eq 978 by eq 979 we get I32 o2x0e0 or 0 aVp 333 X 10 10 3 X 10s 3 In view of eq 978 Ho 01326 333 X 103 or Ho 01326 j3 333 X 103 only these will satisfy Maxwells four equations PRACTICE EXERCISE 97 In air E cos 6 X 107r 3r a Vm r Find j3 and H Answer 02 radm r cos 6 sin 6 X 107 02r ar sin S X llzr2 1207rr cos 6 X 107f 02r Am EXAMPLE 98 In a medium characterized by a 0 x xo eo and E 20 sin 108f j3z a7 Vm calculate 8 and H Solution This problem can be solved directly in time domain or using phasors As in the previous example we find 13 and H by making E and H satisfy Maxwells four equations Method 1 time domain Let us solve this problem the harder wayin time domain It is evident that Gausss law for electric fields is satisfied that is dy 398 Maxwells Equations From Faradays law V X E But V X E dH dt A A A dx dy dz 0 Ey 0 H I V X Edt dEy dEy dz dx Hence 203 cos 108f 3z ax 0 H cos 108r pz dtax s i I3zax 981 It is readily verified that dx showing that Gausss law for magnetic fields is satisfied Lastly from Amperes law V X H CTE 1 E V X H 982 because a 0 But V X H A A A dx dy dz Hr 0 0 dHx dHx cos108 zay 0 where H in eq 981 has been substituted Thus eq 982 becomes 20S 2 E cos10 8r 3zdtay 2O3 2 sin108f Comparing this with the given E we have 20 97 TIMEHARMONIC FIELDS 399 or 108Vtis 10SVIXO 4eo 1082 1082 3 X 10B From eq 981 or 1 2z H sin 108 axAm 3TT V 3 Method 2 using phasors E Im y E av where co Again 10 V X E dy H V X Es or 203 fr Notice that V H 0 is satisfied V X Hs ji E V X H jus Substituting H in eq 984 into eq 985 gives 2 co xe Comparing this with the given Es in eq 983 we have co xe 983 984 985 400 Maxwells Equations or as obtained before From eq 984 2 0 2 3 10 84T X 10 3TT H Im H 1 sin 108f Qz ax Am 3TT as obtained before It should be noticed that working with phasors provides a considerable simplification compared with working directly in time domain Also notice that we have used A Im Asejat because the given E is in sine form and not cosine We could have used A Re Asejo in which case sine is expressed in terms of cosine and eq 983 would be E 20 cos 108 90 av Re EseM or and we follow the same procedure PRACTICE EXERCISE 98 A medium is characterized by a 0 n 2 and s 5eo If H 2 cos jit 3y a Am calculate us and E Answer 2846 X lf rads 4768 cos 2846 X 108f 3v a Vm SUMMARY 1 In this chapter we have introduced two fundamental concepts electromotive force emf based on Faradays experiments and displacement current which resulted from Maxwells hypothesis These concepts call for modifications in Maxwells curl equa tions obtained for static EM fields to accommodate the time dependence of the fields 2 Faradays law states that the induced emf is given by N 1 dt REVIEW QUESTIONS U 401 For transformer emf Vemf and for motional emf Vemf I u X B d 3 The displacement current h h dS where id dD displacement current density is a modification to Amperes circuit dt law This modification attributed to Maxwell predicted electromagnetic waves several years before it was verified experimentally by Hertz 4 In differential form Maxwells equations for dynamic fields are V D Pv VB 0 dt V X H J dt Each differential equation has its integral counterpart see Tables 91 and 92 that can be derived from the differential form using Stokess or divergence theorem Any EM field must satisfy the four Maxwells equations simultaneously 5 Timevarying electric scalar potential Vx y z t and magnetic vector potential AJC y z t are shown to satisfy wave equations if Lorentzs condition is assumed 6 Timeharmonic fields are those that vary sinusoidally with time They are easily ex pressed in phasors which are more convenient to work with Using the cosine refer ence the instantaneous vector quantity AJC y z t is related to its phasor form Asx y z according to Ax y z t Re AX y z eM 91 The flux through each turn of a 100turn coil is t3 2t mWb where t is in seconds The induced emf at t 2 s is a IV b 1 V c 4mV d 04 V e 04 V 402 B Maxwells Equations Increasing B a Decreasing B Figure 913 For Review Question 92 b Decreasing B Increasing B d 92 Assuming that each loop is stationary and the timevarying magnetic field B induces current which of the configurations in Figure 913 are incorrect 93 Two conducting coils 1 and 2 identical except that 2 is split are placed in a uniform mag netic field that decreases at a constant rate as in Figure 914 If the plane of the coils is per pendicular to the field lines which of the following statements is true a An emf is induced in both coils b An emf is induced in split coil 2 c Equal joule heating occurs in both coils d Joule heating does not occur in either coil 94 A loop is rotating about the yaxis in a magnetic field B Ba sin wt ax Wbm2 The voltage induced in the loop is due to a Motional emf b Transformer emf c A combination of motional and transformer emf d None of the above 95 A rectangular loop is placed in the timevarying magnetic field B 02 cos 150irfaz Wbm as shown in Figure 915 Vx is not equal to V2 a True b False Figure 914 For Review Question 93 REVIEW QUESTIONS 403 0 B Figure 915 For Review Question 95 and Problem 910 96 The concept of displacement current was a major contribution attributed to a Faraday b Lenz c Maxwell d Lorentz e Your professor 97 Identify which of the following expressions are not Maxwells equations for timevarying fields a b V D Pv d 4 H d e i B dS 0 e dS dt J 98 An EM field is said to be nonexistent or not Maxwellian if it fails to satisfy Maxwells equations and the wave equations derived from them Which of the following fields in free space are not Maxwellian a H cos x cos 106fav b E 100 cos cot ax c D e10sin105 lOy az d B 04 sin 104fa e H 10 cos 103 ar sinfl f E cos i Vioeo i g B 1 p sin ufaz 404 Maxwells Equations 99 Which of the following statements is not true of a phasor a It may be a scalar or a vector b It is a timedependent quantity c A phasor Vs may be represented as Vo 0 or Voeje where Vo Vs d It is a complex quantity 910 If Ej 10 ej4x ay which of these is not a correct representation of E a Re Esejut b Re Esej c Im E d 10 cos wf jAx ay e 10 sin ut Ax ay Answers 91b 92b d 93a 94c 95a 96c 97a b d g 98b 99ac 910d PRORI FMS conducting circular loop of radius 20 cm lies in the z 0 plane in a magnetic field B 10 cos 377 az mWbm2 Calculate the induced voltage in the loop 92 A rod of length rotates about the zaxis with an angular velocity w If B Boaz calcu late the voltage induced on the conductor 93 A 30cm by 40cm rectangular loop rotates at 130 rads in a magnetic field 006 Wbm2 normal to the axis of rotation If the loop has 50 turns determine the induced voltage in the loop 94 Figure 916 shows a conducting loop of area 20 cm2 and resistance 4 fl If B 40 cos 104faz mWbm2 find the induced current in the loop and indicate its direction 95 Find the induced emf in the Vshaped loop of Figure 917 a Take B 01a Wbm2 and u 2ax ms and assume that the sliding rod starts at the origin when t 0 b Repeat part a if B 05xaz Wbm2 Figure 916 For Problem 94 0 4fi B 1 1 0 PROBLEMS 405 Figure 917 For Problem 95 B 0 V 0 u 0 96 A square loop of side a recedes with a uniform velocity oav from an infinitely long fila ment carrying current along az as shown in Figure 918 Assuming that p po at time t 0 show that the emf induced in the loop at t 0 is Vrmf uoa 2vpp a 97 A conducting rod moves with a constant velocity of 3az ms parallel to a long straight wire carrying current 15 A as in Figure 919 Calculate the emf induced in the rod and state which end is at higher potential 98 A conducting bar is connected via flexible leads to a pair of rails in a magnetic field B 6 cos lOf ax mWbm2 as in Figure 920 If the zaxis is the equilibrium position of the bar and its velocity is 2 cos lOf ay ms find the voltage induced in it 99 A car travels at 120 kmhr If the earths magnetic field is 43 X 105 Wbm2 find the induced voltage in the car bumper of length 16 m Assume that the angle between the earth magnetic field and the normal to the car is 65 910 If the area of the loop in Figure 915 is 10 cm2 calculate Vx and V2 Figure 918 For Problem 96 406 Maxwells Equations 15 A A 20 cm u t 40 cm Figure 919 For Problem 97 911 As portrayed in Figure 921 a bar magnet is thrust toward the center of a coil of 10 turns and resistance 15 fl If the magnetic flux through the coil changes from 045 Wb to 064 Wb in 002 s what is the magnitude and direction as viewed from the side near the magnet of the induced current 912 The cross section of a homopolar generator disk is shown in Figure 922 The disk has inner radius p 2 cm and outer radius p2 10 cm and rotates in a uniform magnetic field 15 mWbm2 at a speed of 60 rads Calculate the induced voltage 913 A 50V voltage generator at 20 MHz is connected to the plates of an air dielectric parallel plate capacitor with plate area 28 cm2 and separation distance 02 mm Find the maximum value of displacement current density and displacement current 914 The ratio JIJd conduction current density to displacement current density is very impor tant at high frequencies Calculate the ratio at 1 GHz for a distilled water p uo e 81e0 a 2 X 103 Sm b sea water p no e 81eo a 25 Sm c limestone p ixo e 5eo j 2 X 104 Sm 915 Assuming that sea water has fi fxa e 81e0 a 20 Sm determine the frequency at which the conduction current density is 10 times the displacement current density in mag nitude Figure 920 For Problem 98 PROBLEMS Figure 921 For Problem 911 407 916 A conductor with crosssectional area of 10 cm carries a conduction current 02 sin l09t mA Given that a 25 X 106 Sm and er 6 calculate the magnitude of the dis placement current density 917 a Write Maxwells equations for a linear homogeneous medium in terms of Es and YLS only assuming the time factor eJu b In Cartesian coordinates write the point form of Maxwells equations in Table 92 as eight scalar equations 918 Show that in a sourcefree region J 0 pv 0 Maxwells equations can be reduced to two Identify the two allembracing equations 919 In a linear homogeneous and isotropic conductor show that the charge density pv satisfies pv 0 dt e 920 Assuming a sourcefree region derive the diffusion equation at shaft brush copper disk Figure 922 For Problem 912 408 axwells Eolations 921 In a certain region J 2yax xzay z3az sin 104r Am nndpvifpvxy0t 0 922 In a chargefree region for which a 0 e eoer and xo H 5cos10 u 4yaAm find a Jd and D b er 923 In a certain region with a 0 x yuo and e 625a0 the magnetic field of an EM wave is H 06 cos I3x cos 108r a Am Find and the corresponding E using Maxwells equations 924 In a nonmagnetic medium E 50 cos109r Sxy 40 sin109 Sxaz Vm find the dielectric constant er and the corresponding H 925 Check whether the following fields are genuine EM fields ie they satisfy Maxwells equations Assume that the fields exist in chargefree regions a A 40 sinco 10ra2 b B coscor 2pa6 P c C f 3p 2 cot jap H a 0 j sin ut d D sin 8 smwt 5rae r 926 Given the total electromagnetic energy W E D H B dv show from Maxwells equations that dW dt f EXHiS E J dv 927 In free space H psin 4ap 2 cos a j cos 4 X 10 t Am find id and E PROBLEMS 409 928 An antenna radiates in free space and H 12 sin 6 cos2ir X lfr 0rag mAm find the corresponding E in terms of 3 929 The electric field in air is given by E ptep Vm find B and J 930 In free space pv 0 J 0 Show that A 2 c o s e a r s i n e ajeJ Awr satisfies the wave equation in eq 952 Find the corresponding V Take c as the speed of light in free space 931 Evaluate the following complex numbers and express your answers in polar form a 4 30 105012 1 J2 b c d 6 7 8 7 3 j42 12 jl 6 10 3620012 932 Write the following timeharmonic fields as phasors a E 4 cosoit 3x 10 ay sincof 3x 20 B b H sin cosut 5rag r c J 6e3x sinojf 2xay 10ecosw 933 Express the following phasors in their instantaneous forms a A 4 3jej0xay 0B c Cs 7 1 j2ejt sin 0a 0 r 934 Given A 4 sin wtax 3 cos wtay and Bs j0zejzax express A in phase form and B in instantaneous form 935 Show that in a linear homogeneous isotropic sourcefree region both Es and Hs must satisfy the wave equation 0 where y2 a2xe and A E or Hs Chapter 10 ELECTROMAGNETIC WAVE PROPAGATION How far you go in life depends on your being tender with the young compas sionate with the aged sympathetic with the striving and tolerant of the weak and the strong Because someday in life you will have been all of these GEORGE W CARVER 101 INTRODUCTION Our first application of Maxwells equations will be in relation to electromagnetic wave propagation The existence of EM waves predicted by Maxwells equations was first in vestigated by Heinrich Hertz After several calculations and experiments Hertz succeeded in generating and detecting radio waves which are sometimes called Hertzian waves in his honor In general waves are means of transporting energy or information Typical examples of EM waves include radio waves TV signals radar beams and light rays All forms of EM energy share three fundamental characteristics they all travel at high velocity in traveling they assume the properties of waves and they radiate outward from a source without benefit of any discernible physical vehicles The problem of radia tion will be addressed in Chapter 13 In this chapter our major goal is to solve Maxwells equations and derive EM wave motion in the following media 1 Free space T 0 s eo JX xo 2 Lossless dielectrics a 0 e eso JX jxrjxo or a sC aie 3 Lossy dielectrics a 0 e EEO fx fxrixo 4 Good conductors a e eo JX ixrfxo or a S we where w is the angular frequency of the wave Case 3 for lossy dielectrics is the most general case and will be considered first Once this general case is solved we simply derive other cases 12 and 4 from it as special cases by changing the values of a e and ix However before we consider wave motion in those different media it is appropriate that we study the characteristics of waves in general This is important for proper understand 410 102 WAVES IN GENERAL 411 ing of EM waves The reader who is conversant with the concept of waves may skip Section 102 Power considerations reflection and transmission between two different media will be discussed later in the chapter 102 WAVES IN GENERAL A clear understanding of EM wave propagation depends on a grasp of what waves are in general A wave is a function of both space and time Wave motion occurs when a disturbance at point A at time to is related to what happens at point B at time t t0 A wave equation as exemplified by eqs 951 and 952 is a partial differential equation of the second order In one dimension a scalar wave equation takes the form of d2E 2 d2E r U r 0 dt2 dz2 101 where u is the wave velocity Equation 101 is a special case of eq 951 in which the medium is source free pv 0 J 0 It can be solved by following procedure similar to that in Example 65 Its solutions are of the form or E fz ut E gz ut Efz ut gz ut 102a 102b 102c where and g denote any function of z ut and z ut respectively Examples of such functions include z ut sin kz ut cos kz ut and eJkzu where k is a constant It can easily be shown that these functions all satisfy eq 101 If we particularly assume harmonic or sinusoidal time dependence eJ0 eq 101 becomes d2E S 0 103 where 3 uu and Es is the phasor form of E The solution to eq 103 is similar to Case 3 of Example 65 see eq 6512 With the time factor inserted the possible solu tions to eq 103 are E 104a 104b 412 B Electromagnetic Wave Propagation and AeiM0z Bejutfiz 104c where A and B are real constants For the moment let us consider the solution in eq 104a Taking the imaginary part of this equation we have E A sin cof 3z 105 This is a sine wave chosen for simplicity a cosine wave would have resulted had we taken the real part of eq 104a Note the following characteristics of the wave in eq 105 1 It is time harmonic because we assumed time dependence ejo to arrive at eq 105 2 A is called the amplitude of the wave and has the same units as E 3 ox 3z is the phase in radians of the wave it depends on time t and space vari able z 4 w is the angular frequency in radianssecond 0 is the phase constant or wave number in radiansmeter Due to the variation of E with both time t and space variable z we may plot as a function of t by keeping z constant and vice versa The plots of Ez t constant and Et z constant are shown in Figure 101a and b respectively From Figure 101a we observe that the wave takes distance X to repeat itself and hence X is called the wave length in meters From Figure 101b the wave takes time T to repeat itself conse quently T is known as the period in seconds Since it takes time T for the wave to travel distance X at the speed u we expect X uT 106a But T lf whereis the frequency the number of cycles per second of the wave in Hertz Hz Hence u X 106b Because of this fixed relationship between wavelength and frequency one can identify the position of a radio station within its band by either the frequency or the wavelength Usually the frequency is preferred Also because a 2TT 107a 107b and 107c 102 WAVES IN GENERAL 413 I A A Y o x V x 3X 2 2X 1 a A 1J 0 2 ITJ T 3r IT 2 Figure 101 Plot of Ez t b with constant z b A sinco z a with constant t we expect from eqs 106 and 107 that 108 Equation 108 shows that for every wavelength of distance traveled a wave undergoes a phase change of 2TT radians We will now show that the wave represented by eq 105 is traveling with a velocity u in the z direction To do this we consider a fixed point P on the wave We sketch eq 105 at times t 0 774 and 772 as in Figure 102 From the figure it is evident that as the wave advances with time point P moves along z direction Point P is a point of constant phase therefore ut j3z constant or dz 109 414 Electromagnetic Wave Propagation Figure 102 Plot of Ez t A sincot 3z at time a t 0 b t T4 c t 772 P moves along z direction with velocity u c t Tj2 which is the same as eq 107b Equation 109 shows that the wave travels with velocity u in the z direction Similarly it can be shown that the wave B sin cof 5z in eq 104b is traveling with velocity u in the z direction In summary we note the following 1 A wave is a function of both time and space 2 Though time 0 is arbitrarily selected as a reference for the wave a wave is without beginning or end 3 A negative sign in ut 3z is associated with a wave propagating in the z di rection forward traveling or positivegoing wave whereas a positive sign indi cates that a wave is traveling in the z direction backward traveling or negative going wave 4 Since sin p sin sin j ir whereas cosi cos p sin j itl2 cos sin p ir sin j cos p if 12 sin p cos j IT cos f 1010a 1010b 1010c lOlOd where p ut ffz With eq 1010 any timeharmonic wave can be represented in the form of sine or cosine 102 WAVES IN GENERAL 415 TABLE 101 Electromagnetic Spectrum EM Phenomena Examples of Uses Approximate Frequency Range Cosmic rays Gamma rays Xrays Ultraviolet radiation Visible light Infrared radiation Microwave waves Radio waves Physics astronomy Cancer therapy Xray examination Sterilization Human vision Photography Radar microwave relays satellite communication UHF television VHF television FM radio Shortwave radio AM radio 1014 GHz and above 101013GHz 108109 GHz 106108 GHz 105106GHz 103104 GHz 3300 GHz 470806 MHz 54216 MHz 326 MHz 5351605 kHz A large number of frequencies visualized in numerical order constitute a spectrum Table 101 shows at what frequencies various types of energy in the EM spectrum occur Frequencies usable for radio communication occur near the lower end of the EM spectrum As frequency increases the manifestation of EM energy becomes dangerous to human beings1 Microwave ovens for example can pose a hazard if not properly shielded The practical difficulties of using EM energy for communication purposes also increase as fre quency increases until finally it can no longer be used As communication methods improve the limit to usable frequency has been pushed higher Today communication satellites use frequencies near 14 GHz This is still far below light frequencies but in the enclosed environment of fiber optics light itself can be used for radio communication2 EXAMPLE 101 The electric field in free space is given by E 50 cos 108r x ay Vm a Find the direction of wave propagation b Calculate 3 and the time it takes to travel a distance of A2 c Sketch the wave at t 0 774 and 772 Solution a From the positive sign in tot 3x we infer that the wave is propagating along This will be confirmed in part c of this example See March 1987 special issue of IEEE Engineering in Medicine and Biology Magazine on Effects of EM Radiation 2See October 1980 issue of IEEE Proceedings on OpticalFiber Communications 416 Electromagnetic Wave Propagation b In free space u c c 3 X 10s or 3 03333 radm If 7 is the period of the wave it takes 7 seconds to travel a distance X at speed c Hence to travel a distance of X2 will take 7 I 2ir K 3 L 4 2 Alternatively because the wave is traveling at the speed of light c X But or t l Hence 6TT 3142 ns 23 X 108 as obtained before c At t OEy 50 cos I3x At t 74 Ey 50 cos co 3JC I 50 cos fix TT2 4co 50sin3x At t 72 EY 50 cos co 0x 50 cos3x it 2co 50 cos fix Ey at r 0 74 72 is plotted against x as shown in Figure 103 Notice that a point P ar bitrarily selected on the wave moves along ax as f increases with time This shows thai the wave travels along ax 103 WAVE PROPAGATION IN LOSSY DIELECTRICS 417 50 sin jix Figure 103 For Example 101 wave travels along ax c t Tl PRACTICE EXERCISE 101 J In free space H 01 cos 2 X 108 kx ay Am Calculate a k A and T b The time tx it takes the wave to travel A8 c Sketch the wave at time tx Answer a 0667 radm 9425 m 3142 ns b 3927 ns c see Figure 104 03 WAVE PROPAGATION IN LOSSY DIELECTRICS As mentioned in Section 101 wave propagation in lossy dielectrics is a general case from which wave propagation in other types of media can be derived as special cases Therefore this section is foundational to the next three sections 418 Electromagnetic Wave Propagation 0 1 Figure 104 For Practice Exercise 101c A lossy dielectric is a medium in which an EM wave loses power as it propagates due to poor conduction In other words a lossy dielectric is a partially conducting medium imperfect dielectric or imperfect conductor with a 0 as distinct from a lossless dielectric perfect or good di electric in which a 0 Consider a linear isotropic homogeneous lossy dielectric medium that is charge free pv 0 Assuming and suppressing the time factor ej Maxwells equations see Table 92 become V E 0 V Hs 0 V X Es junHs Taking the curl of both sides of eq 1013 gives V X V X Es join V X H S Applying the vector identity VX V X A VVA V2A 1011 1012 1013 1014 1015 1016 to the lefthand side of eq 1015 and invoking eqs 1011 and 1014 we obtain VVE V2ES j or V2ES 72ES 0 1017 where 7 jwCff j 1018 103 WAVE PROPAGATION IN LOSSY DIELECTRICS 419 and y is called the propagation constant in per meter of the medium By a similar proce dure it can be shown that for the H field V2HS y2Ks 0 1019 Equations 1017 and 1019 are known as homogeneous vector Helmholtz s equations or simply vector wave equations In Cartesian coordinates eq 1017 for example is equiv alent to three scalar wave equations one for each component of E along ax ay and az Since y in eqs 1017 to 1019 is a complex quantity we may let y a j3 We obtain a and 3 from eqs 1018 and 1020 by noting that Re y2 P2 a2 f and y2 01 a2 ufi VV co From eqs 1021 and 1022 we obtain V 1020 1021 1022 Oi 6 V 2 V v 2 V a cos coe J 2 I J 1023 1024 Without loss of generality if we assume that the wave propagates along az and that Es has only an xcomponent then Es Exszax Substituting this into eq 1017 yields V2 y2Exsz Hence d2Exsz 1025 1026 or 2 2 y2 Exsz 0 dz 1027 420 B Electromagnetic Wave Propagation This is a scalar wave equation a linear homogeneous differential equation with solution see Case 2 in Example 65 EJx Eoeyz Eoeyz 1028 where Eo and Eo are constants The fact that the field must be finite at infinity requires that Eo 0 Alternatively because eiz denotes a wave traveling along az whereas we assume wave propagation along az Eo 0 Whichever way we look at it Eo 0 Inserting the time factor ejo into eq 1028 and using eq 1020 we obtain Efc t Re aJ Re Eoeazeji0zax or Efo i Eoeazcosat j3zax 1029 A sketch of E at times t 0 and t At is portrayed in Figure 105 where it is evident that E has only an xcomponent and it is traveling along the zdirection Having obtained Ez t we obtain Hz t either by taking similar steps to solve eq 1019 or by using eq 1029 in conjunction with Maxwells equations as we did in Example 98 We will even tually arrive at Hz t Re HoeayMft 1030 where H 1031 and 77 is a complex quantity known as the intrinsic impedance in ohms of the medium It can be shown by following the steps taken in Example 98 that V 1032 Figure 105 field with xcomponent traveling along zdirection at times t 0 and t At arrows indicate in stantaneous values of E 103 WAVE PROPAGATION IN LOSSY DIELECTRICS 421 with 1033 where 0 6V 45 Substituting eqs 1031 and 1032 into eq 1030 gives or H eaz cosco pz 0 1034 Notice from eqs 1029 and 1034 that as the wave propagates along az it decreases or attenuates in amplitude by a factor eaz and hence a is known as the attenuation constant or attenuation factor of the medium It is a measure of the spatial rate of decay of the wave in the medium measured in nepers per meter Npm or in decibels per meter dBm An attenuation of 1 neper denotes a reduction to el of the original value whereas an increase of 1 neper indicates an increase by a factor of e Hence for voltages 1 Np 20 log10 e 8686 dB 1035 From eq 1023 we notice that if a 0 as is the case for a lossless medium and free space a 0 and the wave is not attenuated as it propagates The quantity 3 is a measure of the phase shift per length and is called the phase constant or wave number In terms of the wave velocity u and wavelength X are respectively given by see eqs 107b and 108 CO X 2x 0 1036 We also notice from eqs 1029 and 1034 that E and H are out of phase by 0 at any instant of time due to the complex intrinsic impedance of the medium Thus at any time E leads H or H lags E by 6V Finally we notice that the ratio of the magnitude of the con duction current density J to that of the displacement current density Jd in a lossy medium is IX 08 tan I or tan 6 coe 1037 422 Electromagnetic Wave Propagation where tan 6 is known as the loss tangent and d is the loss angle of the medium as illustrated in Figure 106 Although a line of demarcation between good conductors and lossy di electrics is not easy to make tan 6 or 6 may be used to determine how lossy a medium is A medium is said to be a good lossless or perfect dielectric if tan d is very small j SC we or a good conductor if tan 0 is very large a 5 we From the viewpoint of wave propagation the characteristic behavior of a medium depends not only on its consti tutive parameters a e and fx but also on the frequency of operation A medium that is re garded as a good conductor at low frequencies may be a good dielectric at high frequen cies Note from eqs 1033 and 1037 that From eq 1014 V X Hs o jueEs jws 1 E 1038 1039 where 1040a or ec e 1040b and e e s aw sc is called the complex permittivity of the medium We observe that the ratio of e to e is the loss tangent of the medium that is e a tan d e we 1041 In subsequent sections we will consider wave propagation in other types of media which may be regarded as special cases of what we have considered here Thus we will simply deduce the governing formulas from those obtained for the general case treated in this section The student is advised not just to memorize the formulas but to observe how they are easily obtained from the formulas for the general case Jds Figure 106 Loss angle of a lossy medium J J5 oEs 105 PLANE WAVES IN FREE SPACE 423 104 PLANE WAVES IN LOSSLESS DIELECTRICS In a lossless dielectric a C we It is a special case of that in Section 103 except that a 0 e eosr n fiofir Substituting these into eqs 1023 and 1024 gives a 0 3 WVJLE 1 T JXS Also and thus E and H are in time phase with each other 1042 1043a 1043b 1044 i 05 PLANE WAVES IN FREE SPACE This is a special case of what we considered in Section 103 In this case a 0 e eo 1045 This may also be regarded as a special case of Section 104 Thus we simply replace e by eo and k by xo in eq 1043 or we substitute eq 1045 directly into eqs 1023 and 1024 Either way we obtain a 0 3 wVxoso u c X 1046a 1046b where c 3 X 108 ms the speed of light in a vacuum The fact that EM wave travels in free space at the speed of light is significant It shows that light is the manifestation of an EM wave In other words light is characteristically electromagnetic 424 Electromagnetic Wave Propagation By substituting the constitutive parameters in eq 1045 into eq 1033 dv 0 and V oi where rjo is called the intrinsic impedance of free space and is given by 1047 E Eo 3z then H Ho cos ut f3z y coscof 3z 1048a 1048b The plots of E and H are shown in Figure 107a In general if a aH and ak are unit vectors along the E field the H field and the direction of wave propagation it can be shown that see Problem 1014 ak X a aH or X aH Figure 107 a Plot of E and H as func tions of z at t 0 b plot of E and H at z 0 The arrows indicate instantaneous values a E Eo cos oj ax H Ho cos ut ay b 106 PLANE WAVES IN GOOD CONDUCTORS 425 or aE X aH ak 1049 Both E and H fields or EM waves are everywhere normal to the direction of wave prop agation ak That means the fields lie in a plane that is transverse or orthogonal to the di rection of wave propagation They form an EM wave that has no electric or magnetic field components along the direction of propagation such a wave is called a transverse electro magnetic TEM wave Each of E and H is called a uniform plane wave because E or H has the same magnitude throughout any transverse plane defined by z constant The di rection in which the electric field points is the polarization of a TEM wave3 The wave in eq 1029 for example is polarized in the direction This should be observed in Figure 107b where an illustration of uniform plane waves is given A uniform plane wave cannot exist physically because it stretches to infinity and would represent an infinite energy However such waves are characteristically simple but fundamentally important They serve as approximations to practical waves such as from a radio antenna at distances sufficiently far from radiating sources Although our discussion after eq 1048 deals with free space it also applies for any other isotropic medium 06 PLANE WAVES IN GOOD CONDUCTORS This is another special case of that considered in Section 103 A perfect or good conduc tor is one in which a S we so that awe o that is a e so JX fionr Hence eqs 1023 and 1024 become a 13 Also and thus E leads H by 45 If E Eoeazcosat j8z ax 1050 1051a 1051b 1052 1053a 3Some texts define polarization differently 426 P Electromagnetic Wave Propagation then H az cosco z 45 a 1053b Therefore as E or H wave travels in a conducting medium its amplitude is attenuated by the factor eaz The distance 5 shown in Figure 108 through which the wave amplitude decreases by a factor el about 37 is called skin depth or penetration depth of the medium that is or a 1054a The skin depth is a measure of the depth to which an EM wave can penetrate the medium Equation 1054a is generally valid for any material medium For good conductors eqs 1051a and 1054a give 5 1054b The illustration in Figure 108 for a good conductor is exaggerated However for a partially conducting medium the skin depth can be considerably large Note from eqs 1051a 1052 and 1054b that for a good conductor ao a8 1055 Figure 108 Illustration of skin depth 106 PLANE WAVES IN GOOD CONDUCTORS 427 TABLE 102 Skin Frequency Hz Skin depth mm Depth in 10 60 208 86 Copper 100 66 500 299 104 066 66 108 X 103 1010 66 x 104 For copper a 58 X IO7 mhosm fi ft 5 661 vf in mm Also for good conductors eq 1053a can be written as E Eaedh cos otax showing that 5 measures the exponential damping of the wave as it travels through the con ductor The skin depth in copper at various frequencies is shown in Table 102 From the table we notice that the skin depth decreases with increase in frequency Thus E and H can hardly propagate through good conductors The phenomenon whereby field intensity in a conductor rapidly decreases is known as skin effect The fields and associated currents are confined to a very thin layer the skin of the conductor surface For a wire of radius a for example it is a good approximation at high frequencies to assume that all of the current flows in the circular ring of thickness 5 as shown in Figure 109 Skin effect appears in different guises in such problems as attenua tion in waveguides effective or ac resistance of transmission lines and electromagnetic shielding It is used to advantage in many applications For example because the skin depth in silver is very small the difference in performance between a pure silver compo nent and a silverplated brass component is negligible so silver plating is often used to reduce material cost of waveguide components For the same reason hollow tubular con ductors are used instead of solid conductors in outdoor television antennas Effective elec tromagnetic shielding of electrical devices can be provided by conductive enclosures a few skin depths in thickness The skin depth is useful in calculating the ac resistance due to skin effect The resis tance in eq 516 is called the dc resistance that is aS 516 Figure 109 Skin depth at high frequencies 5 SC a 428 Electromagnetic Wave Propagation We define the surface or skin resistance Rs in flm2 as the real part of the 77 for a good conductor Thus from eq 1055 1056 This is the resistance of a unit width and unit length of the conductor It is equivalent to the dc resistance for a unit length of the conductor having crosssectional area 1 X 5 Thus for a given width w and length the ac resistance is calculated using the familiar dc resistance relation of eq 516 and assuming a uniform current flow in the conductor of thickness 6 that is obw w 1057 where S 8w For a conductor wire of radius a see Figure 109 w 2ira so J ac ff27ra6 a fl26 77ra2 Since 6 3C a at high frequencies this shows that ac is far greater than Rdc In general the ratio of the ac to the dc resistance starts at 10 for dc and very low frequencies and in creases as the frequency increases Also although the bulk of the current is nonuniformly distributed over a thickness of 56 of the conductor the power loss is the same as though it were uniformly distributed over a thickness of 6 and zero elsewhere This is one more reason why 5 is referred to as the skin depth EXAMPLE 102 A lossy dielectric has an intrinsic impedance of 200 30 fi at a particular frequency If at that frequency the plane wave propagating through the dielectric has the magnetic field component H 10ecoscofxJaAm find E and a Determine the skin depth and wave polarization Solution The given wave travels along ax so that ak ax aH ay so a a X aH ax x ay az or aE a z 106 PLANE WAVES IN GOOD CONDUCTORS B 429 AlsoWo 10 so H 77 200 rW 200 eJ16 Eo 2000er6 Except for the amplitude and phase difference E and H always have the same form Hence E Re 2000e7rV7Va or E 2e M cosf cot az kVm V 2 6 Knowing that 3 12 we need to determine a Since and KH 1 CT COS 1 coe 1 12 1 1 But tan 2L tan 60 V l Hence we 2 2 lJ V3 or a 4 F 02887 Npm 4 F V3 2V3 and m 5 2 V 3 34641 a The wave has an Ez component hence it is polarized along the zdirection 430 Electromagnetic Wave Propagation PRACTICE EXERCISE 102 A plane wave propagating through a medium with er 8 ixr 2 has E 05 e3 sin108f z ax Vm Determine a 0 b The loss tangent c Wave impedance d Wave velocity e H field Answer a 1374 radm b 05154 c 17772 1363 0 d 7278 X 107 ms e 2leM sin1081 0z 1363ay mAm EXAMPLE 103 In a lossless medium for which q 60ir ixr 1 and H 01 cos cof z ax 05 sin cor zy Am calculate er co and E Solution In this case a 0 a 0 and 3 1 so Xo 12O7T or 120TT 120x er 2 er 4 60TT 2co c or co 1 3 X 108 15 X 108rads From the given H field E can be calculated in two ways using the techniques based on Maxwells equations developed in this chapter or directly using Maxwells equations as in the last chapter Method 1 To use the techniques developed in this chapter we let E H H2 106 PLANE WAVES IN GOOD CONDUCTORS 431 where Hj 01 cos uf z ax and H2 05 sin wt z ay and the corresponding electric field E E E7 where Ej Elo cos cof z ai and E2 E2o sin cof z aEi Notice that although H has components along ax and ay it has no component along the direction of propagation it is therefore a TEM wave ForE b afi a X aHl a X a x a Eo V Hlo 60TT 01 6TT Hence ForE 6x cos bit z av aEl akx aH az X ay ax E2o V H2o 60TT 05 30x Hence E2 30TT sin wt zax Adding E and E2 gives E that is E 9425 sin 15 X 108f z ax 1885 cos 15 X 108 z ay Vm Method 2 We may apply Maxwells equations directly 1 V X H iE s 0 because a 0 But V X H dt JL JL A dx dy dz Hxz Hvz 0 dHy dHx H2o cos bit z ax Hlo sin wf zay where Hlo 01 and2o 05 Hence if W W E VxHi sin wf z a cos cor z a e J eco eco 9425 sincor zax 1885 coswf z a Vm as expected 432 8 Electromagnetic Wave Propagation PRACTICE EXERCISE 103 A plane wave in a nonmagnetic medium has E 50 sin 10 t 2z ay Vm Find a The direction of wave propagation b A and sr c H Answer a along z direction b 3142 m 1592 MHz 36 c 07958 sin108f 2z ax Am EXAMPLE 104 A uniform plane wave propagating in a medium has E 2eaz sin 108f 3z ay Vm If the medium is characterized by er 1 ir 20 and a 3 mhosm find a 3 and H Solution We need to determine the loss tangent to be able to tell whether the medium is a lossy di electric or a good conductor a we 108 X 1 X 10 ro 3393 36TT showing that the medium may be regarded as a good conductor at the frequency of opera tion Hence a 3 4TT X 107 X 201083 12 Also 614 a 614 Npm 3 614 radm 4TT X 10 X 2010s a 8OO7T 12 tan 20 3393 45 TT4 Hence H Hoeaz sin at z 106 PLANE WAVES IN GOOD CONDUCTORS 433 where and aH ak X aE az X ay ax Thus H 691 e614zsin 6142z J ax mAm PRACTICE EXERCISE 104 A plane wave traveling in the direction in a lossy medium er 4 xr 1 cr 102 mhosm has E 30 cos 109r t x4 az Vm at y 0 Find a E at y 1 m 2 ns b The distance traveled by the wave to have a phase shift of 10 c The distance traveled by the wave to have its amplitude reduced by 40 d H at y 2 m t 2 ns Answer a 2787az Vm b 8325 mm c 542 mm d 471a mAm XAMPLE105 A plane wave E Eo cos ut j3z ax is incident on a good conductor at z 0 Find the current density in the conductor Solution Since the current density J CTE we expect J to satisfy the wave equation in eq 1017 that is V2JS T 2JS 0 Also the incident E has only an xcomponent and varies with z Hence J Jxz t ax and l 2 2 5X sx UZ which is an ordinary differential equation with solution see Case 2 of Example 65 7 Aeyz Beyz 434 Electromagnetic Wave Propagation The constant B must be zero because Jsx is finite as z But in a good conductor a we so that a 3 15 Hence and or 7 a jf3 al j Ae 1 j where Jsx 0 is the current density on the conductor surface PRACTICE EXERCISE 105 Due to the current density of Example 105 find the magnitude of the total current through a strip of the conductor of infinite depth along z and width w along y Answer V2 EXAMPLE 106 For the copper coaxial cable of Figure 712 let a 2 mm b 6 mm and t 1 mm Cal culate the resistance of 2 m length of the cable at dc and at 100 MHz Solution Let R Ro Ri where Ro and Rt are the resistances of the inner and outer conductors Atdc aira2 58 X 107TT2 X 1032 2744 mfi aS oirb t2 b2 airt2 2bt 2 58 X 107TT 1 12 X 106 08429 mO Hence Rdc 2744 08429 3587 mfi 107 POWER AND THE POYNTING VECTOR 435 A t 100 MHz Rsl I w o82ira 2KO V o 2 2K X 2 X 103 041 fl TT X 10s X 4r X 10 58 X 107 Since 6 66 xm C t 1 mm H1 2TT for the outer conductor Hence w 2Kb V a 2TT X 6 X 10 01384 fi 3 TT X 10s X 4TT X 58 X 107 Hence Rac 041 01384 05484 U which is about 150 times greater than Rdc Thus for the same effective current i the ohmic loss i2R of the cable at 100 MHz is far greater than the dc power loss by a factor of 150 PRACTICE EXERCISE 106 For an aluminum wire having a diameter 26 mm calculate the ratio of ac to dc re sistance at a 10 MHz b 2 GHz Answer a 2416 b 3417 07 POWER AND THE POYNTING VECTOR As mentioned before energy can be transported from one point where a transmitter is located to another point with a receiver by means of EM waves The rate of such energy transportation can be obtained from Maxwells equations V X E J dt dE dt 1058a 1058b 436 ft Electromagnetic Wave Propagation Dotting both sides of eq 1058b with E gives E V X H oE2 E e ef But for any vector fields A and B see Appendix A 10 V A X B B V X A A V X B Applying this vector identity to eq 1059 letting A H and B E gives dE H V X E V H X E oEz E edt From eq 1058a and thus eq 1060 becomes tdF 2 dt dE2 at Rearranging terms and taking the volume integral of both sides V E X H dv dt oE dv Applying the divergence theorem to the lefthand side gives E X H dS dt v ss v Total power Rate of decrease in Ohmic power leaving the volume energy stored in electric dissipated and magnetic fields 1059 1060 1061 1062 EE2 fiH2 dv I oE2 dv 1063 1064 Equation 1063 is referred to as Poyntings theorem4 The various terms in the equation are identified using energyconservation arguments for EM fields The first term on the righthand side of eq 1063 is interpreted as the rate of decrease in energy stored in the electric and magnetic fields The second term is the power dissipated due to the fact that the medium is conducting a 0 The quantity E X H on the lefthand side of eq 1063 is known as the Poynting vector SP in watts per square meter Wm2 that is 2P E X H 1065 4After J H Poynting On the transfer of energy in the electromagnetic field Phil Trans vol 174 1883 p 343 107 POWER AND THE POYNTINC VECTOR 437 It represents the instantaneous power density vector associated with the EM field at a given point The integration of the Poynting vector over any closed surface gives the net power flowing out of that surface Poyntings theorem stales thit the nel power flowing out of a given volume i i equal to the lime rate of decrease in the energy stored wilhin r minus the conduction losses The theorem is illustrated in Figure 1010 It should be noted that 9s is normal to both E and H and is therefore along the direc tion of wave propagation ak for uniform plane waves Thus ak aE X aH 1049 The fact that 2P points along ak causes 2P to be regarded derisively as a pointing vector Again if we assume that then Ez t Eoe az cos ut f3z ax UZ 0 TT eaz cos at j3z 9 a Power out Figure 1010 Illustration of power balance for EM fields Power in 438 Electromagnetic Wave Propagation and E2 3z 0 77 e2az cos cot fiz cos cot Hz 0J a M e 2az cos 6 cos 2cot 23z 6 a 1066 since cos A cos B cos A 5 cos A B To determine the timeaverage Poynting vector 2avez in Wm2 which is of more practical value than the instantaneous Poynting vector 2Pz t we integrate eq 1066 over the period T 2iru that is dt 1067 It can be shown see Prob 1028 that this is equivalent to 1068 By substituting eq 1066 into eq 1067 we obtain J 1069 The total timeaverage power crossing a given surface S is given by p Of 1070 We should note the difference between 2 Save and Pave SP y z 0 is m e Poynting vector in wattsmeter and is time varying 2PaVe0c y z also in wattsmeter is the time average of the Poynting vector S it is a vector but is time invariant Pave is a total time average power through a surface in watts it is a scalar EXAMPLE 107 In a nonmagnetic medium E 4 sin 2TT X 107 08 a Vm 107 POWER AND THE POYNTING VECTOR 439 Find a er r b The timeaverage power carried by the wave c The total power crossing 100 cm2 of plane 2x y 5 Solution a Since a 0 and 3 coc the medium is not free space but a lossless medium 3 08 co 27r X 107 fx io nonmagnetic e eoer Hence or co V lie co V iioeosr V er 13c 083 X 108 12 r co 2TT X 107 IT sr 1459 8 987 0 sin2cor 3x I UOir 120 f2 102 M 2TJ 2 X IOTT2 81 axmWm2 c On plane 2x y 5 see Example 35 or 85 2a a V5 Hence the total power is Pav 81 X 10X 100 X 1 162 X 105 7245 tW 2ax V5 440 Electromagnetic Wave Propagation PRACTICE EXERCISE 107 In free space H 02 cos uit 3x az Am Find the total power passing through a A square plate of side 10 cm on plane x z 1 b A circular disc of radius 5 cm on plane x 1 Answer a 0 b 5922 mW 108 REFLECTION OF A PLANE WAVE AT NORMAL INCIDENCE So far we have considered uniform plane waves traveling in unbounded homogeneous media When a plane wave from one medium meets a different medium it is partly re flected and partly transmitted The proportion of the incident wave that is reflected or trans mitted depends on the constitutive parameters e ju a of the two media involved Here we will assume that the incident wave plane is normal to the boundary between the media oblique incidence of plane waves will be covered in the next section after we understand the simpler case of normal incidence Suppose that a plane wave propagating along the zdirection is incident normally on the boundary z 0 between medium 1 z 0 characterized by er eu fix and medium 2 z 0 characterized by a2 e2 2 as shown in Figure 1011 In the figure subscripts r and t denote incident reflected and transmitted waves respectively The incident re flected and transmitted waves shown in Figure 1011 are obtained as follows Incident Wave E H is traveling along az in medium 1 If we suppress the time factor eJo and assume that Elsz Eioeyz ax 1071 then Hz Hioez a e av 1072 Reflected Wave En Hr is traveling along z in medium 1 If Eraz Eryz ax 1073 108 REFLECTION OF A PLANE WAVE AT NORMAL INCIDENCE 441 medium 1 ol e H0 incident wave H reflected wave medium 2 o222 H0 transmitted wave 2 0 Figure 1011 A plane wave incident normally on an interface between two different media then KM Hmeyay ewa 1074 where Era has been assumed to be along ax we will consistently assume that for normal in cident E Er and E have the same polarization Transmitted Wave E Ht is traveling along az in medium 2 If then 1075 1076 In eqs 1071 to 1076 Eio Ero and Eto are respectively the magnitudes of the incident reflected and transmitted electric fields at z 0 Notice from Figure 1011 that the total field in medium 1 comprises both the incident and reflected fields whereas medium 2 has only the transmitted field that is Ej E E n H H Hr E2 E H2 Hr At the interface z 0 the boundary conditions require that the tangential components of E and H fields must be continuous Since the waves are transverse E and H fields 442 is Electromagnetic Wave Propagation are entirely tangential to the interface Hence at z 0 Eltan E2tan and Htan H2tan imply that E0 Er0 E0 H0 Hr0 H0 From eqs 1077 and 1078 we obtain Em and Eu Eio Ero Eto Elo Ero 12 E lO 1077 1078 1079 1080 We now define the reflection coefficient T and the transmission coefficient T from eqs 1079 and 1080 as or and li F VF 1081a 1081b or Note that T Eio F rF 1 1 T T 2 Both F and r are dimensionless and may be complex 3 o s jr l 1082a 1082b 1083 The case considered above is the general case Let us now consider a special case when medium 1 is a perfect dielectric lossless O 0 and medium 2 is a perfect con ductor a2 cc For this case r2 0 hence T 1 and T 0 showing that the wave is totally reflected This should be expected because fields in a perfect conductor must vanish so there can be no transmitted wave E2 0 The totally reflected wave combines with the incident wave to form a standing wave A standing wave stands and does not 108 REFLECTION OF A PLANE WAVE AT NORMAL INCIDENCE 443 travel it consists of two traveling waves E and Er of equal amplitudes but in opposite di rections Combining eqs 1071 and 1073 gives the standing wave in medium 1 as But Hence or Thus or Els E Era Eioe yz Eroeyz ax E Eioei0z ez ax E Re EseM E 2Eio sin 3 sin ut ax 1084 1085 1086 By taking similar steps it can be shown that the magnetic field component of the wave is Hi 2Eio cos pz cos ut av 1087 A sketch of the standing wave in eq 1086 is presented in Figure 1012 for t 0 778 774 3778 772 and so on where T 2TTW From the figure we notice that the wave does not travel but oscillates When media 1 and 2 are both lossless we have another special case a 0 a2 In this case and rj2 a r e real a nd so are F and T Let us consider the following cases CASE A If r2 Jji F 0 Again there is a standing wave in medium 1 but there is also a transmit ted wave in medium 2 However the incident and reflected waves have amplitudes that are not equal in magnitude It can be shown that the maximum values of EX j occur at or mr linax 2 WE n 0 1 2 10 444 Electromagnetic Wave Propagation JX Figure 1012 Standing waves E 2Eio sin 3z sin oit x curves 0 1 2 3 4 are respectively at times t 0 778 TIA 378 772 X 2x73 and the minimum values of Et occur at 3zmin 2 1 or 2w 1 23 w u l z 1089 CASE B If r2 r T 0 For this case the locations of Ej maximum are given by eq 1089 whereas those of EX minimum are given by eq 1088 All these are illustrated in Figure 1013 Note that 1 H j minimum occurs whenever there is Ei maximum and vice versa 2 The transmitted wave not shown in Figure 1013 in medium 2 is a purely travel ing wave and consequently there are no maxima or minima in this region The ratio of Ei max to E min or Hj max to Hj min is called the standingwave ratio s that is s Mi IH l r l r 1090 108 REFLECTION OF A PLANE WAVE AT NORMAL INCIDENCE 445 o 0 a 0 Hgure 1013 Standing waves due to reflection at an interface between two lossless media X 2irf3i or s 1 s 1 1091 Since F 1 it follows that 1 s The standingwave ratio is dimensionless and it is customarily expressed in decibels dB as s indB 201og10if 1092 MPLE 108 In free space z 0 a plane wave with H 10 cos 108f 0z ax mAm is incident normally on a lossless medium e 2eo p 8jiio in region z 0 Determine the reflected wave H n Er and the transmitted wave Hr Er Solution This problem can be solved in two different ways Method 1 Consider the problem as illustrated in Figure 1014 For free space 10s c 3 X 108 7o 1207T 446 Electromagnetic Wave Propagation I free space Figure 1014 For Example 108 lossless dielectric For the lossless dielectric medium o 4 ft coVxe wVosoVxr 4 43 Given that H 10 cos 108r 3 ax we expect that where I and Hence Now f io cos 108f X a ax X a ay io 10 E 10rjo cos 108 3z a mVm Thus r Eio E r rj0 cos f 108f z y mVm 108 REFLECTION OF A PLANE WAVE AT NORMAL INCIDENCE 447 from which we easily obtain Hr as Similarly Hr cos 108f z ax mAm F 4 4 r l T or Ew Eio Thus E Eto cos 108f 32z aEi where ai a a r Hence 4 40 3 Er rjocos 108 zaymVm from which we obtain Ht cos 108f zjax mAm Method 2 Alternatively we can obtain Hr and H directly from H using Thus F and T 1 1 0 Hro Hio and to 3 2ro 3 3 10 o H cos 108 j3iz ax mAm 20 H cos 108f P2z ax mAm as previously obtained Notice that the boundary conditions at z 0 namely 40 o E0 Er0 E0 vo cos 108 ay 448 Electromagnetic Wave Propagation and 20 o H0 Hr0 H0 cos 108r ax are satisfied These boundary conditions can always be used to crosscheck E and H PRACTICE EXERCISE 108 A 5GHz uniform plane wave Efa 10 ejl3z ax Vm in free space is incident nor mally on a large plane lossless dielectric slab z 0 having s 4e0 u x0 Find the reflected wave ErJ and the transmitted wave Ets Answer 3333 expOz ax Vm 6667 expjP2z a Vm where p2 Wi 200TT3 EXAMPLE 109 Given a uniform plane wave in air as E 40 cos at Pz ax 30 sin wt z a Vm a FindH b If the wave encounters a perfectly conducting plate normal to the z axis at z 0 find the reflected wave Er and Hr c What are the total E and H fields for z 0 d Calculate the timeaverage Poynting vectors for z 0 and z 0 Solution a This is similar to the problem in Example 103 We may treat the wave as consisting of two waves En and E2 where En 40 cos wf Pz ax E2 30 sin wt 3z ay At atmospheric pressure air has er 10006 1 Thus air may be regarded as free space Let H Hn H2 where H n HiUl cos ait z aHl Eil0 40 120TT 3TT a a t X a a X ax ay Hence Similarly where Hence and 108 REFLECTION OF A PLANE WAVE AT NORMAL INCIDENCE 449 Hn cos ut j3z ay 3TT I2 Hi2o sin ut 3z a2 Ei2o 30 1 o 1207T 47T aj X a az X ay a x H2 sin cor 3z 4TT sin ut j8z ax H cos art 3z av mAm 4TT 3TT This problem can also be solved using Method 2 of Example 103 b Since medium 2 is perfectly conducting 02 that is r 1 T 0 showing that the incident E and H fields are totally reflected F r F F Hence Er 40 cos ut z ax 30 sin ut 3z ay Vm Hr can be found from Er just as we did in part a of this example or by using Method 2 of the last example starting with H Whichever approach is taken we obtain Hr cos ut Sz av sin ut Szax Am iw 4x 450 Electromagnetic Wave Propagation c The total fields in air E E Er and H H Hr can be shown to be standing wave The total fields in the conductor are E2 Er 0 H2 H 0 d For z 0 I E For z 0 1 2 2 T Eioaz Emaz 2 1 2 30 2 240TT 0 402 302az 402 302aJ op E2 2rj2 a 7 0 because the whole incident power is reflected PRACTICE EXERCISE 109 The plane wave E 50 sin ot 5x ay Vm in a lossless medium n 4o e so encounters a lossy medium fi no e 4eo r 01 mhosm normal to the xaxis at x 0 Find a F T and s b ErandHr c ErandH d The timeaverage Poynting vectors in both regions Answer a 08186 1711 02295 3356 10025 b 4093 sin ait 5x 1719 ay Vm 543 sin at 5x 1719 az mAm c 1147 e6UZIsin cor 7826x 3356 ay Vm 1202 e602U M 7826x 401 a mAm d 05469 x Wm2 05469 exp 1204xaxWm2 sin 109 REFLECTION OF A PLANE WAVE AT OBLIQUE INCIDENCE 451 109 REFLECTION OF A PLANE WAVE T OBLIQUE INCIDENCE We now consider a more general situation than that in Section 108 To simplify the analy sis we will assume that we are dealing with lossless media We may extend our analysis to that of lossy media by merely replacing e by sc It can be shown see Problems 1014 and 1015 that a uniform plane wave takes the general form of Er t Eo cosk r cof Re EoeKkrwt 1093 where r xax yay zaz is the radius or position vector and k kxax kyay kzaz is the wave number vector or the propagation vector k is always in the direction of wave propagation The magnitude of k is related to a according to the dispersion relation k 2 k 2 x k k 1094 Thus for lossless media k is essentially the same as 3 in the previous sections With the general form of E as in eq 1093 Maxwells equations reduce to k X E k X H k H 0 k E 0 1095a 1095b 1095c 1095d showing that i E H and k are mutually orthogonal and ii E and H lie on the plane k r kjc kyy kzz constant From eq 1095a the H field corresponding to the E field in eq 1093 is 77 1096 Having expressed E and H in the general form we can now consider the oblique inci dence of a uniform plane wave at a plane boundary as illustrated in Figure 1015a The plane denned by the propagation vector k and a unit normal vector an to the boundary is called the plane of incidence The angle 0 between k and an is the angle of incidence Again both the incident and the reflected waves are in medium 1 while the transmit ted or refracted wave is in medium 2 Let E Eio cos kixx kiyy kizz ust Er Ero cos krxx kny krzz cV E Ero cos ktxx ktyy ktzz ut 1097a 1097b 1097c 452 Electromagnetic Wave Propagation medium 1 ej M 3 sin 6 kiz cos 8 medium a kr 3 cos Br kr 3 b 3 sin 9 r 3 sin 0 z 3 cos 8 Figure 1015 Oblique incidence of a plane wave a illustration of 0 6r and 0 b illustration of the normal and tangential components of k where kh kr and k with their normal and tangential components are shown in Figure 1015b Sincejhetangential component of E must be continuousat the boundary z 0 Ez 0 Erz 0 0 1098 The only way this boundary condition will be satisfied by the waves in eq 1097 for all x and y is that 1 CO Ur O r CO h K JL J ix Krx tx Kx o Kly ftry y y Condition 1 implies that the frequency is unchanged Conditions 2 and 3 require that the tangential components of the propagation vectors be continuous called the phase match ing conditions This means that the propagation vectors k kt and kr must all lie in the pJanejDf incidenceThus by conditions 2 and 3 fe sin i kr sin 6r kj sin 61 k sin 0 1099 10100 109 REFLECTION OF A PLANE WAVE AT OBLIQUE INCIDENCE 453 where 8r is the angle of reflection and 6 is the angle of transmission But for lossless media kt kr 3 co From eqs 1099 and 10101a it is clear that 10101a 10101b 10102 so that the angle of reflection 8r equals the angle of incidence 0 as in optics Also from eqs 10100 and 10101 10103 sin 8i k Hi where u aik is the phase velocity Equation 10103 is the wellknown Snells law which can be written as nx sin 0 n2 sin 0 10104 where nx c e cu and 2 c v ine2 CU2 tne refractive indices of the media Based on these general preliminaries on oblique incidence we will now specifically consider two special cases one with the E field perpendicular to the plane of incidence the other with the E field parallel to it Any other polarization may be considered as a linear combination of these two cases A Parallel Polarization This case is illustrated in Figure 1016 where the E field lies in the xzplane the plane of incidence In medium 1 we have both incident and reflected fields given by Efa ocos 0 a sin 0 az H i jPi sin iz cos 9 Ers Erocoserax merazeJ0xsmfcos p H ro flrfxsin 0z cos 0 e a 10105a 10105b 10106a 10106b where fil co V is Notice carefully how we arrive at each field component The trick in deriving the components is to first get the polarization vector k as shown in Figure 1015b for incident reflected and transmitted waves Once k is known we 454 Electromagnetic Wave Propagation Figure 1016 Oblique incidence with E par allel to the plane of incidence medium 1 z 0 medium 2 define Es such that V Ev 0 or k Es 0 and then Hs is obtained from Hs k E X E a X The transmitted fields exist in medium 2 and are given by Es Mcos 0 ax sin 0 a eusineJcose H i e x s m fl z cos 0 10107a 10107b where f32 o V u2e2 Should our assumption about the relative directions in eqs 10105 to 10107 be wrong the final result will show us by means of its sign Requiring that dr dj and that the tangential components of E and H be continuous at the boundary z 0 we obtain Ei0 Ero cos 0 Eo cos 0t Em Eto Expressing Em and Eta in terms of Eio we obtain Ero 11 COS 0 COS 0 o 7j2 cos 0 rjj cos 0 or and to 2r2 cos 0 Eio 72 cos 0 r cos 0 10108a 10108b 10109a 10109b 10110a 109 REFLECTION OF A PLANE WAVE AT OBLIQUE INCIDENCE 455 or Eo TEia 10110b Equations 10109 and 10110 are called Fresnels equations Note that the equations reduce to eqs 1081 and 1082 when 0 0 0 as expected Since 0 and d are related according to Snells law of eq 10103 eqs 10109 and 10110 can be written in terms of 9j by substituting cos 0 V l sin2 6r V l w2H2sin2 0 From eqs 10109 and 10110 it is easily shown that 10111 Til l fcos6t Vcos 07 10112 From eq 10109a it is evident that it is possible that T 0 because the numerator is the difference of two terms Under this condition there is no reflection Em 0 and the incident angle at which this takes place is called the Brewster angle 0B The Brewster angle is also known as the polarizing angle because an arbitrarily polarized incident wave will be reflected with only the component of E perpendicular to the plane of incidence The Brewster effect is utilized in a laser tube where quartz windows are set at the Brewster angle to control polarization of emitted light The Brewster angle is obtained by setting 0 dB when Tn 0 in eq 10109 that is or r2 2l sin20r Introducing eq 10103 or 10104 gives sin20B 10113 It is of practical value to consider the case when the dielectric media are not only lossless but nonmagnetic as wellthat is fxx JX2 v For this situation eq 10113 becomes sin2 0B 1 sin 0Rl or tan 0B A showing that there is a Brewster angle for any combination of 8 and e2 10114 456 Electromagnetic Wave Propagation B Perpendicular Polarization In this case the E field is perpendicular to the plane of incidence the xzplane as shown in Figure 1017 This may also be viewed as the case where H field is parallel to the plane of incidence The incident and reflected fields in medium 1 are given by p p j73iCtsin0jzcose a Hs cos 6 ax sin 0 a e Hlrs tLrOt Ay Urs cos 6r ax sin 6r az ejl while the transmitted fields in medium 2 are given by E p j32x sin 9z cos 9 cav Elo Hs f cos 6 ax sin 9 az V 10115a 10115b 10116a 10116b 10117a 10117b Notice that in defining the field components in eqs 10115 to 10117 Maxwells equa tions 1095 are always satisfied Again requiring that the tangential components of E and H be continuous at z 0 and setting dr equal to 6h we get p i p p Eo Ero cos dj Elo cos Bt Expressing Ero and Et0 in terms of Eio leads to tLro Eio V2 V2 cos 6 cos 6 r j COS COS 10118a 10118b 10119a E Figure 1017 Oblique incidence with E per pendicular to the plane of incidence 109 REFLECTION OF A PLANE WAVE AT OBLIQUE INCIDENCE 457 or and ro J io Eio ri2 cos 6 vl cos 9 10119b 10120a or Eto 10120b which are the Fresnels equations for perpendicular polarization From eqs 10119 and 10120 it is easy to show that 1 r TL 10121 which is similar to eq 1083 for normal incidence Also when 9 9 0 eqs 10119 and 10120 become eqs 1081 and 1082 as they should For no reflection TL 0 or Er 0 This is the same as the case of total transmis sion TX 1 By replacing 0 with the corresponding Brewster angle 9B we obtain t2 cos 9B rycos 9 or sin20 Incorporating eq 10104 yields sin2 9Bx AM 62 10122 Note that for nonmagnetic media ft A2 AO sin2 0B i n eq 10122 so 9BL does not exist because the sine of an angle is never greater than unity Also if x JX2 and 6 e2 eq 10122 reduces to sin 1 or 10123 Although this situation is theoretically possible it is rare in practice 458 Electromagnetic Wave Propagation EXAMPLE 1010 An EM wave travels in free space with the electric field component E 100e0866va5jaxVm Determine a co and X b The magnetic field component c The time average power in the wave Solution a Comparing the given E with E E eikT E eJkxkykz a it is clear that kx 0 ky 0866 kz 05 Thus But in free space Hence k Vk2 x ky k V08662 052 1 co 2TT k 13 coVi020 C A co kc 3 X 10rads X 2TT 6283 m k b From eq 1096 the corresponding magnetic field is given by Hs k X E iCO 0866ay 05az 4x X 107 X 3 X 108 X 100a re j k r or H 133 av 23 a e m66vl5z mAm c The time average power is 1002 2120TT 0866 av 05 a 1149av 6631 aWm2 109 REFLECTION OF A PLANE WAVE AT OBLIQUE INCIDENCE 459 XAMPLE 1011 PRACTICE EXERCISE 1010 Rework Example 1010 if 10 in free space 5a2 coscof 2y Az Vm Answer a 1342 X 109 rads 1405 m b 2966 cos 1342 X 109f 2y Az ax mAm c 007415 ay 01489 a Wm2 A uniform plane wave in air with E 8 cos at Ax 3z av Vm is incident on a dielectric slab z 0 with fxr 10 er 25 a 0 Find a The polarization of the wave b The angle of incidence c The reflected E field d The transmitted H field Solution a From the incident E field it is evident that the propagation vector is Hence k 4a 3a 5 coVu0e0 5c 15 X 108 rads A unit vector normal to the interface z 0 is az The plane containing k and a is y constant which is the jczplane the plane of incidence Since E is normal to this plane we have perpendicular polarization similar to Figure 1017 b The propagation vectors are illustrated in Figure 1018 where it is clear that tan0 0 5313 kiz 3 Alternatively without Figure 1018 we can obtain 0 from the fact that 0 is the angle between k and an that is cos 0 ak an 3a or 0 5313 460 Electromagnetic Wave Propagation j c An easy way to find E r is to use eq 10116a because we have noticed that this 1 problem is similar to that considered in Section 109b Suppose we are not aware of this I Let j Er Ero cos cor kr r ay which is similar to form to the given E The unit vector ay is chosen in view of the fact that i the tangential component of E must be continuous at the interface From Figure 1018 kr krx ax krz az where krx kr sin 9n krz kr cos 6r But 6r Oj and kr k 5 because both kr and k are in the same medium Hence kr Aax 3az To find Em we need 6t From Snells law sin 6 sin 0 n2 sin 5313 sin 8i 25 or 6 3039 Eio 72 COS 0 IJi COS 0 rj cos 6t cos Figure 10IS Propagation vectors of ExamplelOil 109 REFLECTION OF A PLANE WAVE AT OBLIQUE INCIDENCE 461 where rjl rjo 377 n2 377 2384 Hence and 2384 cos 3513 377 cos 3039 1 2384 cos 5313 377 cos 3039 Em TEio 03898 3112 E 3112 cos 15 X 108f Ax 3zayVm d Similarly let the transmitted electric field be E Eto cos ut k r ay where W 1 c From Figure 1018 k j32 w V 15 X 108 3 X 108 ktx k sin 6 4 kR ktcos6 6819 or k 4ax 6819 az Notice that kix krx ktx as expected Ew 2 72 COS dj Eio i2 cos dj 7 cos 6 2 X 2384 cos 5313 2384 cos 5313 377 cos 3039 0611 The same result could be obtained from the relation T I Hence Eto TLEio 0611 X 8 4888 Ef 4888 cos 15 X 108r Ax 6819z ay 462 Electromagnetic Wave Propagation From E H is easily obtained as 79062384 a cos r k r H 1769 ax 1037 az cos 15 X ft Ax 6819z mAm PRACTICE EXERCISE 1011 If the plane wave of Practice Exercise 1010 is incident on a dielectric medium having a 0 e 4eo x to and occupying z 0 calculate a The angles of incidence reflection and transmission b The reflection and transmission coefficients c The total E field in free space d The total E field in the dielectric e The Brewster angle Answer a 2656 2656 1292 b 0295 0647 c 10 ay 5az cos at 2y 4z 2946a 1473az cos cat 2y 4z Vm d 7055a 1618az cos wf 2y 8718z Vm e 6343 SUMMARY 1 The wave equation is of the form dt2 2d2P u T 0 dz with the solution 4 A sin wf 3z where u wave velocity A wave amplitude co angular frequency 2TT and 3 phase constant Also 3 OJM 2TTX or M fk Xr where X wavelength and T period 2 In a lossy chargefree medium the wave equation based on Maxwells equations is of the form V2AS 72A 0 where As is either Es or Hs and y a jf3 is the propagation constant If we assume Es Exsz x we obtain EM waves of the form Ez t Eoeaz cos cof Pz ax Hz r Hoeaz cos wt 0z 0 av SUMMARY 463 where a attenuation constant j3 phase constant 77 rfln intrinsic imped ance of the medium The reciprocal of a is the skin depth 5 Ia The relationship between 3 w and X as stated above remain valid for EM waves 3 Wave propagation in other types of media can be derived from that for lossy media as special cases For free space set a 0 e sQ fi xo for lossless dielectric media set a 0 e eosr and n jxofxr and for good conductors set a e ea H fio or awe 0 4 A medium is classified as lossy dielectric lossless dielectric or good conductor depend ing on its loss tangent given by tan 6 Js h a coe where ec e je is the complex permittivity of the medium For lossless dielectrics tan0 C 1 for good conductors tan d J 1 and for lossy dielectrics tan 6 is of the order of unity 5 In a good conductor the fields tend to concentrate within the initial distance 6 from the conductor surface This phenomenon is called skin effect For a conductor of width w and length i the effective or ac resistance is awd where 5 is the skin depth 6 The Poynting vector 9 is the powerflow vector whose direction is the same as the di rection of wave propagation and magnitude the same as the amount of power flowing through a unit area normal to its direction f E X H 9ave 12 Re E X H 7 If a plane wave is incident normally from medium 1 to medium 2 the reflection coeffi cient F and transmission coefficient T are given by 12 Eio V2 V The standing wave ratio s is defined as i 1 r s 8 For oblique incidence from lossless medium 1 to lossless medium 2 we have the Fresnel coefficients as rj2cos 6 r cos 0 r2 cos 6 rjt cos 0 II 2j2 cos 6j 12 COS dt Tfj COS dj 464 M Electromagnetic Wave Propagation for parallel polarization and r2 COS 6 7i COS 8t i2 cos 6i ri cos 6 for perpendicular polarization As in optics T 2ry2 COS Oj rj2 cos 6i rjj cos sin i sin 0 02 Total transmission or no reflection F 0 occurs when the angle of incidence 0 is equal to the Brewster angle 101 Which of these is not a correct form of the wave Ex cos ut a cos Pz ut b sin Pz ut TT2 2Kt 2TT c cos I 1 A d Re ew3z e cos 0z ut 102 Identify which of these functions do not satisfy the wave equation a 50eM3z b sinw10z 5t c x 2tf d cos2 50 e sin x cos t f cos 5y 2x 103 Which of the following statements is not true of waves in general a It may be a function of time only b It may be sinusoidal or cosinusoidal c It must be a function of time and space d For practical reasons it must be finite in extent 104 The electric field component of a wave in free space is given by E 10 cos 107f kz av Vm It can be inferred that a The wave propagates along av b The wavelength X 1885 m REVIEW QUESTIONS 465 c The wave amplitude is 10 Vm d The wave number k 033 radm e The wave attenuates as it travels 105 Given that H 05 e rect sin 106 2x a Am which of these statements are incor a a 01 Npm b 0 2 radm c co 106rads d The wave travels along ax e The wave is polarized in the zdirection f The period of the wave is 1 ts 106 What is the major factor for determining whether a medium is free space lossless di electric lossy dielectric or good conductor a Attenuation constant b Constitutive parameters a e f c Loss tangent d Reflection coefficient 107 In a certain medium E 10 cos 108r 3y ax Vm What type of medium is it a Free space b Perfect dielectric c Lossless dielectric d Perfect conductor 108 Electromagnetic waves travel faster in conductors than in dielectrics a True b False 109 In a good conductor E and H are in time phase a True y b False 1010 The Poynting vector physically denotes the power density leaving or entering a given volume in a timevarying field a True b False Answers 101b 102df 103a 104bc 105bf 106c 107c 108b 109b 1010a 466 Electromagnetic Wave Propagation PROBLEMS 101 An EM wave propagating in a certain medium is described by E 25 sin 2TT X 106f 6x a Vm a Determine the direction of wave propagation b Compute the period T the wavelength X and the velocity u c Sketch the wave at t 0 778 774 772 102 a Derive eqs 1023 and 1024 from eqs 1018 and 1020 b Using eq 1029 in conjunction with Maxwells equations show that V y c From part b derive eqs 1032 and 1033 103 At 50 MHz a lossy dielectric material is characterized by e 36e0 p 21to and a 008 Sm If E 6eyx az Vm compute a y b X c u d r e H 104 A lossy material has x 5fio e 2eo If at 5 MHz the phase constant is 10 radm cal culate a The loss tangent b The conductivity of the material c The complex permittivity d The attenuation constant e The intrinsic impedance 105 A nonmagnetic medium has an intrinsic impedance 240 30 0 Find its a Loss tangent b Dielectric constant c Complex permittivity d Attenuation constant at 1 MHz 106 The amplitude of a wave traveling through a lossy nonmagnetic medium reduces by 18 every meter If the wave operates at 10 MHz and the electric field leads the mag netic field by 24 calculate a the propagation constant b the wavelength c the skin depth d the conductivity of the medium 107 Sea water plays a vital role in the study of submarine communications Assuming that for sea water a 4 Sm sr 80 xr 1 and 100 MHz calculate a the phase velocity b the wavelength c the skin depth d the intrinsic impedance 108 In a certain medium with x xo e 4e0 H 2e0Ay sin ir X 108 fiy ax Am find a the wave period T b the wavelength X c the electric field E d the phase difference between E and H PROBLEMS 467 109 In a medium E 16e005x sin 2 X 10st 2x az Vm find a the propagation constant b the wavelength c the speed of the wave d the skin depth 1010 A uniform wave in air has E 10COS2TT X 106f 0zav a Calculate 3 and X b Sketch the wave at z 0 A4 c FindH 1011 The magnetic field component of an EM wave propagating through a nonmagnetic medium p xo is H 25 sin 2 X 108 6x ay mAm Determine a The direction of wave propagation b The permittivity of the medium c The electric field intensity 1012 If H 10 sin oof 4zax mAm in a material for which a 0 ix xo e 4eo cal culate u X and Jd 1013 A manufacturer produces a ferrite material with JX 750xo e 5eo and a lT 6Smatl0MHz a Would you classify the material as lossless lossy or conducting b Calculate j3 and X c Determine the phase difference between two points separated by 2 m d Find the intrinsic impedance 1014 By assuming the timedependent fields E E oe i k r and H Hoekr where k kxax kyy kaz is the wave number vector and r xax ya zaz is the radius vector show that V X E dBdf can be expressed as k X E wH and deduce ak X aE aH 1015 Assume the same fields as in Problem 1014 and show that Maxwells equations in a sourcefree region can be written as k E 0 k H 0 k X E wftH k X H coeE 468 Electromagnetic Wave Propagation From these equations deduce k X a and ak X aw 1016 The magnetic field component of a plane wave in a lossless dielectric is H 30 sin 2ir X 108f 5 az mAm a Ifr lfinde b Calculate the wavelength and wave velocity c Determine the wave impedance d Determine the polarization of the wave e Find the corresponding electric field component f Find the displacement current density 1017 In a nonmagnetic medium E 50 cos 109f 8JC ay 40 sin 109f 8x az Vm find the dielectric constant er and the corresponding H 1018 In a certain medium E 10 cos 2TT X 107r Pxay az Vm If ix 50o e 2e0 and a 0 find 3 and H 1019 Which of the following media may be treated as conducting at 8 MHz a Wet marshy soil e 15eo x xo a 102 Sm b Intrinsic germanium e 16e0 p JXO a 0025 Sm c Sea water e 81eo ji ixo a 25 Sm 1020 Calculate the skin depth and the velocity of propagation for a uniform plane wave at fre quency 6 MHz traveling in polyvinylchloride pr 1 er 4 tan 8V 1 X 102 1021 A uniform plane wave in a lossy medium has a phase constant of 16 radm at 107 Hz and its magnitude is reduced by 60 for every 2 m traveled Find the skin depth and speed of the wave 1022 a Determine the dc resistance of a round copper wire a 58 X 107 Sm jxr 1 er 1 of radius 12 mm and length 600 m b Find the ac resistance at 100 MHz c Calculate the approximate frequency where dc and ac resistances are equal 1023 A 40mlong aluminum a 35 X 107 Sm fir 1 er 1 pipe with inner and outer radii 9 mm and 12 mm carries a total current of 6 sin 106 irf A Find the skin depth and the effective resistance of the pipe 1024 Show that in a good conductor the skin depth 8 is always much shorter than the wave length PROBLEMS 469 1025 Brass waveguides are often silver plated to reduce losses If at least the thickness of silver xo e eo a 61 X 107 Sm must be 55 find the minimum thickness required for a waveguide operating at 12 GHz 1026 A uniform plane wave in a lossy nonmagnetic media has Es 5ax 12aye7Z y 02 34m a Compute the magnitude of the wave at z 4 m b Find the loss in dB suffered by the wave in the interval 0 z 3 m c Calculate the Poynting vector at z 4 t 778 Take co 108 rads 1027 In a nonmagnetic material H 30 cos 2TT X 108f 6x a mAm find a the intrinsic impedance b the Poynting vector c the timeaverage power crossing the surface x 10 y 2 0 z 3 m 1028 Show that eqs 1067 and 1068 are equivalent 1029 In a transmission line filled with a lossless dielectric e 45eo fx ix0 E 40 sin ut 2z ap Vm 1030 find a co and H b the Poynting vector c the total timeaverage power crossing the surface z 1 m 2 mm p 3 mm 0 j 2TT a For a normal incidence upon the dielectricdielectric interface for which Mi M2 icn w e define R and Tas the reflection and transmission coefficients for average powers ie Pravc ve and Pume TPiawe Prove that R l 2 I 2 and T where M and n2 are the reflective indices of the media b Determine the ratio iiin2 so that the reflected and the transmitted waves have the same average power 1031 The plane wave E 30 cosw zax Vm in air normally hits a lossless medium p no e 4eo at z 0 a Find F r and s b Calculate the reflected electric and magnetic fields 1032 A uniform plane wave in air with H 4 sin wf 5x ay Am is normally incident on a plastic region with the parametersx fto e 4e0 andff 0 a Obtain the total electric field in air b Calculate the timeaverage power density in the plastic region c Find the standing wave ratio 470 Electromagnetic Wave Propagation 1033 A plane wave in free space with E 36 cos ut 3x ay Vm is incident normally on an interface at x 0 If a lossless medium with a 0 er 125 exits for x 0 and the reflected wave has H r 12 cos ut 3x a mAm find x2 1034 Region 1 is a lossless medium for which y s 0 x e 4eo whereas region 2 is free space y 0 If a plane wave E 5 cos 108 3y a Vm exists in region 1 find a the total electric field component of the wave in region 2 b the timeaverage Poynting vector in region 1 c the timeaverage Poynting vector in region 2 1035 A plane wave in free space z 0 is incident normally on a large block of material with er 12 xr 3 a 0 which occupies z 0 If the incident electric field is E 30 cos ut z ay Vm find a u b the standing wave ratio c the reflected magnetic field d the average power density of the transmitted wave 1036 A 30MHz uniform plane wave with H 10 sin ut fix az mAm exists in region x 0 having a 0 e 9eo p 4io At x 0 the wave encounters free space Determine a the polarization of the wave b the phase constant 3 c the displacement current density in region x 0 d the reflected and transmitted magnetic fields and e the average power density in each region 1037 A uniform plane wave in air is normally incident on an infinite lossless dielectric mater ial having e 3eo and x xo If the incident wave is E 10 cos ut z av Vm find a X and u of the wave in air and the transmitted wave in the dielectric medium b The incident H field c Tandr d The total electric field and the timeaverage power in both regions 1038 A signal in air z S 0 with the electric field component E 10 sin ut 3z ax Vm hits normally the ocean surface at z 0 as in Figure 1019 Assuming that the ocean surface is smooth and that s 80eo x io a 4 mhosm in ocean determine a co b The wavelength of the signal in air c The loss tangent and intrinsic impedance of the ocean d The reflected and transmitted E field 1039 Sketch the standing wave in eq 1087 at t 0 78 774 378 772 and so on where T 2itlu H PROBLEMS 471 Figure 1019 For Problem 1038 ocean S 80 o U flo T 4 1040 A uniform plane wave is incident at an angle 0 45 on a pair of dielectric slabs joined together as shown in Figure 1020 Determine the angles of transmission 0t and 62 in the slabs 1041 Show that the field Ev 20 sin kj cos kyy az 1042 where k2 x k aj2ioeo can be represented as the superposition of four propagating plane waves Find the corresponding H Show that for nonmagnetic dielectric media the reflection and transmission coefficients for oblique incidence become 2 cos 0 sin 0 r tan tan sin 0r 04 0 sin fit 0 sin 0 4 0 cos 0 0 2 cos 6i sin 6 sin 0 4 0 1043 A parallelpolarized wave in air with E 8a 6a sin cot Ay 3z Vm impinges a dielectric halfspace as shown in Figure 1021 Find a the incidence angle 0 b the time average in air t pt0 e e0 c the reflected and transmitted E fields free space free space Figure 1021 For Problem 1040 472 Electromagnetic Wave Propagation Figure 1021 For Problem 1043 Air E s0 M i E 4 K 1044 In a dielectric medium e 9eo n Mo a plane wave with H 02 cos 109f lex ay Am is incident on an air boundary at z 0 find a 0rand0 b k c The wavelength in the dielectric and air d The incident E e The transmitted and reflected E f The Brewster angle 1045 A plane wave in air with E 8ax 6a 5aj sin wt 3x Ay Vm is incident on a copper slab in y 0 Find u and the reflected wave Assume copper is a perfect conductor Hint Write down the field components in both media and match the boundary conditions 1046 A polarized wave is incident from air to polystyrene with fx no e 26e at Brewster angle Determine the transmission angle Chapter 1 7 TRANSMISSION LINES There is a story about four men named Everybody Somebody Anybody and Nobody There was an important job to be done and Everybody was asked to do it Everybody was sure that Somebody would do it Anybody could have done it but Nobody did it Somebody got angry about that because it was Everybodys job Everybody thought that Anybody could do it and Nobody realized that Everybody wouldnt do it It ended up that Everybody blamed Somebody when actually Nobody did what Anybody could have done ANONYMOUS 11 INTRODUCTION Our discussion in the previous chapter was essentially on wave propagation in unbounded media media of infinite extent Such wave propagation is said to be unguided in that the uniform plane wave exists throughout all space and EM energy associated with the wave spreads over a wide area Wave propagation in unbounded media is used in radio or TV broadcasting where the information being transmitted is meant for everyone who may be interested Such means of wave propagation will not help in a situation like telephone con versation where the information is received privately by one person Another means of transmitting power or information is by guided structures Guided structures serve to guide or direct the propagation of energy from the source to the load Typical examples of such structures are transmission lines and waveguides Waveguides are discussed in the next chapter transmission lines are considered in this chapter Transmission lines are commonly used in power distribution at low frequencies and in communications at high frequencies Various kinds of transmission lines such as the twistedpair and coaxial cables thinnet and thicknet are used in computer networks such as the Ethernet and internet A transmission line basically consists of two or more parallel conductors used to connect a source to a load The source may be a hydroelectric generator a transmitter or an oscillator the load may be a factory an antenna or an oscilloscope respectively Typical transmission lines include coaxial cable a twowire line a parallelplate or planar line a wire above the conducting plane and a microstrip line These lines are portrayed in Figure 111 Notice that each of these lines consists of two conductors in parallel Coaxial cables are routinely used in electrical laboratories and in connecting TV sets to TV antennas Mi crostrip lines similar to that in Figure 11 le are particularly important in integrated circuits where metallic strips connecting electronic elements are deposited on dielectric substrates Transmission line problems are usually solved using EM field theory and electric circuit theory the two major theories on which electrical engineering is based In this 473 474 Transmission Lines e Figure 111 Crosssectional view of typical transmission lines a coaxial line b twowire line c planar line d wire above conducting plane e microstrip line chapter we use circuit theory because it is easier to deal with mathematically The basic concepts of wave propagation such as propagation constant reflection coefficient and standing wave ratio covered in the previous chapter apply here Our analysis of transmission lines will include the derivation of the transmissionline equations and characteristic quantities the use of the Smith chart various practical appli cations of transmission lines and transients on transmission lines 112 TRANSMISSION LINE PARAMETERS It is customary and convenient to describe a transmission line in terms of its line parame ters which are its resistance per unit length R inductance per unit length L conductance per unit length G and capacitance per unit length C Each of the lines shown in Figure 111 has specific formulas for finding R L G and C For coaxial twowire and planar lines the formulas for calculating the values of R L G and C are provided in Table 111 The di mensions of the lines are as shown in Figure 112 Some of the formulas1 in Table 111 were derived in Chapters 6 and 8 It should be noted that 1 The line parameters R L G and C are not discrete or lumped but distributed as shown in Figure 113 By this we mean that the parameters are uniformly distrib uted along the entire length of the line Similar formulas for other transmission lines can be obtained from engineering handbooks or data bookseg M A R Guston Microwave Transmissionline Impedance Data London Van Nos trand Reinhold 1972 112 TRANSMISSION LINE PARAMETERS TABLE 111 Distributed Line Parameters at High Frequencies 475 Parameters R flm LHm G Sm C Fm Coaxial Line TwoWire Line 2x87 La 6 C a c b V b l n 2ir a i n 6 a 2TTE h 8 a i cosh d 7T 2a cosh 2a cosh1 2a Planar Line w8oe 8 0 w ow d BW d w aO 6 j Arn n skin depth of the conductor cosh In if 2a a I 2a 2 For each line the conductors are characterized by ac c ec eo and the homoge neous dielectric separating the conductors is characterized by a fi e 3 G MR R is the ac resistance per unit length of the conductors comprising the line and G is the conductance per unit length due to the dielectric medium separating the conductors 4 The value of L shown in Table 111 is the external inductance per unit length that is L Lext The effects of internal inductance Lm Rlui are negligible as high frequencies at which most communication systems operate 5 For each line G a LC lie and C Hl As a way of preparing for the next section let us consider how an EM wave propagates through a twoconductor transmission line For example consider the coaxial line connect ing the generator or source to the load as in Figure 114a When switch S is closed Figure 112 Common transmission lines a coaxial line b twowire line c planar line 476 Transmission Lines series R and L shunt G and C Figure 113 Distributed parameters of a twoconductor transmission line the inner conductor is made positive with respect to the outer one so that the E field is ra dially outward as in Figure 114b According to Amperes law the H field encircles the current carrying conductor as in Figure 114b The Poynting vector E X H points along the transmission line Thus closing the switch simply establishes a disturbance which appears as a transverse electromagnetic TEM wave propagating along the line This wave is a nonuniform plane wave and by means of it power is transmitted through the line IWV S I generator coaxial line a r load E field H field b Figure 114 a Coaxial line connecting the generator to the load b E and H fields on the coaxial line 113 TRANSMISSION LINE EQUATIONS 477 113 TRANSMISSION LINE EQUATIONS As mentioned in the previous section a twoconductor transmission line supports a TEM wave that is the electric and magnetic fields on the line are transverse to the direction of wave propagation An important property of TEM waves is that the fields E and H are uniquely related to voltage V and current respectively 112 V E d I p H d In view of this we will use circuit quantities V and in solving the transmission line problem instead of solving field quantities E and H ie solving Maxwells equations and boundary conditions The circuit model is simpler and more convenient Let us examine an incremental portion of length Az of a twoconductor transmission line We intend to find an equivalent circuit for this line and derive the line equations From Figure 113 we expect the equivalent circuit of a portion of the line to be as in Figure 115 The model in Figure 115 is in terms of the line parameters R L G and C and may represent any of the twoconductor lines of Figure 113 The model is called the Ltype equivalent circuit there are other possible types see Problem 111 In the model of Figure 115 we assume that the wave propagates along the zdirection from the gen erator to the load By applying Kirchhoffs voltage law to the outer loop of the circuit in Figure 115 we obtain Vz tRAz Iz t L Az dt Vz Az t or Vz Az t Vz t Az RIzt L dlz t dt 113 Izt y A W To generator Vz t o Vz Az t To load GAz CAz r z z Az Figure 115 Ltype equivalent circuit model of a differential length Az of a twoconductor transmission line 478 Transmission Lines Taking the limit of eq 113 as Az 0 leads to dt 114 Similarly applying Kirchoffs current law to the main node of the circuit in Figure 115 gives Iz t Iz Az t A Iz Az t GAz Vz Az t C Az dVz Azt or As A 0 eq 115 becomes dt dt 115 at 116 If we assume harmonic time dependence so that Vz t Re Vsz eJu Iz t Re Isz eJ 117b where Vsz and Isz are the phasor forms of Vz i and Iz t respectively eqs 114 and 116 become dV dz dz In the differential eqs 118 and 119 Vs and Is are coupled To separate them we take the second derivative of Vs in eq 118 and employ eq 119 so that we obtain R juL I3 uQ Vs dz juLG joQ Vs or dz lllOi 113 TRANSMISSION LINE EQUATIONS 479 where 7 a jf3 VR juLG ju 1111 By taking the second derivative of Is in eq 119 and employing eq 118 we get 1112 We notice that eqs 1110 and 1112 are respectively the wave equations for voltage and current similar in form to the wave equations obtained for plane waves in eqs 1017 and 1019 Thus in our usual notations y in eq 1111 is the propagation constant in per meter a is the attenuation constant in nepers per meter or decibels2 per meter and 3 is the phase constant in radians per meter The wavelength X and wave velocity u are re spectively given by X 2ir 1113 fK 1114 The solutions of the linear homogeneous differential equations 1110 and 1112 are similar to Case 2 of Example 65 namely Vsz 1115 and 1116 where Vg Vo 7tt and Io are wave amplitudes the and signs respectively denote wave traveling along z and zdirections as is also indicated by the arrows Thus we obtain the instantaneous expression for voltage as Vz t Re Vsz eM V eaz cos oit fa V eaz cos at 3z 1117 The characteristic impedance Zo of the line is the ratio of positively traveling voltage wave to current wave at any point on the line 2Recall from eq 1035 that 1 Np 8686 dB 480 Transmission Lines Zo is analogous to 77 the intrinsic impedance of the medium of wave propagation By sub stituting eqs 1115 and 1116 into eqs 118 and 119 and equating coefficients of terms eyz and elz we obtain V R joL 1 1118 or R juL RojXo I 1119 where Ro and Xo are the real and imaginary parts of Zo Ro should not be mistaken for R while R is in ohms per meter Ro is in ohms The propagation constant y and the character istic impedance Zo are important properties of the line because they both depend on the line parameters R L G and C and the frequency of operation The reciprocal of Zo is the char acteristic admittance Yo that is Yo 1ZO The transmission line considered thus far in this section is the lossy type in that the conductors comprising the line are imperfect ac and the dielectric in which the con ductors are embedded is lossy a 0 Having considered this general case we may now consider two special cases of lossless transmission line and distortionless line A Lossless Line R 0 G A transmission line is said lo be lossless if the conductors of the line are perfect rt oc and the dielectric medium separating them is lossless a 0 For such a line it is evident from Table 111 that when ac and a 0 i R 0 G 112 This is a necessary condition for a line to be lossless Thus for such a line eq 11201 forces eqs 1111 1114 and 1119 to become a 0 773 ju VLC W 1 P VLC 1121a 1121b 1121c 113 TRANSMISSION LINE EQUATIONS B Distortionless Line RL GC 481 A signal normally consists of a band of frequencies wave amplitudes of different fre quency components will be attenuated differently in a lossy line as a is frequency depen dent This results in distortion A distortionless line is one in which the attenuation constant a is frequency inde pendent while the phase constant i is linearly dependent on frequency From the general expression for a and 3 in eq 1111 a distortionless line results if the line parameters are such that R G LC 1122 Thus for a distortionless line or a VRG 3 u 1123a showing that a does not depend on frequency whereas 0 is a linear function of frequency Also R L G VC K JX or and 1123b u 0 VLC Note that 1123c 1 The phase velocity is independent of frequency because the phase constant lin early depends on frequency We have shape distortion of signals unless a and u are independent of frequency 2 u and Zo remain the same as for lossless lines 482 Transmission Lines TABLE 112 Transmission Line Characteristics Propagation Constant Characteristic Impedance Case 7 a yp Zo Ro jXo General VR joLG jui Lossless 0 jcovLC Distortionless VSG joivLC 3 A lossless line is also a distortionless line but a distortionless line is not necessar ily lossless Although lossless lines are desirable in power transmission telephone lines are required to be distortionless A summary of our discussion is in Table 112 For the greater part of our analysis we shall restrict our discussion to lossless transmission lines EXAMPLE 111 An air line has characteristic impedance of 70 fi and phase constant of 3 radm at 100 MHz Calculate the inductance per meter and the capacitance per meter of the line Solution An air line can be regarded as a lossless line since a 0 Hence R 0 G and a 0 13 LC Dividing eq 1111 by eq 1112 yields or 1111 1112 c 0 2ir X 100 X 10670 682 pFm From eq 1111 R2 OC 702682 X 1012 3342 nHm 113 TRANSMISSION LINE EQUATIONS 483 PRACTICE EXERCISE 111 A transmission line operating at 500 MHz has Zo 80 0 a 004 Npm 3 15 radm Find the line parameters R L G and C Answer 32 0m 382 nHm 5 X 104 Sm 597 pFm XAMPLE 112 A distortionless line has Zo 60 fl a 20 mNpm u 06c where c is the speed of light in a vacuum Find R L G C and X at 100 MHz Solution For a distortionless line RC GL or G RC and hence VRG R L Zo CO 1 or But LC From eq 1122b R a Zo 20 X 10360 12 fim Dividing eq 1121 by eq 1123 results in L A w 333 j j j j M 06 3 X 108 From eq 1122a ce2 400 X IP 6 G 333 uSm fl 12 1121 1122a 1122b 1123 484 Transmission Lines Multiplying eqs 1121 and 1123 together gives or C 1 1 uZ0 06 3 X 108 60 9259 pFm u A 06 3 X 10s z lo m 108 PRACTICE EXERCISE 112 A telephone line has R 30 0km L 100 mHkm G 0 and C 20 jtFkm At 1 kHz obtain a The characteristic impedance of the line b The propagation constant c The phase velocity Answer a 70751367 Q b 2121 X 104 78888 X 103m c 7069 X 105 ms 114 INPUT IMPEDANCE SWR AND POWER Consider a transmission line of length characterized by y and Zo connected to a load ZL as shown in Figure 116 Looking into the line the generator sees the line with the load as an input impedance Zin It is our intention in this section to determine the input impedance the standing wave ratio SWR and the power flow on the line Let the transmission line extend from z 0 at the generator to z at the load First of all we need the voltage and current waves in eqs 1115 and 1116 that is ysz yeTZ Veyz H24 V V Isz e TZ eyz 1125 where eq 1118 has been incorporated To find V and V the terminal conditions must be given For example if we are given the conditions at the input say Vo VZ 0 Iz 0 1126 114 INPUT IMPEDANCE SWR AND POWER 485 y z0 zL FISJUIT U6 a Input impedance due to a line terminated by a load b equivalent circuit for finding Vo and Io in terms of Zm at the input substituting these into eqs 1124 and 1125 results in V V ZJ V O V v O 0 0 VNOZJO 1127a 1127b If the input impedance at the input terminals is Zin the input voltage Vo and the input current Io are easily obtained from Figure 116b as On the other hand if we are given the conditions at the load say VL Vz L Iz Substituting these into eqs 1124 and 1125 gives VL 1128 1129 1130a 1130b 486 Transmission Lines Next we determine the input impedance Zin VszIsz at any point on the line At the generator for example eqs 1124 and 1125 yield Vsz 1131 Substituting eq 1130 into 1131 and utilizing the fact that cosh y yt e J e y sinh y 1132a or we get tanhy sinh yi e7 cosh y e7 7 7 ZL Zo tanh yt Zo ZL tanh yi lossy 1132b 1133 Although eq 1133 has been derived for the input impedance Zin at the generation end it is a general expression for finding Zin at any point on the line To find Zin at a distance V from the load as in Figure 116a we replace t by A formula for calculating the hyper bolic tangent of a complex number required in eq 1133 is found in Appendix A3 For a lossless line y j3 tanh3 j tan and Zo Ro so eq 1133 becomes ZL jZ0 tan Zo jZL tan j lossless 1134 showing that the input impedance varies periodically with distance from the load The quantity 3 in eq 1134 is usually referred to as the electrical length of the line and can be expressed in degrees or radians We now define TL as the voltage reflection coefficient at the load TL is the ratio of the voltage reflection wave to the incident wave at the load that is V 1135 Substituting V and VQ m eq 1130 into eq 1135 and incorporating VL ZJL gives zLzo zL zo 11361 14 INCUT IMPFDANCT SWR AND POWER 487 The voltage reflection coefficient at any point on the line is the ratio of the magni tude of the reflected voltage wave to that of the incident wave That is Tz But z Substituting and combining with eq 1135 we get 2yf 1137 The current reflection coefficient at any point on the line is negative of the voltage reflection coefficient at that point Thus the current reflection coefficient at the load is 1 ey11 e y TL Just as we did for plane waves we define the standing wave ratio s otherwise denoted by SWR as 1 min min L 1138 It is easy to show that max VmaxZo and min VminZo The input impedance Zin in eq 1134 has maxima and minima that occur respectively at the maxima and minima of the voltage and current standing wave It can also be shown that 1139a and linlmin max 1139b As a way of demonstrating these concepts consider a lossless line with characteristic impedance of Zo 50 U For the sake of simplicity we assume that the line is terminated in a pure resistive load ZL 100 0 and the voltage at the load is 100 V rms The condi tions on the line are displayed in Figure 117 Note from the figure that conditions on the line repeat themselves every half wavelength 488 Transmission Lines 5 0 V III 2A 1 A 2 3X 4 jr X 2 n 2 X 4 0 02 radian 0 wavelength Figure 117 Voltage and current wave patterns on a lossless line terminated by a resistive load As mentioned at the beginning of this chapter a transmission is used in transferring power from the source to the load The average input power at a distance from the load is given by an equation similar to eq 1068 that is where the factor is needed since we are dealing with the peak values instead of the rms values Assuming a lossless line we substitute eqs 1124 and 1125 to obtain r ixT i2 Fe23 re23 I 7 Since the last two terms are purely imaginary we have 2 27 1 If 1140 114 INPUT IMPEDANCE SWR AND POWER 489 The first term is the incident power Ph while the second term is the reflected power Pr Thus eq 1140 may be written as P P P rt r rr where Pt is the input or transmitted power and the negative sign is due to the negative going wave since we take the reference direction as that of the voltagecurrent traveling toward the right We should notice from eq 1140 that the power is constant and does not depend on since it is a lossless line Also we should notice that maximum power is de livered to the load when Y 0 as expected We now consider special cases when the line is connected to load ZL 0 ZL o and ZL Zo These special cases can easily be derived from the general case A Shorted Line Z 0 For this case eq 1134 becomes ZL0 jZo tan 3 Also 1141a 1141b We notice from eq 1141a that Zin is a pure reactance which could be capacitive or in ductive depending on the value of The variation of Zin with is shown in Figure 118a B OpenCircuited Line ZL In this case eq 1134 becomes and Zoc lim Zin jZo cot 3 zL j tan j8 rt i 1142a 1142b The variation of Zin with t is shown in Figure 118b Notice from eqs 1141a and 1142a that 1143 C Matched Line ZL ZJ This is the most desired case from the practical point of view For this case eq 1134 reduces to 77 1144a 490 Transmission Lines Inductive Capacitive Inductive Capacitive 118 Input impedance of a lossless line a when shorted b when open and 5 1 1144b that is Vo 0 the whole wave is transmitted and there is no reflection The incident power is fully absorbed by the load Thus maximum power transfer is possible when a transmission line is matched to the load EXAMPLE 113 A certain transmission line operating at co 106 rads has a 8 dBm 1 radm and Zo 60 j40 Q and is 2 m long If the line is connected to a source of 100 V Zs 40 ft and terminated by a load of 20 j50 ft determine a The input impedance b The sendingend current c The current at the middle of the line 114 INPUT IMPEDANCE SWR AND POWER 491 i Solution a Since 1 Np 8686 dB 8 0921 Npm 8686 7 a j5 0921 j m yt 20921 l 184 j2 Using the formula for tanhx jy in Appendix A3 we obtain tanhT 1033 7003929 7 l r Za 60 740 20 75O 60 740X1033 7003929 I 60 740 20 75OlO33 7003929 J Zin 6025 73879 U b The sendingend current is z 0 o From eq 1128 V 10 Kz 0 Zm Zg 6025 j3879 40 93032115mA c To find the current at any point we need V and V But Io Iz 0 93032115mA Vo ZJO 716632770093032115 66671162 V From eq 1127 66671162 60 7400093032115 66871208 V V o ZJO 00518260 At the middle of the line z ill yz 0921 7I Hence the current at this point is 6687 e 60 740 00518e260a9211 60 40 492 Transmission Lines Note that jl is in radians and is equivalent to j573 Thus Uz 2 005l8ej26ae 26ae032lei573 72 l3369 721e 3369 00369e7891 0001805e28361 6673 j34456 raA 3510281 mA PRACTICE EXERCISE 113 A 40mlong transmission line shown in Figure 119 has Vg 15O2Vrms Zo 30 j60 Q and VL 548 Vms If the line is matched to the load calculate a The input impedance Zin b The sendingend current lm and voltage Vm c The propagation constant y Answer a 3060Q b 0U263430 A 75OV c 00101 j02094 lm Zo 30y60 v z 40m iyurc ill For Practice Exercise 113 115 THESMSTH CHART Prior to the advent of digital computers and calculators engineers developed all sorts of aids tables charts graphs etc to facilitate their calculations for design and analysis To reduce the tedious manipulations involved in calculating the characteristics of transmis sion lines graphical means have been developed The Smith chart3 is the most commonly used of the graphical techniques It is basically a graphical indication of the impedance of a transmission line as one moves along the line It becomes easy to use after a small amount of experience We will first examine how the Smith chart is constructed and later 3Devised by Phillip H Smith in 1939 See P H Smith Transmission line calculator Electronics vol 12 pp 2931 1939 and P H Smith An improved transmission line calculator Electronics vol 17 pp 130133318325 1944 115 THE SMITH CHART 493 Figure 1110 Unit circle on which the Smith chart is constructed employ it in our calculations of transmission line characteristics such as TL s and Zin We will assume that the transmission line to which the Smith chart will be applied is lossless Zo Ro although this is not fundamentally required The Smith chart is constructed within a circle of unit radius F 1 as shown in Figure 1110 The construction of the chart is based on the relation in eq 11364 that is zLzo 1145 or r 1146 where Fr and F are the real and imaginary parts of the reflection coefficient F Instead of having separate Smith charts for transmission lines with different charac teristic impedances such as Zo 60100 and 120 fl we prefer to have just one that can be used for any line We achieve this by using a normalized chart in which all impedances are normalized with respect to the characteristic impedance Zo of the particular line under con sideration For the load impedance ZL for example the normalized impedance ZL is given by 1147 1148a Substituting eq 1147 into eqs 1145 and 1146 gives or ZL r jx 1 1148b Whenever a subscript is not attached to F we simply mean voltage reflection coefficient at the load 494 Transmission Lines Normalizing and equating components we obtain i r r r X 2I r r Rearranging terms in eq 1149 leads to 1 r r 1 r 2 and X i Each of eqs 1150 and 1151 is similar to x hf y kf a2 1149a 1149b 1150 1151 1152 which is the general equation of a circle of radius a centered at h k Thus eq 1150 is an rcircle resistance circle with center at TV T radius 1 r 1 0 1 r 1153a 1153b For typical values of the normalized resistance r the corresponding centers and radii of the rcircles are presented in Table 113 Typical examples of the rcircles based on the data in TABLE 113 Radii and Centers of rCircles for Typical Values of r Normalized Resistance r 0 12 1 2 5 V1 r 1 23 12 13 16 0 r 1 r 00 130 12 0 23 0 56 0 10 115 THE SMITH CHART 495 Figure 1111 Typical rcircles for r 005 12 5 o Table 113 are shown in Figure 1111 Similarly eq 1151 is an xcircle reactance circle with center at Tr T 1 radius x 1154a 1154b Table 114 presents centers and radii of the xcircles for typical values of x and Figure 1112 shows the corresponding plots Notice that while r is always positive x can be posi tive for inductive impedance or negative for capacitive impedance If we superpose the rcircles and xcircles what we have is the Smith chart shown in Figure 1113 On the chart we locate a normalized impedance z 2 j for example as the point of intersection of the r 2 circle and the x 1 circle This is point Px in Figure 1113 Similarly z 1 7 05 is located at P2 where the r 1 circle and the x 05 circle intersect Apart from the r and xcircles shown on the Smith chart we can draw the scircles or constant standingwaveratio circles always not shown on the Smith chart which are centered at the origin with s varying from 1 to 00 The value of the standing wave ratio s is TABLE 114 Radii and Centers of xCircles for Typical Value of x Normalized Reactance x Radius V Center 1 x 0 12 1 2 5 oc 2 1 12 15 0 1 1 1 1 1 1 0 2 1 12 15 0 496 Transmission Lines 01 05 Figure 1112 Typical circles for x 0 12 1 2 5 oo 00 Figure 1113 Illustration of the r x and circles on the Smith chart 115 THE SMITH CHART 497 determined by locating where an scircle crosses the Tr axis Typical examples of circles for s 12 3 and are shown in Figure 1113 Since F and s are related according to eq 1138 the circles are sometimes referred to as Fcircles with F varying linearly from 0 to 1 as we move away from the center O toward the periphery of the chart while s varies nonlinearly from 1 to The following points should be noted about the Smith chart 1 At point Psc on the chart r 0 x 0 that is ZL 0 jQ showing that Psc repre sents a short circuit on the transmission line At point Poc r and x or ZL c 70C which implies that Poc corresponds to an open circuit on the line Also at Poc r 0 and x 0 showing that Poc is another location of a short circuit on the line 2 A complete revolution 360 around the Smith chart represents a distance of A2 on the line Clockwise movement on the chart is regarded as moving toward the generator or away from the load as shown by the arrow G in Figure 1114a and b Similarly counterclockwise movement on the chart corresponds to moving toward the load or away from the generator as indicated by the arrow L in Figure 1114 Notice from Figure 1114b that at the load moving toward the load does not make sense because we are already at the load The same can be said of the case when we are at the generator end 3 There are three scales around the periphery of the Smith chart as illustrated in Figure 1114a The three scales are included for the sake of convenience but they are actually meant to serve the same purpose one scale should be sufficient The scales are used in determining the distance from the load or generator in degrees or wavelengths The outermost scale is used to determine the distance on the line from the generator end in terms of wavelengths and the next scale determines the dis tance from the load end in terms of wavelengths The innermost scale is a protrac tor in degrees and is primarily used in determining 6 it can also be used to de termine the distance from the load or generator Since a A2 distance on the line corresponds to a movement of 360 on the chart A distance on the line corresponds to a 720 movement on the chart 720 1155 Thus we may ignore the other outer scales and use the protractor the innermost scale for all our dr and distance calculations Knax occurs where Zin max is located on the chart see eq 1139a and that is on the positive Tr axis or on OPOC in Figure 1114a Vmin is located at the same point where we have Zin min on the chart that is on the negative Tr axis or on OPsc in Figure 1114a Notice that Vmax and Vmin orZjnmax andZinmin are A4 or 180 apart The Smith chart is used both as impedance chart and admittance chart Y 1Z As admittance chart normalized impedance y YIYO g jb the g and b circles correspond to r and xcircles respectively 498 Transmission Lines a C I Genenitoi Transmission line h Load Figure 1114 a Smith chart illustrating scales around the periphery and movements around the chart b corresponding movements along the trans mission line Based on these important properties the Smith chart may be used to determine among other things a T T6r and s b Zin or Ym and c the locations of Vmax and Vmin provided that we are given Zo ZL and the length of the line Some examples will clearly show how we can do all these and much more with the aid of the Smith chart a compass and a plain straightedge 115 THE SMITH CHART 499 EXAMPLE 114 A 30mlong lossless transmission line with Zo 50 Q operating at 2 MHz is terminated with a load ZL 60 j40 0 If u 06c on the line find a The reflection coefficient V b The standing wave ratio s c The input impedance Solution This problem will be solved with and without using the Smith chart Method 1 Without the Smith chart ZL Zo 60 7 4 0 5 0 10 j40 a r ZL Zo 50 j40 50 110 40 0352356 IjT K T 1 03523 c Since u ufi or fi WM 2 X 10630 2TT u 06 3 X 108 Note that fit is the electrical length of the line L 7ZO tan 120 Zo 7ZL tan 5060 740 750tanl20 50 y60 740 tan 120 5 4V3 j6V3 2397 7135 0 Method 2 Using the Smith chart a Calculate the normalized load impedance h 6 0 ZLYO 50 12 7O8 Locate zL on the Smith chart of Figure 1115 at point P where the r 12 circle and the x 08 circle meet To get V at zL extend OP to meet the r 0 circle at Q and measure OP and 0g Since OQ corresponds to T 1 then at P OP OQ 32 cm 91cm 03516 500 Transmission Lines 56 Figure 1115 For Example 114 Note that OP 32 cm and OQ 91 cm were taken from the Smith chart used by the author the Smith chart in Figure 1115 is reduced but the ratio of OPIOQ remains the same Angle 0r is read directly on the chart as the angle between OS and OP that is Thus 6T angle POS 56 T 0351656 b To obtain the standing wave ratio s draw a circle with radius OP and center at O This is the constant s or T circle Locate point S where the circle meets the Fraxis 115 THE SMITH CHART 501 This is easily shown by setting T 0 in eq 1149a The value of r at this point is s that is 5 rforr 1 21 c To obtain Zin first express in terms of X or in degrees u 06 3 X 108 2 X 106 90 m Since X corresponds to an angular movement of 720 on the chart the length of the line corresponds to an angular movement of 240 That means we move toward the generator or away from the load in the clockwise direction 240 on the scircle from point P to point G At G we obtain Hence zin 047 yO035 50047 70035 235 jl15 0 Although the results obtained using the Smith chart are only approximate for engineering purposes they are close enough to the exact ones obtained in Method 1 PRACTICE EXERCISE 114 A 70Q lossless line has s 16 and 0r 300 If the line is 06X long obtain a TZLZin b The distance of the first minimum voltage from the load Answer a 0228 300 805 V336 fi 476 yl75 Q b X6 EXAMPLE 115 A 100 7150C load is connected to a 75fl lossless line Find a T b s c The load admittance YL d Zin at 04X from the load e The locations of Vmax and Vmin with respect to the load if the line is 06X long f Zin at the generator 502 Transmission Lines Solution a We can use the Smith chart to solve this problem The normalized load impedance is ZL 100 150 1332 We locate this at point P on the Smith chart of Figure 1116 At P we obtain Hence 1 OQ 91cm 0r angle POS 40 T 0659 740 Figure 1116 For Example 115 115 THE SMITH CHART 503 Check zo T ZL Zo TooT150 75 0659 4F b Draw the constant scircle passing through P and obtain s 482 Check 1 r 1 0659 S TW 1659 c To obtain YL extend PO to POP and note point P where the constant circle meets POP At P obtain yL 0228 jO35 The load admittance is YoyL 0228 jO35 304 467 mS Check Y ZL 100J150 307 J462 mS d 04X corresponds to an angular movement of 04 X 720 288 on the constant circle From P we move 288 toward the generator clockwise on the circle to reach point R At R Hence Check Xm 03 J063 Zin ZoZm 75 03 y063 225 J4125 0 Y 04A 360 04 144 ZL jZo tan j8 Zo jZL tan 3 75100 J150 l7Ty100 54416525 504 Transmission Lines or Zin 219y476O e 06X corresponds to an angular movement of 06 X 720 432 1 revolution 72 Thus we start from P load end move along the circle 432 or one revolution plus 72 and reach the generator at point G Note that to reach G from P we have passed through point T location of Vmin once and point S location of Vmax twice Thus from the load lstVmax is located at 40 X 0055X 2ndTmax is located at 00555X 0555X and the only Vmin is located at 0055X X4 03055X f At G generator end Zin 1822 Zin 7518 J22 135 jl65 Q This can be checked by using eq 1134 where 06X 216 X We can see how much time and effort is saved using the Smith chart PRACTICE EXERCISE 115 A lossless 60fi line is terminated by a 60 y60fl load a Find T and s If Zin 120 y60 fi how far in terms of wavelengths is the load from the generator Solve this without using the Smith chart b Solve the problem in a using the Smith chart Calculate Zmax and Zin min How far in terms of X is the first maximum voltage from the load Answer a 044726343 2618 1 An n 0 1 2 b 0445762 2612 1 4n 1571 0 2292 Q 00861 X 116 SOME APPLICATIONS OF TRANSMISSION LINES 505 116 SOME APPLICATIONS OF TRANSMISSION LINES Transmission lines are used to serve different purposes Here we consider how transmis sion lines are used for load matching and impedance measurements A QuarterWave Transformer Matching When Zo ZL we say that the load is mismatched and a reflected wave exists on the line However for maximum power transfer it is desired that the load be matched to the trans mission line Zo Z so that there is no reflection F Oors 1 The matching is achieved by using shorted sections of transmission lines We recall from eq 1134 that when t X4 or 3 2TTXX4 TT2 in that is or Zo jZL tan TT2 J zo z L 1 Yin Z L 1156 1157 Thus by adding a X4 line on the Smith chart we obtain the input admittance correspond ing to a given load impedance Also a mismatched load ZL can be properly matched to a line with characteristic im pedance Zo by inserting prior to the load a transmission line X4 long with characteristic impedance Zo as shown in Figure 1117 The X4 section of the transmission line is called a quarterwave transformer because it is used for impedance matching like an ordinary transformer From eq 1156 Zo is selected such that Zin Zo z 1158 Figure 1117 Load matching using a X4 transformer Z Z L 506 Transmission Lines Figure 1118 Voltage standing wave pattern of mismatched load a without a A4 transformer b with a A4 transformer a where Zo Zo and ZL are all real If for example a 1200 load is to be matched to a 75fi line the quarterwave transformer must have a characteristic impedance of V 75 120 95 fi This 95fl quarterwave transformer will also match a 75fl load to a 120fi line The voltage standing wave patterns with and without the X4 transformer are shown in Figure 1118a and b respectively From Figure 1118 we observe that although a standing wave still exists between the transformer and the load there is no standing wave to the left of the transformer due to the matching However the reflected wave or standing wave is eliminated only at the desired wavelength or frequency there will be reflection at a slightly different wavelength Thus the main disadvantage of the quarterwave trans former is that it is a narrowband or frequencysensitive device B SingleStub Tuner Matching The major drawback of using a quarterwave transformer as a linematching device is eliminated by using a singlestub tuner The tuner consists of an open or shorted section of transmission line of length d connected in parallel with the main line at some distance from the load as in Figure 1119 Notice that the stub has the same characteristic imped ance as the main line It is more difficult to use a series stub although it is theoretically fea sible An opencircuited stub radiates some energy at high frequencies Consequently shunt shortcircuited parallel stubs are preferred As we intend that Zin Zo that is zin 1 or yin 1 at point A on the line we first draw the locus y 1 jbr 1 circle on the Smith chart as shown in Figure 1120 If a shunt stub of admittance ys jb is introduced at A then jb jb 1159 Figure 1119 Matching with a singlestub tuner shorted stub 116 SOME APPLICATIONS OF TRANSMISSION LINES 507 Figure 1120 Using the Smith chart to determine and d of a shuntshorted locusofjl6 singlestub tuner 1 circle as desired Since b could be positive or negative two possible values of X2 can be found on the line At A ys jb A and at B ys jb iB as in Figure 1120 Due to the fact that the stub is shorted yL we determine the length d of the stub by finding the distance from Psc at which zL 0 jO to the required stub admittance ys For the stub at A we obtain d dA as the distance from P to A where A corresponds to ys jb located on the periphery of the chart as in Figure 1120 Similarly we obtain d dB as the distance from Psc to B ys jb Thus we obtain d dA and d dB corresponding to A and B respectively as shown in Figure 1120 Note that dA dB A2 always Since we have two possible shunted stubs we normally choose to match the shorter stub or one at a position closer to the load Instead of having a single stub shunted across the line we may have two stubs This is called doublestub matching and allows for the adjustment of the load impedance C Slotted Line Impedance Measurement At high frequencies it is very difficult to measure current and voltage because measuring devices become significant in size and every circuit becomes a transmission line The slotted line is a simple device used in determining the impedance of an unknown load at high frequencies up into the region of gigahertz It consists of a section of an air lossless line with a slot in the outer conductor as shown in Figure 1121 The line has a probe along the E field see Figure 114 which samples the E field and consequently measures the po tential difference between the probe and its outer shield The slotted line is primarily used in conjunction with the Smith chart to determine the standing wave ratio v the ratio of maximum voltage to the minimum voltage and the load impedance ZL The valueof s is read directly on the detection meter when the load is connected To determine ZL we first replace the load by a short circuit and note the locations of voltage minima which are more accurately determined than the maxima because of the sharpness of the turning point on the scale Since impedances repeat every half wavelength any of the minima may be selected as the load reference point We now determine the distance from the selected reference point to the load by replacing the short circuit by the load and noting the locations of voltage minima The distance dis 508 Transmission Lines To Generator To detector Probe slotted line 0 V 50 cm t 1 I I j To load or short l l I I calibrated scale a X2 25 A Load p 1 Short 50 cm b Figure 1121 a Typical slotted line b determining the location of the load Zt and Vmin on the line tance of Vmin toward the load expressed in terms of X is used to locate the position of the load of an scircle on the chart as shown in Figure 1122 We could also locate the load by using which is the distance of Vmin toward the generator Either i or may be used to locate ZL The procedure involved in using the slotted line can be summarized as follows 1 With the load connected read s on the detection meter With the value of s draw the scircle on the Smith chart 2 With the load replaced by a short circuit locate a reference position for zL at a voltage minimum point 3 With the load on the line note the position of Vmin and determine i 4 On the Smith chart move toward the load a distance from the location of V Find ZL at that point scircle i distance toward load Figure 1122 Determining the load imped distance toward generator ance from the Smith chart using the data obtained from the slotted line 116 SOME APPLICATIONS OF TRANSMISSION LINES 509 EXAMPLE 116 With an unknown load connected to a slotted air line 5 2 is recorded by a standing wave indicator and minima are found at 11 cm 19 cm on the scale When the load is re placed by a short circuit the minima are at 16 cm 24 cm If Zo 50 Q calculate X and ZL Solution Consider the standing wave patterns as in Figure 1123a From this we observe that 1 9 1 1 8 cm or X 1 6 c m i 1875 GHz A 16 X 1CT Electrically speaking the load can be located at 16 cm or 24 cm If we assume that the load is at 24 cm the load is at a distance from Vmin where 2 4 1 9 5 cm X 03125 X 16 Figure 1123 Determining ZL using the slotted line a wave pattern b Smith with load c h a r t f o r Example 116 with short 510 Transmission Lines This corresponds to an angular movement of 03125 X 720 225 on the s 2 circle By starting at the location of Vmin and moving 225 toward the load counterclockwise we reach the location of zL as illustrated in Figure 1123b Thus and ZL ZjL 50 14 y075 70 375 fi PRACTICE EXERCISE 116 The following measurements were taken using the slotted line technique with load s 18 Vmax occurred at 23 cm 335 cm with short s Vmax occurred at 25 cm 375 cm If Zo 50 0 determine ZL Answer 325 jll5 fi EXAMPLE 117 Antenna with impedance 40 J30Q is to be matched to a 100fl lossless line with a shorted stub Determine a The required stub admittance b The distance between the stub and the antenna c The stub length d The standing wave ratio on each ratio of the system Solution a ZL 77 40 y30 100 04 j03 Locate zL on the Smith chart as in Figure 1124 and from this draw the scircle so that yL can be located diametrically opposite zL Thus yL 16 jl2 Alternatively we may find yL using zL 100 40 30 16 7I2 Locate points A and B where the scircle intersects the g 1 circle At A ys yl04 and at B ys 104 Thus the required stub admittance is Ys Yoys jM jl04 mS Both j104 mS and jlOA mS are possible values 116 SOME APPLICATIONS OF TRANSMISSION LINES 511 104 Figure 1124 For Example 117 b From Figure 1124 we determine the distance between the load antenna in this case yL and the stub At A AtB 62 39 720 D12 Transmission Lines c Locate points A and B corresponding to stub admittance 7104 and 7104 respec tively Determine the stub length distance from Psc to A and B dA 720 272X 720 X 01222X 03778X Notice that dA dB 05X as expected d From Figure 1124 s 27 This is the standing wave ratio on the line segment between the stub and the load see Figure 1118 1 to the left of the stub because the line is matched and s along the stub because the stub is shorted PRACTICE EXERCISE 117 A 750 lossless line is to be matched to a 100 y80fl load with a shorted stub Calculate the stub length its distance from the load and the necessary stub admit tance Answer A 0093X lB 0272X dA 0126X dB 0374X yl267 mS 117 TRANSIENTS ON TRANSMISSION LINES In our discussion so far we have assumed that a transmission line operates at a single fre quency In some practical applications such as in computer networks pulsed signals may be sent through the line From Fourier analysis a pulse can be regarded as a superposition of waves of many frequencies Thus sending a pulsed signal on the line may be regarded as the same as simultaneously sending waves of different frequencies As in circuit analysis when a pulse generator or battery connected to a transmission line is switched on it takes some time for the current and voltage on the line to reach steady values This transitional period is called the transient The transient behavior just after closing the switch or due to lightning strokes is usually analyzed in the frequency domain using Laplace transform For the sake of convenience we treat the problem in the time domain Consider a lossless line of length and characteristic impedance Zo as shown in Figure 1125a Suppose that the line is driven by a pulse generator of voltage Vg with in ternal impedance Zg at z 0 and terminated with a purely resistive load ZL At the instant t 0 that the switch is closed the starting current sees only Zg and Zo so the initial sit uation can be described by the equivalent circuit of Figure 1125b From the figure the starting current at z 0 t 0 is given by 70 0 1160 117 TRANSIENTS ON TRANSMISSION LINES 513 2 0 b Figure 1125 Transients on a transmission line a a line driven by a pulse generator b the equivalent circuit at z 0 t 0 and the initial voltage is zg z0 g 1161 After the switch is closed waves I Io and V Vo propagate toward the load at the speed 1 u VLC 1162 Since this speed is finite it takes some time for the positively traveling waves to reach the load and interact with it The presence of the load has no effect on the waves before the transit time given by 1163 After ti seconds the waves reach the load The voltage or current at the load is the sum of the incident and reflected voltages or currents Thus to v v vo rLvo i rtvo and li to 1 I l0 Ll0 I V Ll0 where TL is the load reflection coefficient given in eq 1136 that is T ZL Zo 1164 1165 1166 The reflected waves V rLVo and r t 0 travel back toward the generator in addi tion to the waves Vo and o already on the line At time t 2fb the reflected waves have reached the generator so V0 2tO V V TGTLVO TLVO 514 Transmission Lines rrL z 8 a b Figure 1126 Bounce diagram for a a voltage wave and b a current wave or and or V0 2t 02t i r 1167 rLo where FG is the generator reflection coefficient given by Zg Zo 1168 1169 Again the reflected waves from the generator end V TGTLVO and TQTJO prop agate toward the load and the process continues until the energy of the pulse is actually ab sorbed by the resistors Zg and ZL Instead of tracing the voltage and current waves back and forth it is easier to keep track of the reflections using a bounce diagram otherwise known as a lattice diagram The bounce diagram consists of a zigzag line indicating the position of the voltage or current wave with respect to the generator end as shown in Figure 1126 On the bounce diagram the voltage or current at any time may be determined by adding those values that appear on the diagram above that time EXAMPLE 118 For the transmission line of Figure 1127 calculate and sketch a The voltage at the load and generator ends for 0 t 6 fx b The current at the load and generator ends for 0 t 6 s ioo n 12V 117 TRANSIENTS ON TRANSMISSION LINES 1 Figure 1127 For Example 515 z o 50 a ii 108ms 200 n 100 m Solution a We first calculate the voltage reflection coefficients at the generator and load ends Zg Zo 100 50 J G Zg Zo 100 50 3 ZL Zo 200 50 3 200 50 100 The transit time t r 1 us u 108 The initial voltage at the generator end is Z 50 im4V The 4 V is sent out to the load The leading edge of the pulse arrives at the load at t t 1 us A portion of it 435 24 V is reflected back and reaches the generator at t 2tl 2 us At the generator 2413 08 is reflected and the process continues The whole process is best illustrated in the voltage bounce diagram of Figure 1128 V 72 048 016 784 K 784 0096 0032 7968 V 768 016 0096 7936 V 7936 003 002 7986 Figure 1128 Voltage bounce diagram for Example 118 516 Transmission Lines From the bounce diagram we can sketch V0 t and V t as functions of time as shown in Figure 1129 Notice from Figure 1129 that as t the voltages approach an asymptotic value of This should be expected because the equivalent circuits at t 0 and t are as shown in Figure 1130 see Problem 1146 for proof b The current reflection coefficients at the generator and load ends are FG 13 and TL 35 respectively The initial current is V0t Volts 7968 Figure 1129 Voltage not to scale a at the generator end b at the load end 24 IL 08 10 t MS a VHt 768 7936 64 24 048 016 J 8 10 0096 b tins 117 TRANSIENTS ON TRANSMISSION LINES 517 a b Figure 1130 Equivalent circuits for the line in Figure 1127 for a t 0 and b Again 70 t and are easily obtained from the current bounce diagram shown in Figure 1131 These currents are sketched in Figure 1132 Note that 0 V tIZL Hence Figure 1132b can be obtained either from the current bounce diagram of Figure 1131 or by scaling Figure 1129b by a factor of IZL 1200 Notice from Figures 1130b and 1132 that the currents approach an asymptotic value of 12 Z ZL 300 40 mA z r 35 7 416 192 0644032 Figure 1131 Current bounce diagram for Example 118 518 Transmission Lines 0 0 mA 80 48 416 40 16 80mA 32 064 8 10 v 0384 48 a Figure 1132 Current not to scale a at the generator end b at the load end for Example 118 r J f i t l is is PRACTICE EXERCISE 118 Repeat Example 118 if the transmission line is a Shortcircuited b Opencircuited Answer a See Figure 1133 b See Figure 1134 Ms V0t 4V 4V 43 1 49 O t 43 4 160 mA 12445 10667 80 809 803 803 Ms 80 mA 13333 80 1155 809 is Figure 1133 For Practice Exercise 118a 519 520 Transmission Lines t 8V 4V 1 1 1067 43 1 1155 49 1 12V 0 t 0A 0 V0 t tins AV 9333 4V 43 1 4 iiil 49 12V 1 1 0 10 t 80 mA 80 803 8 0 3 809 8 2 9 1 1 803 Figure 1134 For Practice Exercise 118b 117 TRANSIENTS ON TRANSMISSION LINES 521 EXAMPLE 119 A 75fi transmission line of length 60 m is terminated by a 100fi load If a rectangular pulse of width 5 xs and magnitude 4 V is sent out by the generator connected to the line sketch 70 t and t for 0 t 15 xs Take Zg 25 Q and u 01c Solution In the previous example the switching on of a battery created a step function a pulse of in finite width In this example the pulse is of finite width of 5 xs We first calculate the voltage reflection coefficients z o 77 L o The initial voltage and transit time are given by 75 Too t 60 OI 3 X 1O8 2 The time taken by Vo to go forth and back is 2f 4 xs which is less than the pulse dura tion of 5 is Hence there will be overlapping The current reflection coefficients are and r G The initial current 100 40 mA Let i and r denote incident and reflected pulses respectively At the generator end 0 t 5 xs Ir Io 40 mA 4 t 9 40 5714 Ir 5714 2857 t 13 2857 04082 Ir 04082 02041 522 Transmission Lines 12 t 17 7 02041 00292 Ir 00292 00146 and so on Hence the plot of 70 0 versus t is as shown in Figure 1135a 10 mA 40 1 1 1 2 4 31 43 1 6 1 06123 8 2857 714 l I 10 02041 959 04082 05685 12 1 1 hr 14 00146 1 l 1 1 J 00438 00292 Vit Volts 3429 8571 185 a 00306 00176 I l I I I I I I I I L S 1012 14 02143 6246 0228 b mA Figure 1135 For Example 119 not to scale MS MS 01 1 343 1 1 4 n i i i 6 3 8 46 1 10 01 J 76 1 1 14 I I I 117 TRANSIENTS ON TRANSMISSION LINES 523 At the load end 0 t 2 iis V 0 2 t 7 V 3 Vr 3 04296 6 t 11 V 04296 02143 Vr 02143 00306 10 t 14 Vf 00306 00154 Vr 00154 00022 and so on From V t we can obtain t as The plots of V t and t are shown in Figure 1135b and c PRACTICE EXERCISE 119 Repeat Example 119 if the rectangular pulse is replaced by a triangular pulse of Figure 1136 Answer omax 100 mA See Figure 1137 for the current waveforms 10 V Figure 1136 Triangular pulse of Practice Exercise 119 is 524 Transmission Lines 10 t mA 100 Figure il37 Current waves for Practice Exercise 119 1521 8 10 2143 MS a t mA 0 2 4 8571 6 m 8 10 04373 A 12 6122 b 118 MICROSTRIP TRANSMISSION LINES Microstrip lines belong to a group of lines known as parallelplate transmission lines They are widely used in presentday electronics Apart from being the most commonly used form of transmission lines for microwave integrated circuits microstrips are used for circuit components such as filters couplers resonators antennas and so on In comparison with the coaxial line the microstrip line allows for greater flexibility and compactness of design A microstrip line consists of a single ground plane and an open strip conductor sepa rated by dielectric substrate as shown in Figure 1138 It is constructed by the photographic processes used for integrated circuits Analytical derivation of the characteristic properties Strip conductor Figure 1138 Microstrip line Ground plane 118 MICROSTRIP TRANSMISSION LINES 525 of the line is cumbersome We will consider only some basic valid empirical formulas nec essary for calculating the phase velocity impedance and losses of the line Due to the open structure of the microstrip line the EM field is not confined to the di electric but is partly in the surrounding air as in Figure 1139 Provided the frequency is not too high the microstrip line will propagate a wave that for all practical purposes is a TEM wave Because of the fringing the effective relative permittivity eeff is less than the relative permittivity er of the substrate If w is the line width and h is the substrate thick ness an a approximate value of eeff is given by Seff er Sr 2Vl I2hw 1170 The characteristic impedance is given by the following approximate formulas 7 60 h vtA In I V hj 1 1207T wh 1393 0667 In wh 1444 wlh wh 1 j 1171 The characteristic impedance of a wide strip is often low while that of a narrow strip is high Magnetic field Electric field Figure 1139 Pattern of the EM field of a microstrip line Source From D Roddy Microwave Technology 1986 by permission of Prentice Hall 526 Transmission Lines For design purposes if er and Zo are known the ratio wlh necessary to achieve Zo is given by w e2A 2 B 2er where B 1 039 A 60 061 J 60 V 2 er 1 wlh 2 wh 2 1172 ZoVer From the knowledge of eeff and Zo the phase constant and the phase velocity of a wave propagating on the microstrip are given by c c u 1174a 1174b where c is the speed of light in a vacuum The attenuation due to conduction or ohmic loss is in dBm ar 8686 1175 where is the skin resistance of the conductor The attenuation due to dielectric ac8 loss is in dBm ad 273 fceff ler tanfl sr 1 e 1176 eff where uf is the line wavelength and tan 6 alice is the loss tangent of the substrate The total attenuation constant is the sum of the ohmic attenuation constant ac and the di electric attenuation constant ad that is a ac ad 11771 Sometimes ad is negligible in comparison with ac Although they offer an advantage of flexibility and compactness the microstrip lines are not useful for long transmission due to excessive attenuation 118 MICROSTRIP TRANSMISSION LINES 527 EXAMPLE 1110 A certain microstrip line has fused quartz er 38 as a substrate If the ratio of line width to substrate thickness is wlh 45 determine a The effective relative permittivity of the substrate b The characteristic impedance of the line c The wavelength of the line at 10 GHz Solution a For wlh 45 we have a wide strip From eq 1170 48 28 12 b From eq 1171 Z n 1207T c X r c V313145 1393 0667 In 45 1444 9576 fi 3 X 108 eeff 1OUV3131 169 X 102m 169 mm PRACTICE EXERCISE 1110 Repeat Example 1110 for wlh 08 Answer a 275 b 8403 fi c 1809 mm EXAMPLE 1111 At 10 GHz a microstrip line has the following parameters h 1 mm w 08 mm er 66 tan 6 104 ac 58 X 107 Sm Calculate the attenuation due to conduction loss and dielectric loss 528 Transmission Lines Solution The ratio wlh 08 Hence from eqs 1170 and 1171 72 56 60 12 8 08 43 6717 0 The skin resistance of the conductor is 7T X 10 X 109 X 4TT X 10 7 58 X 107 2609 X 102 Qm2 Using eq 1175 we obtain the conduction attenuation constant as 2609 X 102 ac 8686 X 08 X 103 X 6717 4217 dBm To find the dielectric attenuation constant we need X 3 X 10s 10 X 10 9V43 1447 X 10zm Applying eq 1176 we have 4 3492 X 66 X 10 56 X 43 X 1447 X 102 01706 dBm ad 273 X PRACTICE EXERCISE 1111 Calculate the attenuation due to ohmic losses at 20 GHz for a microstrip line con structed of copper conductor having a width of 25 mm on an alumina substrate Take the characteristic impedance of the line as 50 U Answer 2564 dBm SUMMARY 1 A transmission line is commonly described by its distributed parameters R in 0m L in Hm G in Sm and C in Fm Formulas for calculating R L G and C are pro vided in Table 111 for coaxial twowire and planar lines SUMMARY 529 2 The distributed parameters are used in an equivalent circuit model to represent a differ ential length of the line The transmissionline equations are obtained by applying Kirchhoffs laws and allowing the length of the line to approach zero The voltage and current waves on the line are Vz t Veaz cos cor Pz Veaz cos cor 3z Iz t eaz cos cor eaz cos cor fc Zo A showing that there are two waves traveling in opposite directions on the line 3 The characteristic impedance Zo analogous to the intrinsic impedance rj of plane waves in a medium of a line is given by Zn R jaL G jaC and the propagation constant y in per meter is given by 7 a 78 VR joLG The wavelength and wave velocity are x f 4 The general case is that of the lossy transmission line G 0 7 considered earlier For a lossless line 0 G for a distortionless line RIL GIC It is desirable that power lines be lossless and telephone lines be distortionless 5 The voltage reflection coefficient at the load end is defined as L v and the standing wave ratio is i zL zo zL zo rL where ZL is the load impedance 6 At any point on the line the ratio of the phasor voltage to phasor current is the imped ance at that point looking towards the load and would be the input impedance to the line if the line were that long For a lossy line ZL Zo tanh Zo ZL tanh where i is the distance from load to the point For a lossless line a 0 tanh yt 7 tan 3 for a shorted line ZL 0 for an opencircuited line ZL and for a matched line ZL Zo 530 Transmission Lines 7 The Smith chart is a graphical means of obtaining line characteristics such as T s and Zin It is constructed within a circle of unit radius and based on the formula for TL given above For each r and x it has two explicit circles the resistance and reactance circles and one implicit circle the constant circle It is conveniently used in determining the location of a stub tuner and its length It is also used with the slotted line to determine the value of the unknown load impedance 8 When a dc voltage is suddenly applied at the sending end of a line a pulse moves forth and back on the line The transient behavior is conveniently analyzed using the bounce diagrams 9 Microstrip transmission lines are useful in microwave integrated circuits Useful for mulas for constructing microstrip lines and determining losses on the lines have been presented 111 Which of the following statements are not true of the line parameters R L G and C a R and L are series elements b G and C are shunt elements 0 G i d LC tie and RG ae e Both R and G depend on the conductivity of the conductors forming the line f Only R depends explicitly on frequency g The parameters are not lumped but distributed 112 For a lossy transmission line the characteristic impedance does not depend on a The operating frequency of the line b The length of the line c The load terminating the line d The conductivity of the conductors e The conductivity of the dielectric separating the conductors 113 Which of the following conditions will not guarantee a distortionless transmission line a R 0 G b RC GL c Very low frequency range R 55 uL G S aC d Very high frequency range R 3C OJL G C WC 114 Which of these is not true of a lossless line a Zin Za for a shorted line with A8 b Zin yoo for a shorted line with A4 REVIEW QUESTIONS 531 c Zin jZ0 for an open line with X2 d Zin Zo for a matched line e At a halfwavelength from a load Zin ZL and repeats for every halfwavelength thereafter 115 A lossless transmission line of length 50 cm with L 10 UHm C 40 pFm is oper ated at 30 MHz Its electrical length is a 20X b 02X c 108 d 40TT e None of the above 116 Match the following normalized impedances with points A B C D and E on the Smith chart of Figure 1140 i 0 jO iii 0 yl V oc jco vii Zm ii 1 jO iv 0 j vi Zm viii Matched load r 0 117 A 500m lossless transmission line is terminated by a load which is located at P on the Smith chart of Figure 1141 If X 150 m how many voltage maxima exist on the line a 7 b 6 c 5 d 3 e None Figure 1140 For Review Question 116 532 Transmission Lines 180 150 Figure 1141 For Review Question 117 9 0 118 Write true T or false F for each of the following statements a All r and xcircles pass through point Fr F 10 b Any impedance repeats itself every X4 on the Smith chart c An s 2 circle is the same as F 05 circle on the Smith chart d The basic principle of any matching scheme is to eliminate the reflected wave between the source and the matching device e The slotted line is used to determine ZL only f At any point on a transmission line the current reflection coefficient is the recipro cal of the voltage reflection coefficient at that point 119 In an air line adjacent maxima are found at 125 cm and 375 cm The operating fre quency is a 15 GHz b 600 MHz c 300 MHz d 12 GHz 1110 Two identical pulses each of magnitude 12 V and width 2 us are incident at t 0 on a lossless transmission line of length 400 m terminated with a load If the two pulses are separated 3 is similar to the case of Figure 1153 and u 2 X 108 ms when does the contribution to VL i by the second pulse start overlapping that of the first a t 05 ixs b t 2 us c t 5 s d t 55 us e t 6 us Answers lllcde 112bc 113c 114ac 115c 116iDB ii A iiiE ivC vB vi D vii B viii A 117a 118 a T b F c F d T e F f F 119b lllOe PROBLEMS 533 111 An airfilled planar line with w 30 cm d 12 cm t 3 mm has conducting plates PROBLEMS j w i t h x 1Q7 s m C a k u l a t e L c a n d G a t 5 0 0 M H z 112 The copper leads of a diode are 16 mm in length and have a radius of 03 mm They are separated by a distance of 2 mm as shown in Figure 1142 Find the capacitance between the leads and the ac resistance at 10 MHz 113 In Section 113 it was mentioned that the equivalent circuit of Figure 115 is not the only possible one Show that eqs 114 and 116 would remain the same if the IItype and Ttype equivalent circuits shown in Figure 1143 were used 114 A 78fi lossless planar line was designed but did not meet a requirement What fraction of the widths of the strip should be added or removed to get the characteristic impedance of 75 0 r 16 mm Figure 11 42 The diode of Problem 112 RAz Az 0 G az t a wv CAz Kz Az c o b Figure 1141 For Problem 113 a IItype equivalent circuit b Ttype equiva lent circuit 534 Transmission Lines 115 A telephone line has the following parameters R 40 fim G 400 iSm L 02 xHm C 05 nFm a If the line operates at 10 MHz calculate the characteristic impedance Zo and veloc ity u b After how many meters will the voltage drop by 30 dB in the line 116 A distortionless line operating at 120 MHz has R 20 fim L 03 xHm and C 63 pFm a Determine 7 u and Zo b How far will a voltage wave travel before it is reduced to 20 of its initial magnitude c How far will it travel to suffer a 45 phase shift 117 For a lossless twowire transmission line show that 1 a The phase velocity u c LC 120 b The characteristic impedance Zo j cosh Is part a true of other lossless lines 118 A twisted line which may be approximated by a twowire line is very useful in the tele phone industry Consider a line comprised of two copper wires of diameter 012 cm that have a 032cm centertocenter spacing If the wires are separated by a dielectric mater ial with e 35e0 find L C and Zo 119 A lossless line has a voltage wave Vz t Vo sinwr fa Find the corresponding current wave 1110 On a distortionless line the voltage wave is given by V 120 e 0 0 0 2 5 r cos 108r 2 60r00025r cos 108f 2 where is the distance from the load If ZL 300 0 find a a and u b Zo and 1111 a Show that a transmission coefficient may be defined as rL b Find TL when the line is terminated by i a load whose value is nZo ii an open circuit iii a short circuit iv ZL Zo matched line 1112 A coaxial line 56 m long has distributed parameters R 65 fim L 34xHm G 84 mSm and C 215 pFm If the line operates at 2 MHz calculate the charac teristic impedance and the endtoend propagation time delay PROBLEMS 535 Figure 1144 For Problem 1116 i 1113 A lossless transmission line operating at 45 GHz has L 24 xHm and Zo 85 fi Calculate the phase constant j3 and the phase velocity u 1114 A 50fi coaxial cable feeds a 75 J20Q dipole antenna Find T and s 1115 Show that a lossy transmission line of length has an input impedance Zsc Zo tanh yt when shorted and Zoc Zo coth y when open Confirm eqs 1137 and 1139 1116 Find the input impedance of a shortcircuited coaxial transmission line of Figure 1144 if Zn 65 j38 U 7 07 j25 m 08 m 1117 Refer to the lossless transmission line shown in Figure 1145 a Find T and s b De termine Zin at the generator 1118 A quarterwave lossless 100fi line is terminated by a load ZL 210 Q If the voltage at the receiving end is 80 V what is the voltage at the sending end 1119 A 500fi lossless line has VL 10ey25 V ZL 50e30 Find the current at A8 from the load 1120 A 60fi lossless line is connected to a source with Vg 10CT lms and Zg 50 740 fi and terminated with a load j40 0 If the line is 100 m long and 3 025 radm calculate Zin and V at a The sending end b The receiving end c 4 m from the load d 3 m from the source A6 Zo 50 Q 120 n Figure 1145 For Problem 1117 536 Transmission Lines ZL 6035 Figure 1146 For Problem 1122 1121 A lossless transmission line with a characteristic impedance of 75 fi is terminated by a load of 120 Q The length of the line is 125X If the line is energized by a source of 100 V rms with an internal impedance of 50 fi determine a the input impedance and b the magnitude of the load voltage 1122 Three lossless lines are connected as shown in Figure 1146 Determine Zin 1123 Consider the twoport network shown in Figure 1147a The relation between the input and output variables can be written in matrix form as C D12 For the lossy line in Figure 1147b show that the ABCD matrix is cosh yi Zo sinh yi sinh y cosh y 1124 A 50fi lossless line is 42 m long At the operating frequency of 300 MHz the input im pedance at the middle of the line is 80 j60 U Find the input impedance at the genera tor and the voltage reflection coefficient at the load Take u 08c 1125 A 60fi air line operating at 20 MHz is 10 m long If the input impedance is 90 jl50 fi calculate ZL T and s 1126 A 75Q transmission line is terminated by a load of 120 80 fi a Find T and s b Determine how far from the load is the input impedance purely resistive 1127 A 750 transmission line is terminated by a load impedance ZL If the line is 5X8 long calculate Zjn when a ZL j45 U b ZL 25 j65 o a Figure 1147 For Problem 1123 PROBLEMS 537 1128 Determine the normalized input impedance at A8 from the load if a its normalized im pedance is 2 j b its normalized admittance is 02 j05 c the reflection coeffi cient at the load is 03 jOA 1129 A transmission line is terminated by a load with admittance YL 06 4 08Zo Find the normalized input impedance at A6 from the load 1130 An 80ft transmission line operating at 12 MHz is terminated by a load ZL At 22 m from the load the input impedance is 100 j120 ft If M 08c a Calculate TL Zinmax andZmmm b Find ZL s and the input impedance at 28 m from the load c How many Zin max and Zin min are there between the load and the 100 j2Q ft input impedance 1131 An antenna connected to a 150ft lossless line produces a standing wave ratio of 26 If measurements indicate that voltage maxima are 120 cm apart and that the last maximum is 40 cm from the antenna calculate a The operating frequency b The antenna impedance c The reflection coefficient Assume that u c 1132 The observed standingwave ratio on a 100ft lossless line is 8 If the first maximum voltage occurs at 03A from the load calculate the load impedance and the voltage re flection coefficient at the load 1133 A 50fl line is terminated to a load with an unknown impedance The standing wave ratio s 24 on the line and a voltage maximum occurs A8 from the load a Determine the load impedance b How far is the first minimum voltage from the load 1134 A 75fl lossless line is terminated by an unknown load impedance ZL If at a distance 02A from the load the voltage is Vs 2 j V while the current is 10 mA Find ZL and s 1135 Two A4 transformers in tandem are to connect a 50fl line to a 75ft load as in Figure 1148 a Determine the characteristic impedance Zol if Zo2 30 ft and there is no reflected wave to the left of A b If the best results are obtained when Zo2 determine Zol and Zo2 for this case 1136 Two identical antennas each with input impedance 74 0 are fed with three identical 50fi quarterwave lossless transmission lines as shown in Figure 1149 Calculate the input impedance at the source end 538 Transmission Linos zo so n Figure 1148 Double section trans former of Problem 1135 75 n 1137 If the line in the previous problem is connected to a voltage source 120 V with internal impedance 80 fi calculate the average power delivered to either antenna 1138 Consider the three lossless lines in Figure 1150 If Zo 50 fi calculate a Zin looking into line 1 o b Zin looking into line 2 N c Zin looking into line 3 1139 A section of lossless transmission line is shunted across the main line as in Figure 1151 If j X4 2 X8 and 3 7X8 find ym yin2 and y given that Zo 100 0 ZL 200 jl50 fi Repeat the calculations if the shorted section were open 1140 It is desired to match a 50fi line to a load impedance of 60 j50 fi Design a 50fi stub that will achieve the match Find the length of the line and how far it is from the load 1141 A stub of length 012X is used to match a 60fi lossless line to a load If the stub is located at 03X from the load calculate a The load impedance ZL b The length of an alternative stub and its location with respect to the load c The standing wave ratio between the stub and the load 1142 On a lossless line measurements indicate s 42 with the first maximum voltage at X from the load Determine how far from the load a shortcircuited stub should be locatec and calculate its length A4 74 n Figure 1149 For Problems 1136 and 1137 A4 A4 74 O A4 PROBLEMS iS 539 200 Q F i g u r e H 5 0 F o r Problem 1138 1143 A 600 lossless line terminated by load ZL has a voltage wave as shown in Figure 1152 Find s F andZL 1144 The following slottedline measurements were taken on a 500 system With load s 32 adjacent V occurs at 12 cm and32 cm high numbers on the load side with short circuit Vmin occurs at 21 cm Find the operating frequency and the load impedance 1145 A 500 air slotted line is applied in measuring a load impedance Adjacent minima are found at 14 cm and 225 cm from the load when the unknown load is connected and Vmax 095 V and Vmin 045 V When the load is replaced by a short circuit the minima are 32 cm to the load Determine sf T and ZL 1146 Show that for a dc voltage Vg turned on at t 0 see Figure 1130 the asymptotic values t SC of V t and t are Voo and ico 1147 A 60fi lossless line is connected to a 400 pulse generator The line is 6 m long and is terminated by a load of 100 0 If a rectangular pulse of width 5x and magnitude 20 V is sent down the line find V0 t and t for 0 t 10 jus Take u 3 X 108 ms 1148 The switch in Figure 1153 is closed at t 0 Sketch the voltage and current at the right side of the switch for 0 t 6u Take Zo 50 0 and tu 2 xs Assume a lossless transmission line Figure 1151 For Problem 1139 540 Transmission Lines 4V IV I I I 50 45 40 35 30 25 20 15 10 5 0 Zo 60 n Figure 1152 For Problem 1143 K3 1149 For the system shown in Figure 1154 sketch Vl t and f for 0 f 5 s 1150 Refer to Figure 1155 where Zg 25 Q Zo 50 0 ZL 150 Q 150 m u c If at f 0 the pulse shown in Figure 1156 is incident on the line a Draw the voltage and current bounce diagrams b Determine V0 0 V t 10 t and f for 0 f 8 as 1151 A microstrip line is 1 cm thick and 15 cm wide The conducting strip is made of brass oc 11 X 107Sm while the substrate is a dielectric material with er 22 and tan 9 0002 If the line operates at 25 GHz find a Zo and eeff b ac and ad c the distance down the line before the wave drops by 20 dB 1152 A 50fi microstrip line has a phase shift of 45 at 8 GHz If the substrate thickness is h 8 mm with er 46 find a the width of the conducting strip b the length of the microstrip line 1153 An alumina substrate s 96eo of thickness 2 mm is used for the construction of a mi crostrip circuit If the circuit designer has the choice of making the line width to be within 04 to 80 mm what is the range of characteristic impedance of the line 1154 Design a 75fl microstrip line on a 12mm thickduroid er 23 substrate Find the width of the conducting strip and the phase velocity 27 V Z0y 05 Zo Figure 1153 For Problem 1148 PROBLEMS 541 100 v Figure 1154 For Problem 1149 ioo a 200 m z t V Figure 1155 For Problem 1150 15 V 15 V Figure 1156 Two rectangular pulses of Problem 1150 Chapter WAVEGUIDES If a man writes a better book preaches a better sermon or makes a better mouse trap than his neighbor the world will make a beaten path to his door RALPH WALDO EMERSON 121 INTRODUCTION As mentioned in the preceding chapter a transmission line can be used to guide EM energy from one point generator to another load A waveguide is another means of achieving the same goal However a waveguide differs from a transmission line in some respects although we may regard the latter as a special case of the former In the first place a transmission line can support only a transverse electromagnetic TEM wave whereas a waveguide can support many possible field configurations Second at mi crowave frequencies roughly 3300 GHz transmission lines become inefficient due to skin effect and dielectric losses waveguides are used at that range of frequencies to obtain larger bandwidth and lower signal attenuation Moreover a transmission line may operate from dc 0 to a very high frequency a waveguide can operate only above a certain frequency called the cutoff frequency and therefore acts as a highpass filter Thus wave guides cannot transmit dc and they become excessively large at frequencies below mi crowave frequencies Although a waveguide may assume any arbitrary but uniform cross section common waveguides are either rectangular or circular Typical waveguides1 are shown in Figure 121 Analysis of circular waveguides is involved and requires familiarity with Bessel functions which are beyond our scope2 We will consider only rectangular waveguides By assuming lossless waveguides ac a 0 we shall apply Maxwells equations with the appropriate boundary conditions to obtain different modes of wave propagation and the corresponding E and H fields 542 For other tpes of waveguides see J A Seeger Microwave Theory Components and Devices En glewood Cliffs NJ PrenticeHall 1986 pp 128133 2Analysis of circular waveguides can be found in advanced EM or EMrelated texts eg S Y Liao Microwave Devices and Circuits 3rd ed Englewood Cliffs NJ PrenticeHall 1990 pp 119141 122 RECTANGULAR WAVEGUIDES 543 Figure 121 Typical waveguides Circular Rectangular Twist 90 elbow 122 RECTANGULAR WAVEGUIDES Consider the rectangular waveguide shown in Figure 122 We shall assume that the wave guide is filled with a sourcefree pv 0 J 0 lossless dielectric material a 0 and its walls are perfectly conducting ac From eqs 1017 and 1019 we recall that for a lossless medium Maxwells equations in phasor form become kzEs 0 0 121 122 Figure 122 A rectangular waveguide with perfectly conducting walls filled with a lossless material jX T 0 544 Waveguides where k OJVUB 123 and the time factor eJ01t is assumed If we let Exs Eys Ezs and Hxs Hys Hzs each of eqs 121 and 122 is comprised of three scalar Helmholtz equations In other words to obtain E and H fields we have to solve six scalar equations For the zcompo nent for example eq 121 becomes d2Ezs dx2 dy2 dz 124 which is a partial differential equation From Example 65 we know that eq 124 can be solved by separation of variables product solution So we let Ezsx y z Xx Yy Zz 125 where Xx Yy and Zz are functions of y and z respectively Substituting eq 125 into eq 124 and dividing by XYZ gives x r z 2 k2 X Y Z 126 Since the variables are independent each term in eq 126 must be constant so the equa tion can be written as k k y2 k2 127 where k2 k2 and y2 are separation constants Thus eq 126 is separated as X k2 xX 0 128a r k2 yY 0 128b Z T 2Z 0 128c By following the same argument as in Example 65 we obtain the solution to eq 128 as Xx c cos kx c2 sin kyX Yy c3 cos kyy c4 sin kyy Zz c5eyz c6e7Z Substituting eq 129 into eq 125 gives Ezsx y z ci cos kxx c2 sin kXci cos kyy c4 sin kyy c5eyz c6eyz 129a 129b 129c 1210 122 RECTANGULAR WAVEGUIDES 545 As usual if we assume that the wave propagates along the waveguide in the zdirection the multiplicative constant c5 0 because the wave has to be finite at infinity ie Ezsx y z 0 Hence eq 1210 is reduced to Ezsx y z A cos kx A2 sin cos kyy A4 sin kyye 1211 where Aj CiC6 A2 c2c6 and so on By taking similar steps we get the solution of the zcomponent of eq 122 as Hzsx y z Bi cos kpc B2 sin cos kyy B4 sin kyye 1212 Instead of solving for other field component Exs Eys Hxs and Hys in eqs 121 and 122 in the same manner we simply use Maxwells equations to determine them from Ezs and HTS From and V X E y V X H jtoeEs we obtain dE dy dHzs dy dExs dz dHxs dz dEys dx dHy dz dHv dz dx dHz dx 9EX dy dHx jueExs J03flHys dx dy 1213a 1213b 1213c 1213d 1213e 1213f We will now express Exs Eys Hxs and Hys in terms of Ezs and Hzs For Exs for example we combine eqs 1213b and 1213c and obtain dHz 1 fd2Exs d2Ez dy 7C0X dz oxdi 1214 From eqs 1211 and 1212 it is clear that all field components vary with z according to eyz that is plz F 546 Waveguides Hence and eq 1214 becomes dEzs d Exx yEzs j 7 EX dZ dz dHa 1 2 dE jweExs I 7 Exs 7 dy joifi dx or 1 2 2 r 7 dEzs dHzs 7 V Exs jii jun dx dy Thus if we let h2 y2 w2xe y2 k2 E 1 7 dEzs jun dHzs hl dx dy Similar manipulations of eq 1213 yield expressions for Eys Hxs and Hys in terms of Ev and Hzs Thus 1215a 1215b 1215c 1215d Exs EyS M Hys h2 dx 7 dEzs h2 dy jue dEzs h2 dy jae dEzs jan dHz h2 dy jux dHzs h2 dx 7 dHzs h2 dx h2T where h2 y2 k2 k2 x k 1216 Thus we can use eq 1215 in conjunction with eqs 1211 and 1212 to obtain Exs Eys Hxs and Hys From eqs 1211 1212 and 1215 we notice that there are different types of field patterns or configurations Each of these distinct field patterns is called a mode Four dif ferent mode categories can exist namely 1 Ea 0 Hzs TEM mode This is the transverse electromagnetic TEM mode in which both the E and H fields are transverse to the direction of wave propaga tion From eq 1215 all field components vanish for Ezs 0 Hzs Conse quently we conclude that a rectangular waveguide cannot support TEM mode 123 TRANSVERSE MAGNETIC TM MODES 547 Figure 123 Components of EM fields in a rectangular waveguide a TE mode Ez 0 b TM mode Hz 0 2 Ezs 0 Hzs 0 TE modes For this case the remaining components Exs and Eys of the electric field are transverse to the direction of propagation az Under this condition fields are said to be in transverse electric TE modes See Figure 123a 3 Ezs 0 Hzs 0 TM modes In this case the H field is transverse to the direction of wave propagation Thus we have transverse magnetic TM modes See Figure 123b 4 Ezs 0 Hzs 0 HE modes This is the case when neither E nor H field is trans verse to the direction of wave propagation They are sometimes referred to as hybrid modes We should note the relationship between k in eq 123 and j3 of eq 1043a The phase constant 3 in eq 1043a was derived for TEM mode For the TEM mode h 0 so from eq 1216 y2 k2 y a j3 jk that is 3 k For other modes j3 k In the subsequent sections we shall examine the TM and TE modes of propagation sepa rately 23 TRANSVERSE MAGNETIC TM MODES For this case the magnetic field has its components transverse or normal to the direction of wave propagation This implies that we set Hz 0 and determine Ex Ey Ez Hx and Hv using eqs 1211 and 1215 and the boundary conditions We shall solve for Ez and later determine other field components from Ez At the walls of the waveguide the tangential components of the E field must be continuous that is 0 at y 0 y b 0 at Ezs 0 at x 0 0 at x a 1217a 1217b 1217c 1217d 548 Waveguides Equations 1217a and 1217c require that A 0 A3 in eq 1211 so eq 1211 becomes Ea Eo sin kj sin kyy eyz 1218 where Eo A2A4 Also eqs 1217b and 1217d when applied to eq 1218 require that s i n 0 sinkyb O 1219 This implies that kxa rrnr m 1 2 3 1220a kyb nir n 1 2 3 1220b or n7r Ky b 1221 The negative integers are not chosen for m and n in eq 1220a for the reason given in Example 65 Substituting eq 1221 into eq 1218 gives E7 Eo sin fnnrx fniry in cm o c V a I sin b 1222 We obtain other field components from eqs 1222 and 1215 bearing in mind that H7 0 Thus 1223a 1223b 1223c y fnw IT fmirx ny yz l F o sin I I cos I I e T jus Hys v j Eo cos sin 1223d where nir 1224 which is obtained from eqs 1216 and 1221 Notice from eqs 1222 and 1223 that each set of integers m and n gives a different field pattern or mode referred to as TMmn 123 TRANSVERSE MAGNETIC TM MODES 549 mode in the waveguide Integer m equals the number of halfcycle variations in the x direction and integer n is the number of halfcycle variations in the vdirection We also notice from eqs 1222 and 1223 that if m n is 0 0 0 n or m 0 all field com ponents vanish Thus neither m nor n can be zero Consequently TMH is the lowestorder mode of all the TMmn modes By substituting eq 1221 into eq 1216 we obtain the propagation constant 7 mir a nir b 1225 where k u V e as in eq 123 We recall that in general y a j3 In the case of eq 1225 we have three possibilities depending on k or w m and n CASE A cutoff If 1c w jus b 7 0 or a 0 3 The value of w that causes this is called the cutoff angular frequency oc that is 1 U T T I 2 Tmr12 I J U 1226 CASE B evanescent If TOTTT2 Tnir y a In this case we have no wave propagation at all These nonpropagating or attenuating modes are said to be evanescent CASE C propagation If 2 oA mir y a 0 550 Waveguides that is from eq 1225 the phase constant 3 becomes 0 L a nir 1227 This is the only case when propagation takes place because all field components will have the factor eyz ejl3z Thus for each mode characterized by a set of integers m and n there is a correspond ing cutoff frequency fc The cutoff frequency is the operating frequencs below which allcnuaiion occurs and above which propagation lakes place The waveguide therefore operates as a highpass filter The cutoff frequency is obtained fromeq 1226 as 1 2irVue nnr a or fc u N mu N nu 1228 where u phase velocity of uniform plane wave in the lossless dielectric fie medium a 0 fi e filling the waveguide The cutoff wave length is given by or X 1229 Note from eqs 1228 and 1229 that TMn has the lowest cutoff frequency or the longest cutoff wavelength of all the TM modes The phase constant 3 in eq 1227 can be written in terms of fc as wVxsl 123 TRANSVERSE MAGNETIC TM MODES 551 or 1230 i where j3 oilu uVfie phase constant of uniform plane wave in the dielectric medium It should be noted that y for evanescent mode can be expressed in terms of fc namely 1230a The phase velocity up and the wavelength in the guide are respectively given by w 2TT u 1231 The intrinsic wave impedance of the mode is obtained from eq 1223 as y jfi Ex Ey I T M Hy Hx we or TM V 1232 where 17 Vxe intrinsic impedance of uniform plane wave in the medium Note the difference between u 3 and q and u 3 and 77 The quantities with prime are wave characteristics of the dielectric medium unbounded by the waveguide as discussed in Chapter 10 ie for TEM mode For example u would be the velocity of the wave if the waveguide were removed and the entire space were filled with the dielectric The quantities without prime are the wave characteristics of the medium bounded by the wave guide As mentioned before the integers m and n indicate the number of halfcycle variations in the xy cross section of the guide Thus for a fixed time the field configuration of Figure 124 results for TM2 mode for example 552 Waveguides end view n 1 E field H field Figure 124 Field configuration for TM2 mode side view 124 TRANSVERSE ELECTRIC TE MODES In the TE modes the electric field is transverse or normal to the direction of wave propa gation We set Ez 0 and determine other field components Ex Ey Hx Hy and Hz from eqs 1212 and 1215 and the boundary conditions just as we did for the TM modes The boundary conditions are obtained from the fact that the tangential components of the elec tric field must be continuous at the walls of the waveguide that is Exs Exs Eys Eys 0 0 0 0 From eqs 1215 and 1233 the boundary dHzs dy dHzs 0 0 at at at at y 0 y b x 0 x a conditions can be written as at at y0 y b 1233a 1233b 1233c 1233d 1234a 1234b dy dHzs dx dHzs dx 0 0 at at x 0 x a Imposing these boundary conditions on eq 1212 yields mx fmry Hzs Ho cos cos e yz a b J 1234c 1234d 1235 124 TRANSVERSE ELECTRIC TE MODES 553 where Ho BXBT Other field components are easily obtained from eqs 1235 and 1215 as e mrx 1236a 1236b 1236c 1236d where m 0 1 2 3 and n 0 1 2 3 J and 7 remain as defined for the TM modes Again m and n denote the number of halfcycle variations in the xy cross section of the guide For TE32 mode for example the field configuration is in Figure 125 The cutoff frequency fc the cutoff wavelength Xc the phase constant 3 the phase velocity up and the wavelength X for TE modes are the same as for TM modes see eqs 1228 to 1231 For TE modes m ri may be 0 1 or 1 0 but not 0 0 Both m and n cannot be zero at the same time because this will force the field components in eq 1236 to vanish This implies that the lowest mode can be TE10 or TE01 depending on the values of a and b the dimensions of the guide It is standard practice to have a b so that Ia2 1b2 in u u eq 1228 Thus TEi0 is the lowest mode because CTE CTK This mode is TE la Th 2b top view E field field Figure 125 Field configuration for TE32 mode 554 i Waveguides called the dominant mode of the waveguide and is of practical importance The cutoff fre quency for the TEH mode is obtained from eq 1228 as m 1 n 0 Jc to 2a 1237 and the cutoff wavelength for TE0 mode is obtained from eq 1229 as Xt0 2a 1238 Note that from eq 1228 the cutoff frequency for TMn is ua2 b212 2ab which is greater than the cutoff frequency for TE10 Hence TMU cannot be regarded as the dominant mode The dominant mode is the mode with the lowest cutoff frequency or longest cutoff wavelength Also note that any EM wave with frequency fCw or X XC0 will not be propagated in the guide The intrinsic impedance for the TE mode is not the same as for TM modes From eq 1236 it is evident that y jf3 Ex Ey Jfl rTE jry iTx T Ifi 1 or VTE I V 12 J 1239 Note from eqs 1232 and 1239 that rTE and iTM are purely resistive and they vary with frequency as shown in Figure 126 Also note that ITE 1240 Important equations for TM and TE modes are listed in Table 121 for convenience and quick reference 124 TRANSVERSE ELECTRIC TE MODES 555 Figure 126 Variation of wave imped ance with frequency for TE and TM modes TABLE 121 Important Equations for TM and TE Modes TM Modes TE Modes jP frmc fimrx ny pn rm fmirx rny r I Eo cos sin e 7 I Ho cos I sin I e h a J a J b J h b J a b J Exs Eo sin cos e 7Z a J b niry i I cos e a J b Eo sin I sin I I e 1Z a J b jus Ezs 0 Hys yh a a V b ny i e 0 V j nnrx rnry Hzs Ho cos cos x a V b V where 556 Waveguides Fromeqs 1222 1223 1235 and 1236 we obtain the field patterns for the TM and TE modes For the dominant TE0 mode m landn 0 so eq 1235 becomes Hzs Ho cos e JPz In the time domain Hz Re HzseM or Hz Ho cosf Similarly from eq 1236 sin Hx Ho sin a fiz 1241 1242 1243a 1243b 1243c Figure 127 Variation of the field components with x for TE0 mode b 124 TRANSVERSE ELECTRIC TE MODES 557 Figure 128 Field lines for TE10 mode Direction of propagation top view If MiO I x 1 N c E field field Direction of propagation The variation of the E and H fields with x in an xy plane say plane coswf 8z 1 for Hz and plane sinof j8z 1 for Ey and Hx is shown in Figure 127 for the TE10 mode The corresponding field lines are shown in Figure 128 EXAMPLE 121 A rectangular waveguide with dimensions a 25 cm b 1 cm is to operate below 151 GHz How many TE and TM modes can the waveguide transmit if the guide is filled with a medium characterized by a 0 e 4 so 1 Calculate the cutoff frequencies of the modes Solution The cutoff frequency is given by m2 where a 25b or alb 25 and u lie Vr 558 Waveguides Hence c a 3 X 108 425 X 10 Vm2 625M2 or fCmn 3Vm 2 GHz 1211 We are looking for fCnm 151 GHz A systematic way of doing this is to fix m or n and increase the other until fCnm is greater than 151 GHz From eq 1211 it is evident that fixing m and increasing n will quickly give us an fCnm that is greater than 151 GHz ForTE01 mode m 0 n 1 fCm 325 75 GHz TE02 mode m 0n 2Co2 35 15 GHz TE03 modeCm 375 225 GHz Thus for fCmn 151 GHz the maximum n 2 We now fix n and increase m until fCmn is greater than 151 GHz For TE10 mode m 1 n 0 Co 3 GHz TE2o modeC20 6 GHz TE30 modeC3o 9 GHz TE40modeC40 12 GHz TE50 modeCjo 1 5 GHz the same as for TE02 TE60modeC60 18 GHz that is forCn 151 GHz the maximum m 5 Now that we know the maximum m and n we try other possible combinations in between these maximum values ForTE nTM n degenerate modes fCu 3T25 8078 GHz TE21 TM2IC2i 3V1025 96 GHz TE3TM31C31 3Vl525 1172 GHz TE41 TM41C4 3V2225 1414 GHz TE12 TM12Ci 3V26 153 GHz Those modes whose cutoff frequencies are less or equal to 151 GHz will be transmittedthat is 11 TE modes and 4 TM modes all of the above modes except TEi2 TM12 TE60 and TE03 The cutoff frequencies for the 15 modes are illustrated in the line diagram of Figure 129 124 TRANSVERSE ELECTRIC TE MODES S 559 TE4 TE TE30 9 T E 3 TE41TE50TE0 12 15 cGHz TMU TM21 TM31 TM41 Figure 129 Cutoff frequencies of rectangular waveguide with a 25b for Example 121 PRACTICE EXERCISE 121 Consider the waveguide of Example 121 Calculate the phase constant phase veloc ity and wave impedance for TEi0 and TMu modes at the operating frequency of 15 GHz Answer For TE10 3 6156radm u 1531 X 108ms rjJE 1924 0 For TMni3 5294 radm K 178 X 108msrjTM 1588 0 EXAMPLE 122 Write the general instantaneous field expressions for the TM and TE modes Deduce those for TEOi and TM12 modes Solution The instantaneous field expressions are obtained from the phasor forms by using E Re EseJ and H Re Hsejo Applying these to eqs 1222 and 1223 while replacing y and jfi gives the following field components for the TM modes sm iA r j E cos j3 w fmirx fmry AEO sm a cos si b J nvKx fniry in cm I E7 En sin T Eo sm z cos 560 Waveguides H y h2 En cos I1 niry a J sin smat b J 3z Hz 0 Similarly for the TE modes eqs 1235 and 1236 become E mirx j mry Ho cos sin sinuf b a J b J Pz wx fmir mwx frnry r Ho sin cos si h2 I a a J b J 7 0 Ho sin 3 rn7r Hy 2 Ho cos fniry cos sinwr b I a J cos b I j sin y sint 2 Ho cos j jmwx fniry H Ho cos cos cosco pz V a J b J For the TE01 mode we set m 0 n 1 to obtain 12 sin hz b Hy o sin 7T iry Hz Ho cos I I coscof 3z b J For the TM2 mode we set m 1 n 2 to obtain cin cir 3 TIA X A 27ry Ex j I o cos sin sincof 3z a b cos I I sir TTX Ez Eo sin sin 2iry cosof V a J b J 124 TRANSVERSE ELECTRIC TE MODES 561 Hr o sin I cos sincof 3z oe fir Ix 2wy Hy r EO cos sin smut y h2 aj a J V b where PRACTICE EXERCISE 122 An airfilled 5 by 2cm waveguide has Ezs 20 sin 40irx sin 50ry e3 Vm at 15 GHz a What mode is being propagated b Find 8 c Determine EyIEx Answer a TM2i b 2413 radm c 125 tan 40wx cot 50ry EXAMPLE 123 1 In a rectangular waveguide for which a 15 cm 08 cm a 0 fi JXO and e 4eo Hx 2 sin cos sin T X 10nt 0z Am Determine a The mode of operation b The cutoff frequency c The phase constant 3 d The propagation constant y e The intrinsic wave impedance 77 Solution a It is evident from the given expression for Hx and the field expressions of the last example that m 1 n 3 that is the guide is operating at TMI3 or TE13 Suppose we 562 U Waveguides choose TM13 mode the possibility of having TE13 mode is left as an exercise in Practice Exercise 123 b Hence c fcmn 2 u fiB fca 1 4 V 15 x icr22 08 x icr2 r22 V0444 1406 X 102 2857 GHz LJ fc 100 co 2TT 7T X 10 or 50 GHz 0 3 X 10s d y j0 yl71881m 2857 50 171881 radm e V 1547 12 377 I 285712 50 PRACTICE EXERCISE 123 Repeat Example 123 if TEn mode is assumed Determine other field components for this mode Answer fc 2857 GHz 0 171881 radm 8 IJTE 22969 fi 25841 cos sin sinw fa Vm a J b J Ev 4594 sin cos J sincor fa Vm a J b J 0 1125 cos sin TTJ 3ry I sin I a V b J sinaf 8z Am 796 cos cos b J cos at fa Am 125 WAVE PROPAGATION IN THE GUIDE 563 125 WAVE PROPAGATION IN THE GUIDE Examination of eq 1223 or 1236 shows that the field components all involve the terms sine or cosine of miai or nirlby times eyz Since sin 6 eje ei6 2 cos 6 eje e jB 1244a 1244b a wave within the waveguide can be resolved into a combination of plane waves reflected from the waveguide walls For the TE0 mode for example 1245 c HjJ 2x The first term of eq 1245 represents a wave traveling in the positive zdirection at an angle tan 1246 with the zaxis The second term of eq 1245 represents a wave traveling in the positive zdirection at an angle 6 The field may be depicted as a sum of two plane TEM waves propagating along zigzag paths between the guide walls at x 0 and x a as illustrated in Figure 1210a The decomposition of the TE0 mode into two plane waves can be ex tended to any TE and TM mode When n and m are both different from zero four plane waves result from the decomposition The wave component in the zdirection has a different wavelength from that of the plane waves This wavelength along the axis of the guide is called the waveguide wave length and is given by see Problem 1213 X X 1247 where X uf As a consequence of the zigzag paths we have three types of velocity the medium ve locity u the phase velocity up and the group velocity ug Figure 1210b illustrates the re lationship between the three different velocities The medium velocity u 1Vxe is as 564 Waveguides Figure 1210 a Decomposition of TE10 mode into two plane waves b relationship between u up and a wave path ID explained in the previous sections The phase velocity up is the velocity at which loci of constant phase are propagated down the guide and is given by eq 1231 that is 7T d248a or Up cos e 1248b This shows that up u since cos 6 1 If u c then up is greater than the speed of light in vacuum Does this violate Einsteins relativity theory that messages cannot travel faster than the speed of light Not really because information or energy in a waveguide generally does not travel at the phase velocity Information travels at the group velocity which must be less than the speed of light The group velocity ug is the velocity with which the resultant repeated reflected waves are traveling down the guide and is given by 1249a or uo u cos 6 u 1249b 126 POWER TRANSMISSION AND ATTENUATION 565 Although the concept of group velocity is fairly complex and is beyond the scope of this chapter a group velocity is essentially the velocity of propagation of the wavepacket en velope of a group of frequencies It is the energy propagation velocity in the guide and is always less than or equal to u From eqs 1248 and 1249 it is evident that upug u2 1250 This relation is similar to eq 1240 Hence the variation of up and ug with frequency is similar to that in Figure 126 for rTE and rjTM EXAMPLE 124 A standard airfilled rectangular waveguide with dimensions a 8636 cm b 4318 cm is fed by a 4GHz carrier from a coaxial cable Determine if a TE10 mode will be propa gated If so calculate the phase velocity and the group velocity Solution For the TE10 mode fc u 11a Since the waveguide is airfilled u c 3 X 108 Hence fc 3 X 10 1737 GHz 2 X 8636 X 102 As 4 GHz fc the TE10 mode will propagate u 3 X 108 V l fjff V l 173742 333 X 108 ms 16 g 9 X 10 j 333 X 108 2702 X 108 ms PRACTICE EXERCISE 124 Repeat Example 124 for the TM n mode Answer 125 X 108 ms 7203 X 107 ms 126 POWER TRANSMISSION AND ATTENUATION To determine power flow in the waveguide we first find the average Poynting vector from eq 1068 1251 566 Waveguides In this case the Poynting vector is along the zdirection so that 1 Ea2 Eys2 2V 1252 where rj rjTE for TE modes or 77 TM for TM modes The total average power trans mitted across the cross section of the waveguide is at JC 1253 dy dx 0 Jy0 Of practical importance is the attenuation in a lossy waveguide In our analysis thus far we have assumed lossless waveguides a 0 ac for which a 0 7 j3 When the dielectric medium is lossy a 0 and the guide walls are not perfectly con ducting ac 00 there is a continuous loss of power as a wave propagates along the guide According to eqs 1069 and 1070 the power flow in the guide is of the form P P e 2az 1254 In order that energy be conserved the rate of decrease in Pave must equal the time average power loss PL per unit length that is P L dPa dz or fl In general ac ad 1255 1256 where ac and ad are attenuation constants due to ohmic or conduction losses ac 00 and dielectric losses a 0 respectively To determine ad recall that we started with eq 121 assuming a lossless dielectric medium a 0 For a lossy dielectric we need to incorporate the fact that a 0 All our equations still hold except that 7 jj3 needs to be modified This is achieved by replacing e in eq 1225 by the complex permittivity of eq 1040 Thus we obtain mir frnr2 2 1257 126 POWER TRANSMISSION AND ATTENUATION 567 where ec e je s j 1258 CO Substituting eq 1258 into eq 1257 and squaring both sides of the equation we obtain 2 2 7 ad 2 f t A l S fiir Equating real and imaginary parts a T 1259a 2adf3d coxa or ad Assuming that ad 3d az d j3z d 3J so eq 1259a gives a 1259b 1260 which is the same as 3 in eq 1230 Substituting eq 1260 into eq 1259b gives 1261 where rj Vxe The determination of ac for TMmn and TEmn modes is time consuming and tedious We shall illustrate the procedure by finding ac for the TE10 mode For this mode only Ey Hx and Hz exist Substituting eq 1243a into eq 1253 yields a f b dxdy x0 Jy 2ir r dy sin dx a 1262 ave The total power loss per unit length in the walls is yo yb Pi X0 O 1263 568 Waveguides since the same amount is dissipated in the walls y 0 and y b or x 0 and x a For the wall y 0 TJC j Hxsl Hzszdx a 2 2 7TJC H2 nsin2 H z o cos1 dx 1264 1 7T where Rs is the real part of the intrinsic impedance tc of the conducting wall From eq 1056 1 ar8 1265 where 5 is the skin depth Rs is the skin resistance of the wall it may be regarded as the re sistance of 1 m by 5 by 1 m of the conducting material For the wall x 0 C I Hzszdy H2 ody RJbHl 1266 Substituting eqs 1264 and 1266 into eq 1263 gives 1267 Finally substituting eqs 1262 and 1267 into eq 1255 2 2 2ir r ar 1268a It is convenient to express ac in terms of and fc After some manipulations we obtain for the TE10 mode 2RS k L 1268b 127 WAVEGUIDE CURRENT AND MODE EXCITATION 569 By following the same procedure the attenuation constant for the TEm modes n 0 can be obtained as 1269 r md for the TMmn fc J 2 l 1 I modes as c TM OH 2 2 2 m bid fViblaf 2 n2 m2 m2 kV n2 f 2 f 1270 The total attenuation constant a is obtained by substituting eqs 1261 and 1269 or 1270 into eq 1256 127 WAVEGUIDE CURRENT AND MODE EXCITATION For either TM or TE modes the surface current density K on the walls of the waveguide may be found using K an X H 1271 where an is the unit outward normal to the wall and H is the field intensity evaluated on the wall The current flow on the guide walls for TE10 mode propagation can be found using eq 1271 with eqs 1242 and 1243 The result is sketched in Figure 1211 The surface charge density ps on the walls is given by ps an D an eE where E is the electric field intensity evaluated on the guide wall 1272 Figure 1211 Surface current on guide walls for TE10 mode 570 Waveguides J a TEn mode 0 a 0 bTMM mode Figure 1212 Excitation of modes in a rectangular waveguide A waveguide is usually fed or excited by a coaxial line or another waveguide Most often a probe central conductor of a coaxial line is used to establish the field intensities of the desired mode and achieve a maximum power transfer The probe is located so as to produce E and H fields that are roughly parallel to the lines of E and H fields of the desired mode To excite the TE10 mode for example we know from eq 1243a that Ey has maximum value at x ail Hence the probe is located at x a2 to excite the TEIO mode as shown in Figure 1212a where the field lines are similar to those of Figure 128 Sim ilarly the TMii mode is launched by placing the probe along the zdirection as in Figure EXAMPLE 125 An airfilled rectangular waveguide of dimensions a 4 cm b 2 cm transports energy in the dominant mode at a rate of 2 mW If the frequency of operation is 10 GHz determine the peak value of the electric field in the waveguide Solution The dominant mode for a b is TE10 mode The field expressions corresponding to this mode m 1 n 0 are in eq 1236 or 1243 namely Exs 0 Eys jE0 sin where 127 WAVEGUIDE CURRENT AND MODE EXCITATION I I 571 Jc 2a 24 X 102 V 377 c 4067 1 375 J V L io From eq 1253 the average power transmitted is P r r M ave L L 2 Hence dy 4r 2 t 44067 X 2 X 1Q3 ab 8 X 10 En 6377 Vm 4067 PRACTICE EXERCISE 125 In Example 125 calculate the peak value Ho of the magnetic field in the guide if a 2 cm b 4 cm while other things remain the same Answer 6334 mAm EXAMPLE 126 A copperplated waveguide ac 58 X 107 Sm operating at 48 GHz is supposed to deliver a minimum power of 12 kW to an antenna If the guide is filled with polystyrene a 1017 Sm e 255eo and its dimensions are a 42 cm b 26 cm calculate the power dissipated in a length 60 cm of the guide in the TE10 mode Solution Let Pd power loss or dissipated Pa power delivered to the antenna Po input power to the guide so that P0 Pd Pa Fromeq 1254 D D 2xz 572 Waveguides Hence Pa 2az or Now we need to determine a from From eq 1261 Paelaz 1 a ad ac or Since the loss tangent 1017 ue q 109 2x X 48 X 109 X X 255 36TT then 147 X 1017 1 lossless dielectric medium 2361 1879 X 108ms Br 2a 2 X 42 X 10 1017 X 2361 2234 GHz 223412 L 48 J ad 1334 X 1015Npm For the TE10 mode eq 1268b gives ar V2 If 05 k LJ 127 WAVEGUIDE CURRENT AND MODE EXCITATION 573 where Hence ac8 1808 X 102Q 1 hrfiJ hrX 48 X 109 X 4ir X 10 7 58 X 107 2 X l i a 26 X1O2X 236 4218 X 103Npm 1 1 234 Note that ad C ac showing that the loss due to the finite conductivity of the guide walls is more important than the loss due to the dielectric medium Thus a ad ac ac 4218 X 103 Npm and the power dissipated is 6089 W Pd Pae i 12 x ioV x 4 2 1 8 x l o x a 6 1 PRACTICE EXERCISE 126 A brass waveguide ac 11 X 107 mhosm of dimensions a 42 cm b 15 cm is filled with Teflon er 26 a 1015 mhosm The operating frequency is 9 GHz For the TE10 mode a Calculate xd and ac b What is the loss in decibels in the guide if it is 40 cm long Answer a 1206 X 10iJ Npm 1744 X 10zNpm b 00606 dB EXAMPLE 127 Sketch the field lines for the TMn mode Derive the instantaneous expressions for the surface current density of this mode Solution From Example 122 we obtain the fields forTMn mode m 1 n 1 as Ey Ti i j Eo sin c o s y sincof j3z 574 Waveguides Ez Eo sin I I sin I I coswr 3z x a J b J we TT I x iry Hx j I o sin I I cos I sinwf h bj a I b we vr y tf a E TXX Va 0 For the electric field lines dy E a firx y r 7 tan cot cfx x b a I b For the magnetic field lines dy Hy b TTX dx Hx a a J iry b J Notice that EyExHyHx 1 showing that electric and magnetic field lines are mutu ally orthogonal This should also be observed in Figure 1213 where the field lines are sketched The surface current density on the walls of the waveguide is given by K an X H a X Hx Hy 0 At x 0 an ax K Hy0 y z t az that is we fir At x a an a x K Hya y z i az or Tr we ir 1 a Figure 1213 Field lines for TMU mode for Example 127 E field H field 128 WAVEGUIDE RESONATORS 575 At y 0 an ay K Hxx 0 z t az or Aty ban ay K Hxx b z t az or COS 7T TTX K I Eo sin smatf j3z az h bj a I3z az PRACTICE EXERCISE 127 Sketch the field lines for the TE n mode Answer See Figure 1214 The strength of the field at any point is indicated by the density of the lines the field is strongest or weakest where the lines are closest together or farthest apart 128 WAVEGUIDE RESONATORS Resonators are primarily used for energy storage At high frequencies 100 MHz and above the RLC circuit elements are inefficient when used as resonators because the di mensions of the circuits are comparable with the operating wavelength and consequently unwanted radiation takes place Therefore at high frequencies the RLC resonant circuits end view side view U J i E field field top view Figure 1214 For Practice Exercise 127 for TEn mode 576 Waveguides are replaced by electromagnetic cavity resonators Such resonator cavities are used in kly stron tubes bandpass filters and wave meters The microwave oven essentially consists of a power supply a waveguide feed and an oven cavity Consider the rectangular cavity or closed conducting box shown in Figure 1215 We notice that the cavity is simply a rectangular waveguide shorted at both ends We therefore expect to have standing wave and also TM and TE modes of wave propagation Depending on how the cavity is excited the wave can propagate in the x y or zdirection We will choose the zdirection as the direction of wave propagation In fact there is no wave propagation Rather there are standing waves We recall from Section 108 that a standing wave is a combination of two waves traveling in opposite directions A TM Mode to z For this case Hz 0 and we let EJx y z Xx Yy Zz 1273 be the production solution of eq 121 We follow the same procedure taken in Section 122 and obtain Xx C cos kjX c2 sin kpc Yy c3 cos kyy c4 sin kyy Zz c5 cos kzz c6 sin kzz where k2 k2 x k k u2ixe The boundary conditions are z 0 at x 0 a Ez 0 at 06 Ey 0Ex 0 at z 0 c 1274a 1274b 1274c 1275 1276a 1276b 1276c Figure 1215 Rectangular cavity 128 WAVEGUIDE RESONATORS 577 As shown in Section 123 the conditions in eqs 127a b are satisfied when cx 0 c3 and nvK a y b 1277 where m 1 2 3 n 1 2 3 To invoke the conditions in eq 1276c we notice that eq 1214 with Hzs 0 yields d2Exs d2Ezs dz dz Similarly combining eqs 1213a and 1213d with Hzs 0 results in ycoe ys From eqs 1278 and 1279 it is evident that eq 1276c is satisfied if 0 at z 0 c dz This implies that c6 0 and sin kzc 0 sin pir Hence 1278 1279 1280 1281 where p 0 1 2 3 Substituting eqs 1277 and 1281 into eq 1274 yields 1282 where Eo c2c4c5 Other field components are obtained from eqs 1282 and 1213 The phase constant 3 is obtained from eqs 1275 1277 and 1281 as 1283 Since 32 co2i from eq 1283 we obtain the resonant frequency fr 2irfr ur fie or u fr r w i a 2 r 12 1 I r r 1284 578 Waveguides The corresponding resonant wavelength is u fr llm VU r J 2 In L f 1 UJ 9 1285 From eq 1284 we notice that the lowestorder TM mode is TM110 B TE Mode to z In this case Ez 0 and Hzs bt cos sin 3 cos kyy b4 sin kyy 5 cos kzz sin kzz The boundary conditions in eq 1276c combined with eq 1213 yields at z 0 c Hzs dx dy 0 at x 0 a 0 at 0b 1286 1287a 1287b 1287c Imposing the conditions in eq 1287 on eq 1286 in the same manner as for TM mode to z leads to 1288 where m 0 1 2 3 n 0 1 2 3 and p 1 2 3 Other field com ponents can be obtained from eqs 1213 and 1288 The resonant frequency is the same as that of eq 1284 except that m or n but not both at the same time can be zero for TE modes The reason why m and n cannot be zero at the same time is that the field components will be zero if they are zero The mode that has the lowest resonant frequency for a given cavity size a b c is the dominant mode If a b c it implies that Ia lb lie and hence the dominant mode is TE101 Note that for a b c the resonant frequency of TMU0 mode is higher than that for TE101 mode hence TE101 is dominant When different modes have the same resonant frequency we say that the modes are degenerate one mode will dominate others depending on how the cavity is excited A practical resonant cavity has walls with finite conductivity ac and is therefore capable of losing stored energy The quality factor Q is a means of determining the loss 128 WAVEGUIDE RESONATORS 579 The qiialitx factor is also a measure of I he bandwidth ol the cavity resonator It may be defined as Time average energy stored Energy loss per cycle of oscillation W W 1289 where T 1 the period of oscillation PL is the time average power loss in the cavity and W is the total time average energy stored in electric and magnetic fields in the cavity Q is usually very high for a cavity resonator compared with that for an RLC resonant circuit By following a procedure similar to that used in deriving ac in Section 126 it can be shown that the quality factor for the dominant TE01 is given by3 GTE 1 0 1 52ba3 a2 c 3 c2abc acia fc 1290 where 5 is the skin depth of the cavity walls EXAMPLE 128 An airfilled resonant cavity with dimensions a 5 cm b 4 cm and c 10 cm is made of copper oc 58 X 107 mhosm Find a The five lowest order modes b The quality factor for TE1Oi mode Solution a The resonant frequency is given by m where u c 3For the proof see S V Marshall and G G Skitek Electromagnetic Concepts and Applications 3rd ed Englewood Cliffs NJ PrenticeHall 1990 pp 440442 580 S Waveguides Hence 3 X lf m 5 X 10 4 X 10 2 10 X 10 2 15V004m2 006252 0012 GHz Since c a b or 1c Ia 1 the lowest order mode is TE101 Notice that TMioi and TE1Oo do not exist because m 12 3 n 12 3 and p 0 1 2 3 for the TM modes and m 0 1 2 n 0 1 2 and p 123 for the TE modes The resonant frequency for the TE1Oi mode is frm 15V004 0 001 3335 GHz The next higher mode is TE011 TM011 does not exist with fron 15V0 00625 001 404 GHz The next mode is TE102 TM02 does not exist with frim 15V004 0 004 4243 GHz The next mode is TM110 TE110 does not exist with fruo 15V004 00625 0 48 GHz The next two modes are TE i n and TM n degenerate modes with frni 15V004 00625 001 5031 GHz The next mode is TM103 with frm 15V004 0 009 5408 GHz Thus the five lowest order modes in ascending order are TE101 335 GHz TEon 404 GHz TE102 4243 GHz TM U 0 48 GHz TEiorTM i n 5031 GHz b The quality factor for TE01 is given by 2TEI01 a2 c2 abc S2ba c3 aca2 c2 25 100 200 X 102 58125 1000 5025 100 616 61 VV335 X 109 4TT X 107 58 X 107 61 14358 SUMMARY 581 PRACTICE EXERCISE 128 If the resonant cavity of Example 128 is filled with a lossless material xr 1 er 3 find the resonant frequency fr and the quality factor for TE101 mode Answer 1936 GHz 1093 X 104 SUMMARY 1 Waveguides are structures used in guiding EM waves at high frequencies Assuming a lossless rectangular waveguide ac o a 0 we apply Maxwells equations in ana lyzing EM wave propagation through the guide The resulting partial differential equa tion is solved using the method of separation of variables On applying the boundary conditions on the walls of the guide the basic formulas for the guide are obtained for different modes of operation 2 Two modes of propagation or field patterns are the TMmn and TEmn where m and n are positive integers For TM modes m 1 2 3 and n 1 2 3 and for TE modes m 0 1 2 and n 0 1 2 n mz0 3 Each mode of propagation has associated propagation constant and cutoff frequency The propagation constant y a jfl does not only depend on the constitutive pa rameters e x a of the medium as in the case of plane waves in an unbounded space it also depends on the crosssectional dimensions a b of the guide The cutoff fre quency is the frequency at which y changes from being purely real attenuation to purely imaginary propagation The dominant mode of operation is the lowest mode possible It is the mode with the lowest cutoff frequency If a b the dominant mode is TE10 4 The basic equations for calculating the cutoff frequency fc phase constant 13 and phase velocity u are summarized in Table 121 Formulas for calculating the attenuation con stants due to lossy dielectric medium and imperfectly conducting walls are also pro vided 5 The group velocity or velocity of energy flow ug is related to the phase velocity up of the wave propagation by upug u2 where u 1vxs is the medium velocityie the velocity of the wave in the di electric medium unbounded by the guide Although up is greater than u up does not exceed u 6 The mode of operation for a given waveguide is dictated by the method of exci tation 7 A waveguide resonant cavity is used for energy storage at high frequencies It is nothing but a waveguide shorted at both ends Hence its analysis is similar to that of a wave guide The resonant frequency for both the TE and TM modes to z is given by m 582 Waveguides For TM modes m 1 2 3 n 1 2 3 and p 0 1 2 3 and for TE modes m 0123 n 0 1 2 3 and p 1 2 3 m n 0 If a b c the dominant mode one with the lowest resonant frequency is TE1Oi 8 The quality factor a measure of the energy loss in the cavity is given by 2 121 At microwave frequencies we prefer waveguides to transmission lines for transporting EM energy because of all the following except that a Losses in transmission lines are prohibitively large b Waveguides have larger bandwidths and lower signal attenuation c Transmission lines are larger in size than waveguides d Transmission lines support only TEM mode 122 An evanscent mode occurs when a A wave is attenuated rather than propagated b The propagation constant is purely imaginary c m 0 n so that all field components vanish d The wave frequency is the same as the cutoff frequency 123 The dominant mode for rectangular waveguides is a TE b TM n c TE1Oi d TE10 124 The TM10 mode can exist in a rectangular waveguide a True b False 125 For TE30 mode which of the following field components exist a Ex b Ey c Ez d Hx e Hv PROBLEMS 583 126 If in a rectangular waveguide for which a 2b the cutoff frequency for TE02 mode is 12 GHz the cutoff frequency for TMH mode is a 3 GHz b 35GHz c 12 GHz d 6A GHz e None of the above 127 If a tunnel is 4 by 7 m in cross section a car in the tunnel will not receive an AM radio signal eg 10 MHz a True b False 128 When the electric field is at its maximum value the magnetic energy of a cavity is a At its maximum value b At V 2 of its maximum value c At p of its maximum value V2 d At 12 of its maximum value e Zero 129 Which of these modes does not exist in a rectangular resonant cavity a TE110 b TEQH c TM110 d TMm 1210 How many degenerate dominant modes exist in a rectangular resonant cavity for which a b c a 0 b 2 c 3 d 5 e oo Answers 121c 122a 123d 124b 125bd 126b 127a 128e 129a 1210c PROBLEMS I n o w m a t a rectangular waveguide does not support TM10 and TM01 modes b Explain the difference between TEmn and TMmn modes 584 Waveguides 122 A 2cm by 3cm waveguide is filled with a dielectric material with er 4 If the wave guide operates at 20 GHz with TMU mode find a cutoff frequency b the phase con stant c the phase velocity 123 A 1cm X 2cm waveguide is filled with deionized water with er 81 If the operating frequency is 45 GHz determine a all possible propagating modes and their cutoff fre quencies b the intrinsic impedance of the highest mode c the group velocity of the lowest mode 124 Design a rectangular waveguide with an aspect ratio of 3 to 1 for use in the k band 18265 GHz Assume that the guide is air filled 125 A tunnel is modeled as an airfilled metallic rectangular waveguide with dimensions a 8 m and b 16 m Determine whether the tunnel will pass a a 15MHz AM broadcast signal b a 120MHz FM broadcast signal 126 In an airfilled rectangular waveguide the cutoff frequency of a TE10 mode is 5 GHz whereas that of TEOi mode is 12 GHz Calculate a The dimensions of the guide b The cutoff frequencies of the next three higher TE modes c The cutoff frequency for TEn mode if the guide is filled with a lossless material having er 225 and ir 127 An airfilled hollow rectangular waveguide is 150 m long and is capped at the end with a metal plate If a short pulse of frequency 72 GHz is introduced into the input end of the guide how long does it take the pulse to return to the input end Assume that the cutoff frequency of the guide is 65 GHz 128 Calculate the dimensions of an airfilled rectangular waveguide for which the cutoff fre quencies for TM n and TE03 modes are both equal to 12 GHz At 8 GHz determine whether the dominant mode will propagate or evanesce in the waveguide 129 An airfilled rectangular waveguide has crosssectional dimensions a 6 cm and b 3 cm Given that E 5 sin sin A cos 1012f 0z Vm V a V b calculate the intrinsic impedance of this mode and the average power flow in the guide 1210 In an airfilled rectangular waveguide a TE mode operating at 6 GHz has Ey 5 sm2irxa coswyb smat 12z Vm Determine a the mode of operation b the cutoff frequency c the intrinsic imped ance d Hx PROBLEMS 585 1211 In an airfilled rectangular waveguide with a 2286 cm and b 1016 cm the ycomponent of the TE mode is given by Ey sin27rxa sin107r X 1010r j3z Vm find a the operating mode b the propagation constant 7 c the intrinsic impedance V 1212 For the TM mode derive a formula for the average power transmitted down the guide 1213 a Show that for a rectangular waveguide x X 1 b For an airfilled waveguide with a 2b 25 cm operating at 20 GHz calculate up and X for TE n and TE2i modes 1214 A 1cm X 3cm rectangular airfilled waveguide operates in the TE2 mode at a fre quency that is 20 higher than the cutoff frequency Determine a the operating fre quency b the phase and group velocities 1215 A microwave transmitter is connected by an airfilled waveguide of cross section 25 cm X 1 cm to an antenna For transmission at 11 GHz find the ratio of a the phase velocity to the medium velocity and b the group velocity to the medium velocity 1216 A rectangular waveguide is filled with polyethylene s 225eo and operates at 24 GHz If the cutoff frequency of a certain TE mode is 16 GHz find the group velocity and intrinsic impedance of the mode 1217 A rectangular waveguide with cross sections shown in Figure 1216 has dielectric dis continuity Calculate the standing wave ratio if the guide operates at 8 GHz in the domi nant mode 1218 Analysis of circular waveguide requires solution of the scalar Helmholtz equation in cylindrical coordinates namely V2EZS k2Ezs 0 25 cm 5 cm Figure 1216 For Problem 1217 fio so fio 225so 586 Waveguides or 1 d f dEzs 1 d2Ea d2Ezs 2 p dp V dp p 30 3z By assuming the product solution Ezsp z Rp t Zz show that the separated equations are Z k Z 0 i o where pR A2 p2 kl R 0 t 2 t 2 2 1219 For TE01 mode xs Y Ho sinirybe 7 Eys Find and Pa 1220 A 1cm X 2cm waveguide is made of copper ac 58 X 107 Sm and filled with a dielectric material for which e 26eo i po ad 1CT4 Sm If the guide operates at 9 GHz evaluate ac and ad for a TE10 and b TMU 1221 A 4cmsquare waveguide is filled with a dielectric with complex permittivity ec 16eol 7IO4 and is excited with the TM2i mode If the waveguide operates at 10 above the cutoff frequency calculate attenuation ad How far can the wave travel down the guide before its magnitude is reduced by 20 1222 If the walls of the square waveguide in the previous problem are made of brass ac 15 X 10 Sm find ac and the distance over which the wave is attenuated by 30 1223 A rectangular waveguide with a 2b 48 cm is filled with teflon with er 211 and loss tangent of 3 X 104 Assume that the walls of the waveguide are coated with gold 7C 41 X 107 Sm and that a TE10 wave at 4 GHz propagates down the waveguide find a ad b ctc 1224 A rectangular brass ac 137 X 107 Sm waveguide with dimensions a 225 cm and b 15 cm operates in the dominant mode at frequency 5 GHz If the waveguide is filled with teflon pr 1 er 211 a 0 determine a the cutoff frequency for the dominant mode b the attenuation constant due to the loss in the guide walls 1225 For a square waveguide show that attenuation ac is minimum for TE1 mode when 2962 c PROBLEMS 587 1226 The attenuation constant of a TM mode is given by At what frequency will a be maximum 1227 Show that for TE mode to z in a rectangular cavity jwixmir fmirx Iniry pirz Eys A HO sin cos sin h a J a b J c Find Hxs 1228 For a rectangular cavity show that pirz c o s for TM mode to z Determine Eys 1229 In a rectangular resonant cavity which mode is dominant when a a b c b a b c c a c b 1230 For an airfilled rectangular cavity with dimensions a 3 cm b 2 cm c 4 cm determine the resonant frequencies for the following modes TE o n TE101 TMno and TM1U List the resonant frequencies in ascending order 1231 A rectangular cavity resonator has dimensions a 3 cm b 6 cm and c 9 cm If it is filled with polyethylene e 25e0 find the resonant frequencies of the first five lowestorder modes 1232 An airfilled cubical cavity operates at a resonant frequency of 2 GHz when excited at the TE1Oi mode Determine the dimensions of the cavity 1233 An airfilled cubical cavity of size 32 cm is made of brass jc 137 X 107 Sm Cal culate a the resonant frequency of the TE101 mode b the quality factor at that mode 1234 Design an airfilled cubical cavity to have its dominant resonant frequency at 3 GHz 1235 An airfilled cubical cavity of size 10 cm has E 200 sin 30TTX sin 30iry cos 6 X l09r a Vm Find H Chapter 13 ANTENNAS The Ten Commandments of Success 1 Hard Work Hard work is the best investment a man can make 2 Study Hard Knowledge enables a man to work more intelligently and effec tively 3 Have Initiative Ruts often deepen into graves 4 Love Your Work Then you will find pleasure in mastering it 5 Be Exact Slipshod methods bring slipshod results 6 Have the Spirit of Conquest Thus you can successfully battle and overcome difficulties 7 Cultivate Personality Personality is to a man what perfume is to the flower 8 Help and Share with Others The real test of business greatness lies in giving opportunity to others 9 Be Democratic Unless you feel right toward your fellow men you can never be a successful leader of men 10 In all Things Do Your Best The man who has done his best has done every thing The man who has done less than his best has done nothing CHARLES M SCHWAB 131 INTRODUCTION Up until now we have not asked ourselves how EM waves are produced Recall that elec tric charges are the sources of EM fields If the sources are time varying EM waves prop agate away from the sources and radiation is said to have taken place Radiation may be thought of as the process of transmitting electric energy The radiation or launching of the waves into space is efficiently accomplished with the aid of conducting or dielectric struc tures called antennas Theoretically any structure can radiate EM waves but not all struc tures can serve as efficient radiation mechanisms An antenna may also be viewed as a transducer used in matching the transmission line or waveguide used in guiding the wave to be launched to the surrounding medium or vice versa Figure 131 shows how an antenna is used to accomplish a match between the line or guide and the medium The antenna is needed for two main reasons efficient radiation and matching wave impedances in order to minimize reflection The antenna uses voltage and current from the transmission line or the EM fields from the waveguide to launch an EM wave into the medium An antenna may be used for either transmitting or receiving EM energy 588 131 INTRODUCTION 589 EM wave Generator Transmission line Antenna Surrounding medium Figure 131 Antenna as a matching device between the guiding struc ture and the surrounding medium Typical antennas are illustrated in Figure 132 The dipole antenna in Figure 132a consists of two straight wires lying along the same axis The loop antenna in Figure 132b consists of one or more turns of wire The helical antenna in Figure 132c consists of a wire in the form of a helix backed by a ground plane Antennas in Figure 132ac are called wire antennas they are used in automobiles buildings aircraft ships and so on The horn antenna in Figure 132d an example of an aperture antenna is a tapered section of waveguide providing a transition between a waveguide and the surroundings Since it is conveniently flush mounted it is useful in various applications such as aircraft The parabolic dish reflector in Figure 132e utilizes the fact that EM waves are reflected by a conducting sheet When used as a transmitting antenna a feed antenna such as a dipole or horn is placed at the focal point The radiation from the source is reflected by the dish acting like a mirror and a parallel beam results Parabolic dish antennas are used in communications radar and astronomy The phenomenon of radiation is rather complicated so we have intentionally delayed its discussion until this chapter We will not attempt a broad coverage of antenna theory our discussion will be limited to the basic types of antennas such as the Hertzian dipole the halfwave dipole the quarterwave monopole and the small loop For each of these types we will determine the radiation fields by taking the following steps 1 Select an appropriate coordinate system and determine the magnetic vector poten tial A 2 Find H from B tH V X A 3 Determine E from V X H e or E iH X as assuming a lossless medium dt a 0 4 Find the far field and determine the timeaverage power radiated using dS where ve Re E X H Note that Pnd throughout this chapter is the same as Pme in eq 1070 590 Antennas a dipole b loop c helix d pyramidal horn Radiating dipole Reflector e parabolic dish reflector Figure 132 Typical antennas 132 HERTZIAN DIPOLE By a Hertzian dipole we mean an infinitesimal current element dl Although such a current element does not exist in real life it serves as a building block from which the field of a practical antenna can be calculated by integration Consider the Hertzian dipole shown in Figure 133 We assume that it is located at the origin of a coordinate system and that it carries a uniform current constant throughout the dipole I Io cos cot From eq 954 the retarded magnetic vector potential at the field point P due to the dipole is given by A Awr 131 132 HERTZIAN DIPOLE 591 Figure 133 A Hertzian dipole carrying current I Io cos cot where is the retarded current given by Io cos a t Io cos bit 3r u J 132 Re IoejMM where 3 tow 2TTA and u 1Vxe The current is said to be retarded at point P because there is a propagation time delay rlu or phase delay 3r from O to P By substitut ing eq 132 into eq 131 we may write A in phasor form as 133 Azs A e Transforming this vector in Cartesian to spherical coordinates yields A Ars A6s A where A n A z s cos 8 Affs Azs sin 6 But B H V X As hence we obtain the H field as 0 Iodl H sin 0 r e j3 4x lr r Hrs 0 Ss We find the E field using V X H e dWdt or V X Hs jueEs u fl j 7 3 r E 0 r r 134 135a 135b 136a 136b 136c 592 Hi Antennas where V A close observation of the field equations in eqs 135 and 136 reveals that we have terms varying as 1r3 1r2 and 1r The 1r3 term is called the electrostatic field since it corresponds to the field of an electric dipole see eq 482 This term dominates over other terms in a region very close to the Hertzian dipole The 1r term is called the induc tive field and it is predictable from the BiotSavart law see eq 73 The term is impor tant only at near field that is at distances close to the current element The 1r term is called the far field or radiation field because it is the only term that remains at the far zone that is at a point very far from the current element Here we are mainly concerned with the far field or radiation zone j3r 5 1 or 2irr S X where the terms in 1r3 and 1r2 can be neglected in favor of the 1r term Thus at far field 4irr sin 6 e V Ers 0 I37a I37b Note from eq 137a that the radiation terms of Hs and E9s are in time phase and orthog onal just as the fields of a uniform plane wave Also note that nearzone and farzone fields are determined respectively to be the inequalities 3r C I and f3r I More specifically we define the boundary between the near and the far zones by the value of r given by 2d2 r 138 where d is the largest dimension of the antenna The timeaverage power density is obtained as 1 2Pave Re Es X H Re E6s H ar 139 Substituting eq 137 into eq 139 yields the timeaverage radiated power as dS to Jeo 3 2 T T 2 327r2r2 sin2 6 r2 sin 6 dd dj 1310 2TT sin 6 dO 132 HERTZIAN DIPOLE 593 But sin 6d6 1 cosz 0 dcos 9 cos30 cos i and 02 4TT2X2 Hence eq 1310 becomes rad dl 3 L X If free space is the medium of propagation rj 120TT and 1311a 1311b This power is equivalent to the power dissipated in a fictitious resistance rad by current I Io cos cot that is rad rms rad or 1 1312 where rms is the rootmeansquare value of From eqs 1311 and 1312 we obtain OP r z rad 1 o 11 rad ZV 1313a or 1313b The resistance Rmd is a characteristic property of the Hertzian dipole antenna and is called its radiation resistance From eqs 1312 and 1313 we observe that it requires anten nas with large radiation resistances to deliver large amounts of power to space For example if dl X20 Rrad 2 U which is small in that it can deliver relatively small amounts of power It should be noted that rad in eq 1313b is for a Hertzian dipole in free space If the dipole is in a different lossless medium rj Vxe is substituted in eq 1311a and rad is determined using eq 1313a Note that the Hertzian dipole is assumed to be infinitesimally small dl S 1 or dl X10 Consequently its radiation resistance is very small and it is in practice difficult to match it with a real transmission line We have also assumed that the dipole has a 594 Antennas uniform current this requires that the current be nonzero at the end points of the dipole This is practically impossible because the surrounding medium is not conducting However our analysis will serve as a useful valid approximation for an antenna with dl s X10 A more practical and perhaps the most important antenna is the halfwave dipole considered in the next section 133 HALFWAVE DIPOLE ANTENNA The halfwave dipole derives its name from the fact that its length is half a wavelength A2 As shown in Figure 134a it consists of a thin wire fed or excited at the mid point by a voltage source connected to the antenna via a transmission line eg a twowire line The field due to the dipole can be easily obtained if we consider it as consisting of a chain of Hertzian dipoles The magnetic vector potential at P due to a differential length dl dz of the dipole carrying a phasor current Is Io cos fiz is 1314 Transmission line Dipole antenna Current distribution Figure 134 A halfwave dipole cos 3z t a 133 HALFWAVE DIPOLE ANTENNA 595 Notice that to obtain eq 1314 we have assumed a sinusoidal current distribution because the current must vanish at the ends of the dipole a triangular current distribution is also possible see Problem 134 but would give less accurate results The actual current distribution on the antenna is not precisely known It is determined by solving Maxwells equations subject to the boundary conditions on the antenna but the procedure is mathe matically complex However the sinusoidal current assumption approximates the distribu tion obtained by solving the boundaryvalue problem and is commonly used in antenna theory If r S as explained in Section 49 on electric dipoles see Figure 421 then r r z cos i or Thus we may substitute r r in the denominator of eq 1314 where the magnitude of the distance is needed For the phase term in the numerator of eq 1314 the dif ference between fir and ftr is significant so we replace r by r z cos 6 and not r In other words we maintain the cosine term in the exponent while neglecting it in the de nominator because the exponent involves the phase constant while the denominator does not Thus W4 4irr A4 1315 j8z cos e cos fiz dz A4 From the integral tables of Appendix A8 eaz cos bz dz eaz a cos bz b sin bz Applying this to eq 1315 gives Azs nloejl3rejl3zcose UP cos 0 cos 3z ff sin ffz A4 A4 1316 Since 0 2xX or 3 X4 TT2 and cos2 0 1 sin2 0 eq 1316 becomes A f en0 13 e 0 ft 1 3 1 7 Awrfi sin 0 Using the identity eJX ec 2 cos x we obtain cos 6 1318 txloe ircos I c o s I 2Trrj3sin2 6 596 Antennas We use eq 134 in conjunction with the fact that B HS V X As and V X H ycoeEs to obtain the magnetic and electric fields at far zone discarding the 1r3 and 1r2 terms as 1319 Notice again that the radiation term of Hs and Es are in time phase and orthogonal Using eqs 139 and 1319 we obtain the timeaverage power density as cos2 cos 6 1320 8TTV sin2 The timeaverage radiated power can be determined as 2 COS2 2K fw I 2 COS2 I COS 0 8x2r2 sin2 r2 sin 0 d6 dj 2TT 1321 sin i J rT COS I COS I sde o sm0 where t 120TT has been substituted assuming free space as the medium of propagation Due to the nature of the integrand in eq 1321 TT2 COS COS 6 sine cosl cos I de I de sin 0 J0 Jitl2 This is easily illustrated by a rough sketch of the variation of the integrand with d Hence 602 IT c o s i sin I 1322 133 HALFWAVE DIPOLE ANTENNA S 597 Changing variables u cos 6 and using partial fraction reduces eq 1322 to COS 2 TT u2 du 307 2 1 2 1 COS KU r COS KU 2 2 j du du 1 U 0 1 u 1323 Replacing 1 u with v in the first integrand and 1 u with v in the second results in rad 302 302 sin27TV dv L0 2 Sin 2 7TV 2 sin TTV dv 1324 Changing variables w irv yields 2TT sin w dw 15 15 2 1 COS 2 4 6 8 1325 w2 w4 w6 w8 since cos w l H 1 Integrating eq 1325 term by term and 2 4 6 8 evaluating at the limit leads to 2 f 2TT2 2TT4 2r6 2TT8 1 5 L 22 44 66 88 3656 ll 1326 The radiation resistance Rrad for the halfwave dipole antenna is readily obtained from eqs 1312 and 1326 as 1327 598 Antennas Note the significant increase in the radiation resistance of the halfwave dipole over that of the Hertzian dipole Thus the halfwave dipole is capable of delivering greater amounts of power to space than the Hertzian dipole The total input impedance Zin of the antenna is the impedance seen at the terminals of the antenna and is given by in 1328 where Rin Rmd for lossless antenna Deriving the value of the reactance Zin involves a complicated procedure beyond the scope of this text It is found that Xin 425 0 so Zin 73 y425 0 for a dipole length X2 The inductive reactance drops rapidly to zero as the length of the dipole is slightly reduced For 0485 X the dipole is resonant with Xin 0 Thus in practice a X2 dipole is designed such that Xin approaches zero and Zin 73 0 This value of the radiation resistance of the X2 dipole antenna is the reason for the standard 750 coaxial cable Also the value is easy to match to transmission lines These factors in addition to the resonance property are the reasons for the dipole antennas popularity and its extensive use 134 QUARTERWAVE MONOPOLE ANTENNA Basically the quarterwave monopole antenna consists of onehalf of a halfwave dipole antenna located on a conducting ground plane as in Figure 135 The monopole antenna is perpendicular to the plane which is usually assumed to be infinite and perfectly conduct ing It is fed by a coaxial cable connected to its base Using image theory of Section 66 we replace the infinite perfectly conducting ground plane with the image of the monopole The field produced in the region above the ground plane due to the X4 monopole with its image is the same as the field due to a X2 wave dipole Thus eq 1319 holds for the X4 monopole However the integration in eq 1321 is only over the hemispherical surface above the ground plane ie 0 d TT2 because the monopole radiates only through that surface Hence the monopole radiates only half as much power as the dipole with the same current Thus for a X4 monopole 18282 1329 and IP ad Figure 135 The monopole antenna Image Infinite conducting ground plane 135 SMALL LOOP ANTENNA 599 or Rmd 365 0 1330 By the same token the total input impedance for a A4 monopole is Zin 365 2125 12 135 SMALL LOOP ANTENNA The loop antenna is of practical importance It is used as a directional finder or search loop in radiation detection and as a TV antenna for ultrahigh frequencies The term small implies that the dimensions such as po of the loop are much smaller than X Consider a small filamentary circular loop of radius po carrying a uniform current Io cos co as in Figure 136 The loop may be regarded as an elemental magnetic dipole The magnetic vector potential at the field point P due to the loop is A 1331 where 7 7O cos cor 3r Re loeji ISr Substituting 7 into eq 1331 we obtain A in phasor form as ejfir e Ait L r 1332 Evaluating this integral requires a lengthy procedure It can be shown that for a small loop po SC r can be replaced by r in the denominator of eq 1332 and As has only f component given by s AK 1 jreir sin 6 1333 Figure 136 The small loop antenna N Transmission line 600 Antennas where S wpl loop area For a loop with N turns S Nirpl Using the fact that Bs xHs VX A and V X H S jusEs we obtain the electric and magnetic fields from eq 1333 as Ai sin I 1334a 2m 4TTT 3r3 sin 0 J r 2 3rJ ra Eds Hfs 0 1334b 1334c 1334d Comparing eqs 135 and 136 with eq 1334 we observe the dual nature of the field due to an electric dipole of Figure 133 and the magnetic dipole of Figure 136 see Table 82 also At far field only the 1r term the radiation term in eq 1334 remains Thus at far field 4irr 18 sin 6 e r2 sin o e or 1335a Hrs 0 1335b where 77 120TT for free space has been assumed Though the far field expressions in eq 1335 are obtained for a small circular loop they can be used for a small square loop with one turn S a with Af turns S Na2 or any small loop provided that the loop di mensions are small d A10 where d is the largest dimension of the loop It is left as an exercise to show that using eqs 1313a and 1335 gives the radiation resistance of a small loop antenna as 1336 135 SMALL LOOP ANTENNA M 601 EXAMPLE 131 A magnetic field strength of 5 Am is required at a point on 6 TT2 2 km from an antenna in air Neglecting ohmic loss how much power must the antenna transmit if it is a A Hertzian dipole of length X25 b A halfwave dipole c A quarterwave monopole d A 10turn loop antenna of radius po X20 Solution a For a Hertzian dipole 7o3 dl sin 6 051 A 4irr where dl X25 or 0 dl Hence 5 X 1T6 4TT 2 X 103 105 or Io 05 A H 40TT2 I X 158 mW 40x2052 252 b For a X2 dipole 5 x o cos I cos 2irr sin 6 1 2TT2 X or 207T mA rad 1220TTZ X 1073 144 mW 602 B Antennas c For a X4 monopole as in part b d For a loop antenna 2 o 20TT mA l2I2 0Rmd 1220TT2 X 1063656 72 mW S r X2 sin 8 For a single turn S Kpo For Vturn S Nwp0 Hence or 5 X io 6 2 X 103 L X 10 IOTT2 LPO 4053 mA I X 103 320 7T6 X 100 io l 1923fi Zrad orad 40532 X 106 1923 158 mW PRACTICE EXERCISE 131 A Hertzian dipole of length X100 is located at the origin and fed with a current of 025 sin 108f A Determine the magnetic field at a r X50 30 b r 200X 6 60 Answer a 02119 sin 10s 205 a0 mAm b 02871 sin l08t 90 a0 135 SMALL LOOP ANTENNA 603 EXAMPLE 132 An electric field strength of 10 uVm is to be measured at an observation point 6 ir2 500 km from a halfwave resonant dipole antenna operating in air at 50 MHz a What is the length of the dipole b Calculate the current that must be fed to the antenna c Find the average power radiated by the antenna d If a transmission line with Zo 75 0 is connected to the antenna determine the stand ing wave ratio Solution c 3 X 108 a The wavelength X r 6 m 50 X 106 Hence the length of the halfdipole is 3 m b From eq 1319 rJo cos cos 6 2wr sin 6 or 2irr sin 9 ro cos I cos 6 2 j 10 X 10 62TT500 X 103 1 120irl c 8333 mA Rmd 73 Q 83332 X 106 X 73 d 2535 mW F ZL Zin in this case z Zo 73 y425 75 2 y425 73 y425 75 42559269 153981602 148 y425 027637667 s 1 r 1 02763 r 1 02763 1763 604 Antennas PRACTICE EXERCISE 132 Repeat Example 132 if the dipole antenna is replaced by a X4 monopole Answer a 15m b 8333 mA c 1268 mW d 2265 136 ANTENNA CHARACTERISTICS Having considered the basic elementary antenna types we now discuss some important characteristics of an antenna as a radiator of electromagnetic energy These characteristics include a antenna pattern b radiation intensity c directive gain d power gain A Antenna Patterns An antenna pattern or radiation pattern is a ihrceclimensional plot of iis radia tion ai fur field When the amplitude of a specified component of the E field is plotted it is called the field pattern or voltage pattern When the square of the amplitude of E is plotted it is called the power pattern A threedimensional plot of an antenna pattern is avoided by plotting sepa rately the normalized ES versus 0 for a constant 4 this is called an Eplane pattern or ver tical pattern and the normalized ES versus t for 8 TT2 called the Hplanepattern or horizontal pattern The normalization of ES is with respect to the maximum value of the so that the maximum value of the normalized ES is unity For the Hertzian dipole for example the normalized iSj is obtained from eq 137 as sin0 1337 which is independent of t From eq 1337 we obtain the plane pattern as the polar plot of j8 with 8 varying from 0 to 180 The result is shown in Figure 137a Note that the plot is symmetric about the zaxis 8 0 For the fplane pattern we set 8 TT2 SO that0 1 which is circle of radius 1 as shown in Figure 137b When the two plots of Figures 137a and b are combined we have a threedimensional field pattern of Figure 137c which has the shape of a doughnut A plot of the timeaverage power 2Pave 2Pave for a fixed distance r is the power pattern of the antenna It is obtained by plotting separately 2Pave versus 8 for constant j and Save versus 4 for constant 8 For the Hertzian dipole the normalized power pattern is easily obtained from eqs 1337 or 139 as 20 sin2 0 1338 which is sketched in Figure 138 Notice that Figures 137b and 138b show circles because fi8 is independent of j and that the value of OP in Figure 138a is the relative 136 ANTENNA CHARACTERISTICS B 605 a c Figure 137 Field patterns of the Hertzian dipole a normalized plane or vertical pattern 4 constant 0 b normalized ffplane or horizontal pattern 6 TT2 C threedimensional pattern Polar axis a b Figure 138 Power pattern of the Hertzian dipole a 4 constant 0 b 6 constant T2 606 Antennas average power for that particular 6 Thus at point Q 0 45 the average power is one half the maximum average power the maximum average power is at 6 TT2 B Radiation Intensity The radiation intensity of an antenna is defined as me 0 r2 ga 1339 From eq 1339 the total average power radiated can be expressed as Sin 6 dd d Udj sin dddt 1340 2ir fir U6 dU o Jeo where dQ sin 9 dd df is the differential solid angle in steradian sr Hence the radiation intensity U6 f is measured in watts per steradian Wsr The average value of Ud j is the total radiated power divided by 4TT sr that is rrad 4T 1341 C Directive Gain Besides the antenna patterns described above we are often interested in measurable quan tities such as gain and directivity to determine the radiation characteristics of an antenna The directive gain i06 of itn unlenna is a measure of the concentration of the ra diated power in a particular direction e p It may be regarded as the ability of the antenna to direct radiated power in a given direc tion It is usually obtained as the ratio of radiation intensity in a given direction 6 f to the average radiation intensity that is 1342 136 ANTENNA CHARACTERISTICS 607 By substituting eq 1339 into eq 1342 0 may be expressed in terms of directive gain as 1343 ave Airr The directive gain GJfi j depends on antenna pattern For the Hertzian dipole as well as for A2 dipole and X4 monopole we notice from Figure 138 that 2Pave is maximum at 6 7r2 and minimum zero at 6 0 or TT Thus the Hertzian dipole radiates power in a direction broadside to its length For an isotropic antenna one that radiates equally in all directions Gd 1 However such an antenna is not a physicality but an ideality The directivity I of an antenna is ihe ratio of the maximum radiation intensity to the average radiaiion intensity Obviously D is the maximum directive gain Gd max Thus D Gd max 1344a or D Prad 1344b D 1 for an isotropic antenna this is the smallest value D can have For the Hertzian dipole G6j 15 sin2 0 D 15 For the A2 dipole 1345 Gdd where i 120x rad 73 fi and DM rad IT COS COS I sin0 1346 1347 D Power Gain Our definition of the directive gain in eq 1342 does not account for the ohmic power loss P of the antenna Pt is due to the fact that the antenna is made of a conductor with 608 Antennas finite conductivity As illustrated in Figure 139 if Pin is the total input power to the antenna Pin 1348 where 7in is the current at the input terminals and R is the loss or ohmic resistance of the antenna In other words Pin is the power accepted by the antenna at its terminals during the radiation process and Prad is the power radiated by the antenna the difference between the two powers is P the power dissipated within the antenna We define the power gain Gp6 j of the antenna as 1349 The ratio of the power gain in any specified direction to the directive gain in that direction is referred to as the radiation efficiency vr of the antennas that is GP Vr Introducing eq 1348 leads to Vr Prad Vad Rf 1350 For many antennas rr is close to 100 so that GP Gd It is customary to express direc tivity and gain in decibels dB Thus DdB 101og0 G dB 10 log10 G 1351a 1351b It should be mentioned at this point that the radiation patterns of an antenna are usually measured in the far field region The far field region of an antenna is commonly taken to exist at distance r rmin where 2dz 1352 Figure 139 Relating Pm P and Prad Prad 136 ANTENNA CHARACTERISTICS 609 and d is the largest dimension of the antenna For example d I for the electric dipole antenna and d 2p0 for the small loop antenna EXAMPLE 133 Show that the directive gain of the Hertzian dipole is Gd0 15 sin2 6 and that of the halfwave dipole is cos cos 6 Gd9t 1 6 4 sin Solution From eq 1342 f 4TT20 f 6 d a For the Hertzian dipole 4TT sin2 6 sin3 6 d6 dj 4TT sin2 6 2TT 43 15 sin2 6 as required b For the halfwave dipole 4TT COS cos sin2 2lr rir cos I cos 6 I dO df G9 t From eq 1326 the integral in the denominator gives 27r12188 Hence G8 0 4TT cos2 cos 9 sin20 12188 164 cos I cos I sin20 as required 610 Antennas PRACTICE EXERCISE 133 Calculate the directivity of a The Hertzian monopole b The quarterwave monopole Answer a 3 b 328 EXAMPLE 134 Determine the electric field intensity at a distance of 10 km from an antenna having a di rective gain of 5 dB and radiating a total power of 20 kW Solution or From eq 1343 But Hence 5 GddB 101og10Grf 05 log10 Gd GdPrad lO05 3162 at u ave op ave 4irr E 2V 1207T316220 X 103 E 2irr2 2TT10 X 1032 Es 01948 Vm PRACTICE EXERCISE 134 A certain antenna with an efficiency of 95 has maximum radiation intensity of 05 Wsr Calculate its directivity when a The input power is 04 W b The radiated power is 03 W Answer a 1653 b 2094 EXAMPLE 135 136 ANTENNA CHARACTERISTICS The radiation intensity of a certain antenna is 2 sin d sin3 0 0 0 TT 0 0 TT 611 U8 0 0 elsewhere Determine the directivity of the antenna Solution The directivity is defined as D ua From the given U 2 1 1 4TT J 2TT 9 i o eo sin 0 sin j sin 0 dj s i n jdt o 1 cos 20 d0 1 cosz 0 rfcos A 27r 4 2 4 sin 2TT2 27r2j3j 3 COS f I cos i o l 3 Hence Z 13 6 PRACTICE EXERCISE 135 Evaluate the directivity of an antenna with normalized radiation intensity fsin 0 0 0 TT2 0 0 2TT 0 otherwise Answer 2546 612 B Antennas 137 ANTENNA ARRAYS In many practical applications eg in an AM broadcast station it is necessary to design antennas with more energy radiated in some particular directions and less in other direc tions This is tantamount to requiring that the radiation pattern be concentrated in the di rection of interest This is hardly achievable with a single antenna element An antenna array is used to obtain greater directivity than can be obtained with a single antenna element An antenna array is a group of radiating elements arranged so us to produce some particular radiation characteristics It is practical and convenient that the array consists of identical elements but this is not fundamentally required We shall consider the simplest case of a twoelement array and extend our results to the more complicated general case of an Nelement array Consider an antenna consisting of two Hertzian dipoles placed in free space along the zaxis but oriented parallel to the taxis as depicted in Figure 1310 We assume that the dipole at 0 0 d2 carries current Ils I0cx and the one at 0 0 d2 carries current hs 40 where a is the phase difference between the two currents By varying the spacing d and phase difference a the fields from the array can be made to interfere con structively add in certain directions of interest and interfere destructively cancel in other directions The total electric field at point P is the vector sum of the fields due to the individual elements If P is in the far field zone we obtain the total electric field at P from eq 137a as Is COS0 1353 Note that sin 6 in eq 137a has been replaced by cos 6 since the element of Figure 133 is zdirected whereas those in Figure 1310 are xdirected Since P is far from the array Figure 1310 A twoelement array i 9 62 and ae ae we use 137 ANTENNA ARRAYS S 613 i In the amplitude we can set rx r r2 but in the phase d rx r cos I r2 r r cos i 1354a 1354b Thus eq 1353 becomes 4x r a2j 4r r cos cos L2 cos 1355 Comparing this with eq 137a shows that the total field of an array is equal to the field of single element located at the origin multiplied by an array factor given by AF 2 cos tacos 8 u eja2 1356 Thus in general the far field due to a twoelement array is given by E total E due to single element at origin X array factor 1357 Also from eq 1355 note that cos d is the radiation pattern due to a single element whereas the normalized array factor cosl28Jcos 6 a is the radiation pattern of the array if the elements were isotropic These may be regarded as unit pattern and group pattern respectively Thus the resultant pattern is the product of the unit pattern and the group pattern that is Resultant pattern Unit pattern X Group pattern 1358 This is known as pattern multiplication It is possible to sketch almost by inspection the pattern of an array by pattern multiplication It is therefore a useful tool in the design of an array We should note that while the unit pattern depends on the type of elements the array is comprised of the group pattern is independent of the element type so long as the spacing d and phase difference a and the orientation of the elements remain the same Let us now extend the results on the twoelement array to the general case of an N element array shown in Figure 1311 We assume that the array is linear in that the ele ments are spaced equally along a straight line and lie along the zaxis Also we assume that the array is uniform so that each element is fed with current of the same magnitude but of progressive phase shift a that is Ils O012s Iou I3s 7o2q and so on We are mainly interested in finding the array factor the far field can easily be found from eq 614 Antennas Figure 1311 An iVelement uniform linear array d cos 6 1357 once the array factor is known For the uniform linear array the array factor is the sum of the contributions by all the elements Thus AF 1 eJ4r ej2p eAN14 where 3d cos 6 a 1359 1360 In eq 1360 fi 2xX d and a are respectively the spacing and interelement phase shift Notice that the righthand side of eq 1359 is a geometric series of the form 1 x x2 x3 Hence eq 1359 becomes 1 x AF 1 1361 1362 which can be written as AF ejN42 sin Aty2 sin l2 1363 The phase factor eJN ln would not be present if the array were centered about the origin Neglecting this unimportant term 1364 AF b fid cos 6 a 137 ANTENNA ARRAYS 615 Note that this equation reduces to eq 1356 when N las expected Also note the fol lowing 1 AF has the maximum value of TV thus the normalized AF is obtained by dividing AF by N The principal maximum occurs when J 0 that is 0 fid cos 6 a or 2 AF has nulls or zeros when AF 0 that is Nip cos 0 a Yd br k 12 3 1365 1366 where k is not a multiple of N 3 A broadside array has its maximum radiation directed normal to the axis of the array that is p 0 90 so that a 0 4 An endfire array has its maximum radiation directed along the axis of the array that is p 0 B so that a These points are helpful in plotting AF For N23 and 4 the plots of AF are sketched in Figure 1312 EXAMPLE 136 For the twoelement antenna array of Figure 1310 sketch the normalized field pattern when the currents are a Fed in phase a 0 d A2 b Fed 90 out of phase a TT2 d A4 Solution The normalized field of the array is obtained from eqs 1355 to 1357 as cos 6 cos 0d cos 8 a a If a 0 d A213d TT Hence A 2 1 resultant pattern cos0 1 unit X pattern cos cos 6 1 group pattern The sketch of the unit pattern is straightforward It is merely a rotated version of that in Figure 137a for the Hertzian dipole and is shown in Figure 1313a To sketch a 616 m Antennas 0 TT2 IT 3TT2 2TT Figure 1312 Array factor for uniform linear array 108 TVS group pattern requires that we first determine its nulls and maxima For the nulls or zeros K i IT 3f cos cos 0 0 cos 0 2 2 2 2 or For the maxima or 0 0 180 cos cos 0 1 cos 0 0 0 90 unit pattern a x X 137 ANTENNA ARRAYS 617 group pattern b resultant pattern c Figure 1313 For Example 136a field patterns in the plane containing the axes of the elements The group pattern is as shown in Figure 1312b It is the polar plot obtained by sketching for0 0 5 10 15 360 and incorporating the nulls and cos cos 0 maxima at 0 0 180 and 0 90 respectively Multiplying Figure 1313a with Figure 1313b gives the resultant pattern in Figure 1313c It should be observed that the field patterns in Figure 1313 are in the plane containing the axes of the elements Note that 1 In the yzplane which is normal to the axes of the elements the unit pattern 1 is a circle see Figure 137b while the group pattern remains as in Figure 1313b there fore the resultant pattern is the same as the group pattern in this case 2 In the xyplane 0 7r2 SO the unit pattern vanishes while the group pattern 1 is a circle b If a TT2 d A4 and fid A 4 2 cos c o s 0 1 I group pattern I i resultant unit X pattern pattern The unit pattern remains as in Figure 1313a For the group pattern the null occurs when COS j 1 COS 6 0 1 COS 0 y y or cos 8 1 0 0 The maxima and minima occur when cos 1 cos 0 0 sin 0 sin 1 cos 0 0 dd I 4 J 4 sin0 O0 0 180 618 Antennas x X unit pattern a group pattern b resultant pattern c Figure 1314 For Example 136b field patterns in the plane containing the axes of the elements and sin 1 cos 6 0 cos I 1 or 6 180 Each field pattern is obtained by varying 0 0 5 10 15 180 Note that 8 180 corresponds to the maximum value of AF whereas d 0 corresponds to the null Thus the unit group and resultant patterns in the plane containing the axes of the el ements are shown in Figure 1314 Observe from the group patterns that the broadside array a 0 in Figure 1313 is bidirectional while the endfire array a 3d in Figure 1314 is unidirectional PRACTICE EXERCISE 136 Repeat Example 136 for cases when a a IT d A2 b a TT2 d A4 Answer See Figure 1315 EXAMPLE 137 Consider a threeelement array that has current ratios 121 as in Figure 1316a Sketch the group pattern in the plane containing the axes of the elements Solution For the purpose of analysis we split the middle element in Figure 1316a carrying current 270 into two elements each carrying current 10 This results in four elements instead of three as shown in Figure 1316b If we consider elements 1 and 2 as a group and ele ments 3 and 4 as another group we have a twoelement array of Figure 1316c Each 137 ANTENNA ARRAYS 619 x X x X a b Figure 1315 For Practice Exercise 136 0 2IlQ O Figure 1316 For Example 137 a a threeelement array i J with current ratios 121 b and c equivalent twoelement X2 X 2 a arrays 3 2 b 4 12 34 c 620 Antennas group is a twoelement array with d X2 a 0 that the group pattern of the two element array or the unit pattern for the threeelement array is as shown in Figure 1313b The two groups form a twoelement array similar to Example 136a with d X2 a 0 so the group pattern is the same as that in Figure 1313b Thus in this case both the unit and group patterns are the same pattern in Figure 1313b The resultant group pattern is obtained in Figure 1317c We should note that the pattern in Figure 1317c is not the resultant pattern but the group pattern of the threeelement array The re sultant group pattern of the array is Figure 1317c multiplied by the field pattern of the element type An alternative method of obtaining the resultant group pattern of the threeelement array of Figure 1316 is following similar steps taken to obtain eq 1359 We obtain the normalized array factor or the group pattern as AFn 4 J 2 2el e1 c o s where yj fid cos d a if the elements are placed along the zaxis but oriented parallel to 2TT X the xaxis Since a 0 d X2 fid x X 2 AFn AFn I resultant group pattern cos cos 6 cos cos i 4 unit pattern X cos cos 0 4 group pattern The sketch of these patterns is exactly what is in Figure 1317 If two threeelement arrays in Figure 1316a are displaced by X2 we obtain a four element array with current ratios 1331 as in Figure 1318 Two of such fourelement Figure 1317 For Example 137 obtain ing the resultant group pattern of the threeelement array of Figure 1316a unit pattern a roup pattern b resultant group pattern 0 138 EFFECTIVE AREA AND THE FRIIS EQUATION 621 30 X2 0 Figure 1318 A fourelement array with current ratios 1331 for Practice Exercise 137 arrays displaced by X2 give a fiveelement array with current ratios 14641 Contin uing this process results in an element array spaced X2 and N lX2 long whose current ratios are the binomial coefficients Such an array is called a linear binomial army PRACTICE EXERCISE 137 a Sketch the resultant group pattern for the fourelement array with current ratios 1331 shown in Figure 1318 b Derive an expression for the group pattern of a linear binomial array of N ele ments Assume that the elements are placed along the zaxis oriented parallel to the taxis with spacing d and interelement phase shift a Answer a See Figure 1319 b c o s where if fid cos d a Figure 1319 For Practice Exercise 137a 138 EFFECTIVE AREA AND THE FRIIS EQUATION In a situation where the incoming EM wave is normal to the entire surface of a receiving antenna the power received is Pr 1367 But in most cases the incoming EM wave is not normal to the entire surface of the antenna This necessitates the idea of the effective area of a receiving antenna The concept of effective area or effective aperture receiving cross section of an antenna is usually employed in the analysis of receiving antennas The effective area A of u receiving antenna is the ratio of the timeaverage power received Pr or delivered to he load to be strict to the timeaverage power density of the incident wave at the antenna 622 Antennas That is Pr Ob J avp 1368 From eq 1368 we notice that the effective area is a measure of the ability of the antenna to extract energy from a passing EM wave Let us derive the formula for calculating the effective area of the Hertzian dipole acting as a receiving antenna The Thevenin equivalent circuit for the receiving antenna is shown in Figure 1320 where Voc is the opencircuit voltage induced on the antenna termi nals Zin 7rad jXin is the antenna impedance and ZL RL jXL is the external load impedance which might be the input impedance to the transmission line feeding the antenna For maximum power transfer ZL Zn and XL Xin The timeaverage power delivered to the matched load is therefore rad v 1369 8 D ra For the Hertzian dipole Rmd S0ir2dlX2 and yoc Edl where E is the effective field strength parallel to the dipole axis Hence eq 1369 becomes Pr E2 640TT2 The timeaverage power at the antenna is ave 2TJ0 240TT Inserting eqs 1370 and 1371 in eq 1368 gives 3X2 X2 A L 5 1370 1371 or 1372 Figure 1320 Thevenin equivalent of a receiving antenna 138 EFFECTIVE AREA AND THE FRIIS EQUATION 623 where D 15 is the directivity of the Hertzian dipole Although eq 1372 was derived for the Hertzian dipole it holds for any antenna if D is replaced by GJfi j Thus in general 1373 Now suppose we have two antennas separated by distance r in free space as shown in Figure 1321 The transmitting antenna has effective area Aet and directive gain Gdt and transmits a total power P Prlii The receiving antenna has effective area of Aer and di rective gain Gdn and receives a total power of Pr At the transmitter 4rU P or op ave p 1374 By applying eqs 1368 and 1373 we obtain the timeaverage power received as P Op A C r r ave er dr Substituting eq 1374 into eq 1375 results in 1375 1376 This is referred to as the Friis transmission formula It relates the power received by one antenna to the power transmitted by the other provided that the two antennas are separated by r 2d2l where d is the largest dimension of either antenna see eq 1352 There fore in order to apply the Friis equation we must make sure that the two antennas are in the far field of each other Transmitter Receiver H r Figure 1321 Transmitting and receiving antennas in free space 624 Antennas EXAMPLE 138 Find the maximum effective area of a A2 wire dipole operating at 30 MHz How much power is received with an incident plane wave of strength 2 mVm Solution c 3 X 108 A T 10m 30 X 106 Gd6 0raax 164 102 164 1305 m2 p Op A A V 2 X 1 0 1640 240TT 1305 7162 nW PRACTICE EXERCISE 138 Determine the maximum effective area of a Hertzian dipole of length 10 cm operat ing at 10 MHz If the antenna receives 3 iW of power what is the power density of the incident wave Answer 1074 m2 2793 MWm2 EXAMPLE 139 The transmitting and receiving antennas are separated by a distance of 200 A and have di rective gains of 25 and 18 dB respectively If 5 mW of power is to be received calculate the minimum transmitted power Solution Given that Gdt dB 25 dB 10 log10 Gdt Gdt 1025 31623 Similarly Gdr dB 18 db or Gdr 10 631 139 THE RADAR EQUATION 625 Using the Friis equation we have or P P Pr GdrGdt J P 47rr12 5 x 103 1583 W J GdrG dt 4TT X 200 X X 1 631X31623 PRACTICE EXERCISE 139 An antenna in air radiates a total power of 100 kW so that a maximum radiated elec tric field strength of 12 mVm is measured 20 km from the antenna Find a its di rectivity in dB b its maximum power gain if rr Answer a 334 dB b 2117 139 THE RADAR EQUATION Radars are electromagnetic devices used for detection and location of objects The term radar is derived from the phrase radio detection and ranging In a typical radar system shown in Figure 1322a pulses of EM energy are transmitted to a distant object The same antenna is used for transmitting and receiving so the time interval between the trans mitted and reflected pulses is used to determine the distance of the target If r is the dis k Target a Figure 1322 a Typical radar system b simplification of the system in a for calculating the target cross section a b 626 Antennas tance between the radar and target and c is the speed of light the elapsed time between the transmitted and received pulse is 2rc By measuring the elapsed time r is determined The ability of the target to scatter or reflect energy is characterized by the scattering cross section a also called the radar cross section of the target The scattering cross section has the units of area and can be measured experimentally The scattering cross section is the equivalent area intercepting that amount ol power that when scattering isotropicall produces at the radar a power density which is equal to thai scattered or reflected by the actual target That is lim 4irr2 or 3 a lim 4xr2 9 1377 where SP is the incident power density at the target T while 3 is the scattered power density at the transreceiver O as in Figure 1322b From eq 1343 the incident power density 2P at the target Tis op op d p J i ave 9 rad 4TIT The power received at transreceiver O is 1378 or Aer 1379 Note that 2P and 9 are the timeaverage power densities in wattsm2 and Prad and Pr are the total timeaverage powers in watts Since Gdr Gdt Gd and Aer Aet Ae substi tuting eqs 1378 and 1379 into eq 1377 gives a 4irr22 1 Gd or AeaGdPmd 4irr22 1380a 1380b 139 THE RADAR EQUATION 627 TABLE 131 Designations of Radar Frequencies Designation UHF L S C X Ku K Millimeter Frequency 3001000 MHz 10002000 MHz 2000000 MHz 40008000 MHz 800012500 MHz 12518 GHz 18265 GHz 35 GHz From eq 1373 Ae 2GJAi Hence 1381 This is the radar transmission equation for free space It is the basis for measurement of scattering cross section of a target Solving for r in eq 1381 results in 1382 Equation 1382 is called the radar range equation Given the minimum detectable power of the receiver the equation determines the maximum range for a radar It is also useful for obtaining engineering information concerning the effects of the various parameters on the performance of a radar system The radar considered so far is the monostatic type because of the predominance of this type of radar in practical applications A bistatic radar is one in which the transmitter and receiver are separated If the transmitting and receiving antennas are at distances rx and r2 from the target and Gdr Gdt eq 1381 for bistatic radar becomes GdtGdr 4TT rad 1383 Radar transmission frequencies range from 25 to 70000 MHz Table 131 shows radar frequencies and their designations as commonly used by radar engineers EXAMPLE 1310 An Sband radar transmitting at 3 GHz radiates 200 kW Determine the signal power density at ranges 100 and 400 nautical miles if the effective area of the radar antenna is 9 m2 With a 20m2 target at 300 nautical miles calculate the power of the reflected signal at the radar 628 Antennas Solution The nautical mile is a common unit in radar communications 1 nautical mile nm 1852 m c 3 X 108 3 X 10 r X2 et 01 01m 9 3600r For r 100 nm 1852 X 105 m ad 3600TT X 200 X 103 4TIT2 4TT1852 2X 1010 5248 mWm2 For r 400 nm 4 1852 X 105 m 5248 42 0328 mWm2 Aea Gd Prad Using eq 1380b where r 300 nm 5556 X 105 m 9 X 20 X 36007T X 200 X 103 4TT X 555622 X 1020 The same result can be obtained using eq 1381 2706 X 1014W PRACTICE EXERCISE 1310 A Cband radar with an antenna 18 m in radius transmits 60 kW at a frequency of 6000 MHz If the minimum detectable power is 026 mW for a target cross section of 5 m2 calculate the maximum range in nautical miles and the signal power density at half this range Assume unity efficiency and that the effective area of the antenna is 70 of the actual area Answer 06309 nm 50090 Wm2 SUMMARY 629 SUMMARY 1 We have discussed the fundamental ideas and definitions in antenna theory The basic types of antenna considered include the Hertzian or differential length dipole the halfwave dipole the quarterwave monopole and the small loop 2 Theoretically if we know the current distribution on an antenna we can find the re tarded magnetic vector potential A and from it we can find the retarded electromag netic fields H and E using H V X E T H X a The farzone fields are obtained by retaining only lr terms 3 The analysis of the Hertzian dipole serves as a stepping stone for other antennas The radiation resistance of the dipole is very small This limits the practical usefulness of the Hertzian dipole 4 The halfwave dipole has a length equal to X2 It is more popular and of more practi cal use than the Hertzian dipole Its input impedance is 73 J425 fi 5 The quarterwave monopole is essentially half a halfwave dipole placed on a con ducting plane 6 The radiation patterns commonly used are the field intensity power intensity and ra diation intensity patterns The field pattern is usually a plot of ES or its normalized form flft The power pattern is the plot of 2Pave or its normalized form20 7 The directive gain is the ratio of U9 f to its average value The directivity is the maximum value of the directive gain 8 An antenna array is a group of radiating elements arranged so as to produce some particular radiation characteristics Its radiation pattern is obtained by multiply ing the unit pattern due to a single element in the group with the group pattern which is the plot of the normalized array factor For an TVelement linear uniform array AF where j 13d cos 9 a 0 2X d spacing between the elements and a in terelement phase shift 9 The Friis transmission formula characterizes the coupling between two antennas in terms of their directive gains separation distance and frequency of operation 10 For a bistatic radar one in which the transmitting and receiving antennas are sepa rated the power received is given by 4TT r J aPnad For a monostatic radar r r2 r and Gdt Gdr 630 Antennas 131 An antenna located in a city is a source of radio waves How much time does it take the wave to reach a town 12000 km away from the city a 36 s b 20 us c 20 ms d 40 ms e None of the above 132 In eq 1334 which term is the radiation term a 1rterm b lr2term c IIr term d All of the above 133 A very small thin wire of length X100 has a radiation resistance of a 0 G b 008 G c 79 G d 790 0 134 A quarterwave monopole antenna operating in air at frequency 1 MHz must have an overall length of a X b 300 m c 150 m d 75 m e sC X 135 If a small singleturn loop antenna has a radiation resistance of 004 G how many turns are needed to produce a radiation resistance of 1 G a 150 b 125 c 50 d 25 e 5 REVIEW QUESTIONS 631 136 At a distance of 8 km from a differential antenna the field strength is 12 iVm The field strength at a location 20 km from the antenna is a 75xVm b 30xVm c 48xVm d 192zVm 137 An antenna has fmax 10 Wsr lave 45 Wsr and ir 95 The input power to the antenna is a 2222 W b 1211 W c 5555 W d 5952 W 138 A receiving antenna in an airport has a maximum dimension of 3 m and operates at 100 MHz An aircraft approaching the airport is 12 km from the antenna The aircraft is in the far field region of the antenna a True b False 139 A receiving antenna is located 100 m away from the transmitting antenna If the effective area of the receiving antenna is 500 cm2 and the power density at the receiving location is 2 mWm2 the total power received is a lOnW b 100 nW c 1xW d 10 W e 100 W 1310 Let R be the maximum range of a monostatic radar If a target with radar cross section of 5 m2 exists at R2 what should be the target cross section at 3R2 to result in an equal signal strength at the radar a 00617 m2 b 0555 m2 c 15 m2 d 45 m2 e 405 m2 Answers 13Id 132a 133b 134d 135e 136c 137d 138a 139e 1310e 632 Antennas PROBLEMS I 131 The magnetic vector potential at point Pr 8 j due to a small antenna located at the origin is given by 50 eBr A where r2 x2 y2 z2 Find Er 6 j t and Hr d j i at the far field 132 A Hertzian dipole at the origin in free space has di 20 cm and 7 1 0 cos 2irl07t A find E6s at the distant point 100 0 0 133 A 2A source operating at 300 MHz feeds a Hertzian dipole of length 5 mm situated at the origin Find Es and H at 10 30 90 134 a Instead of a constant current distribution assumed for the short dipole of Section 132 assume a triangular current distribution 7 7O I 1 j shown in Figure 1323 Show that rad 2 0 7TZ I which is onefourth of that in eq 1313 Thus Rmd depends on the current distribu tion b Calculate the length of the dipole that will result in a radiation resistance of 05 0 135 An antenna can be modeled as an electric dipole of length 5 m at 3 MHz Find the radia tion resistance of the antenna assuming a uniform current over its length 136 A halfwave dipole fed by a 500 transmission line calculate the reflection coefficient and the standing wave ratio 137 A 1mlong car radio antenna operates in the AM frequency of 15 MHz How much current is required to transmit 4 W of power Figure 1323 Short dipole antenna with triangular current distri bution for Problem 134 PROBLEMS 633 138 a Show that the generated far field expressions for a thin dipole of length carrying si nusoidal current Io cos z are 3rCos Yc0StJ cos y 2wr sin 8 Hint Use Figure 134 and start with eq 1314 b On a polar coordinate sheet plot fifi in part a for X 3X2 and 2X 139 For Problem 134 a Determine E and Hs at the far field b Calculate the directivity of the dipole 1310 An antenna located on the surface of a flat earth transmits an average power of 200 kW Assuming that all the power is radiated uniformly over the surface of a hemisphere with the antenna at the center calculate a the timeaverage Poynting vector at 50 km and b the maximum electric field at that location 1311 A 100turn loop antenna of radius 20 cm operating at 10 MHz in air is to give a 50 mVm field strength at a distance 3 m from the loop Determine a The current that must be fed to the antenna b The average power radiated by the antenna 1312 Sketch the normalized Efield and field patterns for a A halfwave dipole b A quarterwave monopole 1313 Based on the result of Problem 138 plot the vertical field patterns of monopole antennas of lengths 3X2 X 5X8 Note that a 5X8 monopole is often used in practice 1314 In free space an antenna has a farzone field given by where 3 wVxoeo Determine the radiated power 1315 At the far field the electric field produced by an antenna is E s ej3r cos 6 cos j az Sketch the vertical pattern of the antenna Your plot should include as many points as possible 634 Antennas 1316 For an Hertzian dipole show that the timeaverage power density is related to the radia tion power according to 15 sin20 4irr 1317 At the far field an antenna produces 2 sin 6 cos 4 ave ar Wm2 0 6 x 0 x2 Calculate the directive gain and the directivity of the antenna 1318 From Problem 138 show that the normalized field pattern of a fullwave X antenna is given by cosx cos 6 1 sin0 Sketch the field pattern 1319 For a thin dipole A16 long find a the directive gain b the directivity c the effec tive area d the radiation resistance 1320 Repeat Problem 1319 for a circular thin loop antenna A12 in diameter 1321 A halfwave dipole is made of copper and is of diameter 26 mm Determine the effi ciency of the dipole if it operates at 15 MHz Hint Obtain R from RRdc a28 see Section 106 1322 Find Cave tmax and D if a Uifi 4 sin2 20 0 0 x 0 0 2TT b Uifi t 4 esc2 20 TT3 0 x2 0 j x c U6 4 2 sin2 0 sin2 j 0 d x 0 t x 1323 For the following radiation intensities find the directive gain and directivity a U6 4 sin2 0 0 0 x 0 j 2x b U6 t 4 sin2 0 cos 2 0 O 0 T T 0 0 TT c Uifi t 10 cos2 0 sin2 42 0 0 x 0 f x2 1324 In free space an antenna radiates a field 4TIT at far field Determine a the total radiated power b the directive gain at 0 60 1325 Derive Es at far field due to the twoelement array shown in Figure 1324 Assume that the Hertzian dipole elements are fed in phase with uniform current o cos cot PROBLEMS 635 Figure 1324 Twoelement array of Problem 1325 y 1326 An array comprises two dipoles that are separated by one wavelength If the dipoles are fed by currents of the same magnitude and phase a Find the array factor b Calculate the angles where the nulls of the pattern occur c Determine the angles where the maxima of the pattern occur d Sketch the group pattern in the plane containing the elements 1327 An array of two elements that are fed by currents that are 180 out of phase with each other Plot the group pattern if the elements are separated by a d A4 b d X2 1328 Sketch the group pattern in the xzplane of the twoelement array of Figure 1310 with a d A a all b d A4 a 3TT4 c d 3A4 a 0 1329 An antenna array consists of N identical Hertzian dipoles uniformly located along the z axis and polarized in the direction If the spacing between the dipole is A4 sketch the group pattern when a N 2 b N 4 1330 Sketch the resultant group patterns for the fourelement arrays shown in Figure 1325 X 2 X2 a 12 l0 Figure 1325 For Problem 1330 x2 12 X4 I jit 12 X4 b 3ir2 X4 636 Antennas 1331 For a 10turn loop antenna of radius 15 cm operating at 100 MHz calculate the effective area at 30 j 90 1332 An antenna receives a power of 2 xW from a radio station Calculate its effective area if the antenna is located in the far zone of the station where E 50 mVm 1333 a Show that the Friis transmission equation can be written as r AerAet b Two halfwave dipole antennas are operated at 100 MHz and separated by 1 km If 80 W is transmitted by one how much power is received by the other 1334 The electric field strength impressed on a halfwave dipole is 3 mVm at 60 MHz Cal culate the maximum power received by the antenna Take the directivity of the halfwave dipole as 164 1335 The power transmitted by a synchronous orbit satellite antenna is 320 W If the antenna has a gain of 40 dB at 15 GHz calculate the power received by another antenna with a gain of 32 dB at the range of 24567 km 1336 The directive gain of an antenna is 34 dB If the antenna radiates 75 kW at a distance of 40 km find the timeaverage power density at that distance 1337 Two identical antennas in an anechoic chamber are separated by 12 m and are oriented for maximum directive gain At a frequency of 5 GHz the power received by one is 30 dB down from that transmitted by the other Calculate the gain of the antennas in dB 1338 What is the maximum power that can be received over a distance of 15 km in free space with a 15GHz circuit consisting of a transmitting antenna with a gain of 25 dB and a re ceiving antenna with a gain of 30 dB The transmitted power is 200 W 1339 An Lband pulse radar with a common transmitting and receiving antenna having a di rective gain of 3500 operates at 1500 MHz and transmits 200 kW If the object is 120 km from the radar and its scattering cross section is 8 m2 find a The magnitude of the incident electric field intensity of the object b The magnitude of the scattered electric field intensity at the radar c The amount of power captured by the object d The power absorbed by the antenna from the scattered wave 1340 A transmitting antenna with a 600 MHz carrier frequency produces 80 W of power Find the power received by another antenna at a free space distance of 1 km Assume both an tennas has unity power gain 1341 A monostable radar operating at 6 GHz tracks a 08 m2 target at a range of 250 m If the gain is 40 dB calculate the minimum transmitted power that will give a return power of 2tW PROBLEMS 637 1342 In the bistatic radar system of Figure 1326 the groundbased antennas are separated by 4 km and the 24 m2 target is at a height of 3 km The system operates at 5 GHz For Gdt of 36 dB and Gdr of 20 dB determine the minimum necessary radiated power to obtain a return power of 8 X 1012W Scattered wave Receiving antenna Target a Figure 1326 For Problem 1342 3 km Transmitting antenna Chapter 14 MODERN TOPICS The future has several names For the weak it is the impossible For the faith hearted it is the unknown For the thoughtful and valiant it is ideal VICTOR HUGO 141 INTRODUCTION Besides wave propagation transmission lines waveguides and antennas there are several other areas of applications of EM These include microwaves electromagnetic interfer ence and compatibility fiber optics satellite communication bioelectromagnetics electric machines radar meteorology and remote sensing Due to space limitation we shall cover the first three areas in this chapter microwaves electromagnetic interference and compat ibility and fiber optics Since these topics are advanced only an introductory treatment of each topic will be provided Our discussion will involve applying the circuit concepts learned in earlier courses and the EM concepts learned in earlier chapters 142 MICROWAVES At the moment there are three means for carrying thousands of channels over long dis tances a microwave links b coaxial cables and c fiber optic a relatively new tech nology to be covered later Microwaves arc IM wines whose IrequiMicies rnngo from approximately MX MH to I KM Gil For comparison the signal from an AM radio station is about 1 MHz while that from an FM station is about 100 MHz The higher frequency edge of microwaves borders on the optical spectrum This accounts for why microwaves behave more like rays of light than ordinary radio waves You may be familiar with microwave appliances such as the mi crowave oven which operates at 24 GHz the satellite television which operates at about 4 GHz and the police radar which works at about 22 GHz Features that make microwaves attractive for communications include wide available bandwidths capacities to carry information and directive properties of short wavelengths Since the amount of information that can be transmitted is limited by the available band 638 142 MICROWAVES 639 width the microwave spectrum provides more communication channels than the radio and TV bands With the ever increasing demand for channel allocation microwave communi cations has become more common A microwave system1 normally consists of a transmitter including a microwave os cillator waveguides and a transmitting antenna and a receiver subsystem including a re ceiving antenna transmission line or waveguide microwave amplifiers and a receiver A microwave network is usually an interconnection of various microwave components and devices There are several microwave components and variations of these components Common microwave components include Coaxial cables which are transmission lines for interconnecting microwave compo nents Resonantors which are usually cavities in which EM waves are stored Waveguide sections which may be straight curved or twisted Antennas which transmit or receive EM waves efficiently Terminators which are designed to absorb the input power and therefore act as one ports Attenuators which are designed to absorb some of the EM power passing through it and thereby decrease the power level of the microwave signal Directional couplers which consist of two waveguides and a mechanism for cou pling signals between them Isolators which allow energy flow only in one direction Circulators which are designed to establish various entryexit points where power can either be fed or extracted Filters which suppress unwanted signals andor separate signals of different fre quencies The use of microwaves has greatly expanded Examples include telecommunications radio astronomy land surveying radar meteorology UHF television terrestrial mi crowave links solidstate devices heating medicine and identification systems We will consider only four of these 1 Telecommunications the transmission of analog or digital information from one point to another is the largest application of microwave frequencies Microwaves propa gate along a straight line like a light ray and are not bent by the ionosphere as are lower fre quency signals This makes communication satellites possible In essence a communica tion satellite is a microwave relay station that is used to link two or more groundbased transmitters and receivers The satellite receives signals at one frequency repeats or am plifies it and transmits it at another frequency Two common modes of operation for satel lite communication are portrayed in Figure 141 The satellite provides a pointtopoint For a comprehensive treatment of microwaves see D M Pozar Microwave Engineering New York John Wiley 2nd 1998 640 Modern Topics Satellite a Pointtopoint link via satellite microwave Satellite Multiple receivers Multiple receivers b Broadcast link via satellite microwave Figure 141 Satellite communications configurations Source W Stallings Data and Computer Communications 5th ed Upper Saddle River NJ Pren tice Hall 1977 p 90 142 MICROWAVES 641 link in Figure 141 a while it is being used to provide multiple links between one ground based transmitter and several groundbased receivers in Figure 141b 2 Radar Systems Radar systems provided the major incentive for the development of microwave technology because one obtains better resolution for radar instruments at higher frequencies Only the microwave region of the spectrum could provide the required resolution with antennas of reasonable size The ability to focus a radiated wave sharply is what makes microwaves so useful in radar applications Radar is used to detect aircraft guide supersonic missiles observe and track weather patterns and control flight traffic at airports It is also used in burglar alarms garagedoor openers and police speed detectors 3 Heating Microwave energy is more easily directed controlled and concentrated than lowfrequency EM waves Also various atomic and molecular resonances occur at microwave frequencies creating diverse application areas in basic science remote sensing and heating methods The heating properties of microwave power are useful in a wide variety of commercial and industrial applications The microwave oven shown in Figure 142 is a typical example When the magnetron oscillates microwave energy is ex tracted from the resonant cavities The reflections from the stationary walls and the motion of the stirring fan cause the microwave energy to be well distributed Thus the microwave enables the cooking process to be fast and even Besides cooking microwave heating properties are used in physical diathermy and in drying potato chips paper cloth etc A microwave circuit consists of microwave components such as sources transmission lines waveguides attenuators resonators circulators and filters One way of analyzing such as a circuit is to relate the input and output variables of each component Several sets of parameters can be used for relating input and output variables but at high frequen cies such as microwave frequencies where voltage and current are not well defined Sparameters are often used to analyze microwave circuits The scattering or parameters are defined in terms of wave variables which are more easily measured at microwave fre quencies than voltage and current Stirrer fan Waveguide Metal cavity Magnetron tube Power supply Figure 142 Microwave oven Source N Schlager ed How Products are Made Detroit MI Gale Research Inc 1994 p 289 642 Modern Topics Consider the twoport network shown in Figure 143 The traveling waves are related to the scattering parameters according to b1 Sua1 Sna2 or in matrix form SU S2 52 S22 141 142 where ax and a2 represent the incident waves at ports 1 and 2 respectively while b and b2 represent the reflected waves as shown in Figure 143 For the S matrix the offdiagonal terms represent voltage wave transmission coefficients while the diagonal terms represent reflection coefficients If the network is reciprocal it will have the same transmission char acteristics in either direction ie If the network is symmetrical then S22 For a matched twoport the reflection coefficients are zero and Sn S22 0 143 144 145 The input reflection coefficient can be expressed in terms of the Sparameters and the load ZL as where b c a S22i L 146 147 O O O b2 o Figure 143 A twoport network 142 MICROWAVES 643 Similarly the output reflection coefficient with Vg 0 can be expressed in terms of the generator impedance Zg and the Sparameters as r a2 148 vgo where zgzo 8 zK z o 149 EXAMPLE 141 The following Sparameters are obtained for a microwave transistor operating at 25 GHz Su 085300 512 00756 521 168120 5 U O854O0 Determine the input reflection coefficient when ZL Zo 75 0 Solution From Eq 147 Hence using Eq 146 leads to T Sn 085730 PRACTICE EXERCISE 141 For an hybrid coupler the VSWRs for the input and output ports are respectively given as 1 S S Calculate s and so for the following scattering matrix 04 J06 j06 02 Answer 2333 15 644 B Modern Topics 143 ELECTROMAGNETIC INTERFERENCE AND COMPATIBILITY Every electronic device is a source of radiated electromagnetic fields called radiated emis sions These are often an accidental byproduct of the design Electromagnetic interference KMI is the degradation in the performance of a device clue to the fields making up the electromagnetic environment The electromagnetic environment consists of various apparatuses such as radio and TV broadcast stations radar and navigational aids that radiate EM energy as they operate Every electronic device is susceptible to EMI Its influence can be seen all around us The results include ghosts in TV picture reception taxicab radio interference with police radio systems power line transient interference with personal computers and selfoscilla tion of a radio receiver or transmitter circuit Electromagnetic compatibility liMCj is achieved when a device functions satis factorily without introducing intolerable disturbances to the electromagnetic envi ronment or to other devices in its neighborhood EMC2 is achieved when electronic devices coexist in harmony such that each device func tions according to its intended purpose in the presence of and in spite of the others EMI is the problem that occurs when unwanted voltages or currents are present to influence the performance of a device while EMC is the solution to the problem The goal of EMC is to ensure system or subsystem compatibility and this is achieved by applying proven design techniques the use of which ensures a system relatively free of EMI problems EMC is a growing field because of the everincreasing density of electronic circuits in modern systems for computation communication control etc It is not only a concern to electrical and computer engineers but to automotive engineers as well The increasing ap plication of automotive electronic systems to improve fuel economy reduce exhaust emis sions ensure vehicle safety and provide assistance to the driver has resulted in a growing need to ensure compatibility during normal operation We will consider the sources and characteristics of EMI Later we will examine EMI control techniques A Source and Characteristics of EMI First let us classify EMI in terms of its causes and sources The classification will facilitate recognition of sources and assist in determining means of control As mentioned earlier any electronic device may be the source of EMI although this is not the intention of the de signer The cause of the EMI problem may be either within the system in which case it is termed an intrasystem problem or from the outside in which case it is called an intersys 2 For an indepth treatment of EMC see C R Paul Introduction to Electromagnetic Compatibility New York John Wiley 1992 Microwave relay link 143 ELECTROMAGNETIC INTERFERENCE A N D COMPATIBILITY M 645 Aircraft Ship E Emitters of Interference S Susceptible Equipment Figure 144 Typical examples of intersystem EMI problems Source JIN Violette et al Electro magnetic Compatibility Handbook New York Van Nostrand Reinhold 1987 p 4 tern problem Figure 144 shows intersystem EMI problems The term emitter is com monly used to denote the source of EMI while the term susceptor is used to designate a victim device Tables 141 and 142 present typical causes of both intrasystem and inter system problems Both intrasystem and intersystem EMI generally can be controlled by the system design engineer by following some design guidelines and techniques For in TABLE141 Intrasystem EMI Causes Emitters Susceptors Power supplies Radar transmitters Mobile radio transmitters Fluorescent lights Car ignition systems Relays Radar receivers Mobile radio receivers Ordnance Car radio receivers 646 Modem Topics TABLE 142 Intersystem EMI Causes Emitters Susceptors Lightning strokes Computers Power Lines Radar transmitters Police radio transmitters Fluorescent lights Aircraft transmitters Radio receivers TV sets Heart pacers Aircraft navigation systems Taxicab radio receivers Industrial controls Ship receivers trasystem EMI problems for example the design engineer may apply proper grounding and wiring arrangements shielding of circuits and devices and filtering The sources of EMI can be classified as natural or artificial manmade The origins of EMI are basically undesired conducted emissions voltages andor currents or radiated emissions electric andor magnetic fields Conducted emissions are those currents that are carried by metallic paths the units power cord and placed on the common power network where they may cause interference with other devices that are connected to this network Radiated emissions concern the electric fields radiated by the device that may be received by other electronic devices causing interference in those devices Figure 145 il lustrates the conceptual difference between conducted and radiated paths No single operating agency has jurisdiction over all systems to dictate actions neces sary to achieve EMC Thus EMC is usually achieved by industrial association voluntary regulation governmentenforced regulation and negotiated agreements between the af fected parties Frequency plays a significant role in EMC Frequency allocations and as signments are made according to the constraints established by international treaties The Radio Regulations resulting from such international treaties are published by the Interna Conducted interference Power cables Figure 145 Differences between conducted and radiated emissions 143 ELECTROMAGNETIC INTERFERENCE A N D COMPATIBILITY 647 tional Telecommunication Union ITU The Federal Communications Commission FCC has the authority over radio and wire communications in the United States The FCC has set limits on the radiated and conducted emissions of electronic devices including elec tronic typewriters calculators televisions printers modems and personal computers It is illegal to market an electronic device in the United States unless its radiated and conducted emissions have been measured and do not exceed the limits of FCC regulations Therefore any electronic device designed today that is designed without incorporating EMC design principles will probably fail to comply with the FCC limits B EMI Control Techniques To control or suppress EMI the three common means employed in the design process are grounding shielding and filtering Although each technique has a distinct role in system design proper grounding may sometimes minimize the need for shielding and filtering also proper shielding may minimize the need for filtering Therefore we discuss the three techniques grounding shielding and filtering in that order Grounding Grounding is the establishment of an electrically conductive path between two points to connect electrical and electronic elements of a system to one another or to some reference point which may be designated the ground An ideal ground plane is a zeropotential zeroimpedance body that can be used as a reference for all signals in associated circuitry and to which any undesired current can be transferred for the elimination of its effects The purpose of the floating ground is to isolate circuits or equipment electrically from a common ground plane This type of grounding technique may cause a hazard The singlepoint grounding is used to minimize the effects of facility ground currents The multiplepoint grounding minimizes ground lead lengths The ground plane might be a ground wire that is carried throughout the system or a large conductive body Bonding is the establishment of a lowimpedance path between two metal surfaces Grounding is a circuit concept while bonding denotes the physical implementation of that concept The purpose of a bond is to make a structure homogeneous with respect to the flow of electrical currents thus avoiding the development of potentials between the metal lic parts since such potentials may result in EMI Bonds provide protection from electrical shock power circuit current return paths and antenna ground plane connections and also minimize the potential difference between the devices They have the ability to carry large fault current There are two types of bond direct and indirect bonds The direct bond is a metalto metal contact between the elements connected while the indirect bond is a contact through the use of conductive jumpers The dc resistance Rdc of a bond is often used as an indication of bond quality It is given by aS 1410 648 Modern Topics where is the length of the bond a is its conductivity and S is its crosssectional area As frequency increases the bond resistance increases due to skin effect Thus the ac resistance Rac is given as adw 1411 where w is the width of the bond and 5 is the skin depth Bonding effectiveness can be expressed as the difference in dB between the induced voltages on an equipment case with and without the bond trap Shielding The purpose of shielding is to confine radiated energy to a specific region or to prevent ra diated energy from entering a specific region Shields may be in the form of partitions and boxes as well as in the form of cable and connector shields Shield types include solid nonsolid eg screen and braid as is used on cables In all cases a shield can be characterized by its shielding effectiveness The shielding effec tiveness SE is defined as SE101og 10 incident power density transmitted power density 1412 where the incident power density is the power density at a measuring point before a shield is installed and the transmitted power is the power density at the same point after the shield is in place In terms of the field strengths the shielding effectiveness may also be defined as the ratio of the field Et transmitted through to the inside to the incident field Et Thus SE is given by SE 20 logl0 1413 For magnetic fields SE 20 log10 Ht 1414 For example aluminum has a 35 X 107 Sm e eo jt xo an aluminum sheet at 100 MHz has an SE of 100 dB at a thickness of 001 mm Since analuminum sheet for a computer cabinet is much thicker than this an aluminum case is considered a highly effec tive shield A cabinet that effectively shields the circuits inside from external fields is also highly effective in preventing radiation from those circuits to the external world Because of the effective shield radiated emission from the computer system is caused by openings in the cabinet such as cracks holes from disc drives etc and from wires that penetrate the cabinet such as power cord and cables to external devices 144 OPTICAL FIBER 649 Filtering An electrical filter is a network of lumped or distributed constant resistors inductors and capacitors that offers comparatively little opposition to certain frequencies while blocking the passage of other frequencies Filter provides the means whereby levels of conducted interference are substantially reduced The most significant characteristic of a filter is the insertion loss it provides as a func tion of frequency Insertion loss IL is defined as L 201og 1 0 1415 where V1 is the output voltage of a signal source with the filter in the circuit and V2 is the output voltage of the signal source without the use of the filter Lowpass filters are com monly used in EMC work The insertion loss for the lowpass filters is given by IL 10 log10 1 F2 dB 1416 irfRC for capacitive filter where k irfLR for inductive filter andis the frequency 144 OPTICAL FIBER In the mid 1970s it was recognized that the existing copper technology would be unsuit able for future communication networks In view of this the telecommunication industry invested heavily in research into optical fibers Optical fiber provides an attractive alterna tive to wire transmission lines such are twisted pair and coaxial cable or coax Optical fiber3 has the following advantages over copper Bandwidth It provides a very high capacity for carrying information It has suffi cient bandwidth that bitserial transmission can be used thereby considerably re ducing the size cost and complexity of the hardware Attenuation It provides low attenuation and is therefore capable of transmitting over a long distance without the need of repeaters Noise susceptibility It neither radiates nor is affected by electromagnetic interfer ence The immunity from EMI is due to the fact that there are no metal parts so that there can be no conduction currents 3 There are several excellent books that can provide further exposition on optical fiber See for example S L W Meardon The Elements of Fiber Optics Englewood Cliffs NJ RegentsPrentice Hall 1993 650 Modern Topics Security It is more secure from malicious interception because it is not easy to tap a fiberoptic cable without interrupting communication Cost The cost of optical fibers has fallen considerably over the past few years and will continue to fall So is the cost of related components such as optical transmit ters and receivers These impressive advantages of fiber optics over electrical media have made it a popular transmission medium in recent times Although optical fiber is more expensive and is used mainly for pointtopoint links there has been a rapid changeover from coax and twisted pair to optical fibers for telecommunication systems instrumentation cable TV networks industrial automation and data transmission systems An optical fiber is a dielectric waveguide operating at optical frequency Optical frequencies are on the order of 100 THz As shown in Figure 146 an optical fiber consists of three concentric cylindrical sections the core the cladding and the jacket The core consists of one or more thin strands made of glass or plastic The cladding is the glass or plastic coating surrounding the core which may be stepindex or gradedindex In the stepindex core the refractive index is uniform but undergoes an abrupt change at the corecladding interface while the gradedindex core has a refractive index that varies with the radial distance from the center of the fiber The jacket surrounds one or a bundle of cladded fibers The jacket is made of plastic or other materials to protect against moisture crushing etc A ray of light entering the core will be internally reflected when incident in the denser medium and the angle of incidence is greater than a critical value Thus a light ray is re flected back into the original medium and the process is repeated as light passes down the core This form of propagation is multimode referring to the variety of angles that will reflect as shown in Figure 147 It causes the signal to spread out in time and limits the rate Core N Jacket4 Cladding Light at less than critical angle is absorbed in jacket Angle of incidence Angle of reflection Figure 146 Optical fiber 144 OPTICAL FIBER 651 a Multimode Absorptive jacket Cladding b Multimode graded index c Single mode Figure 147 Optical fiber transmission modes Source W Stallings Local and Metropolitan Area Networks 4th ed New York Macmillan 1993 p 85 652 Modern Topics at which data can be accurately received By reducing the radius of the core a singlemode propagation occurs This eliminates distortion A fiberoptic system is similar to a conventional transmission system As shown in Figure 148 a fiberoptic system consists of a transmitter a transmission medium and a re ceiver The transmitter accepts and converts input electrical signals in analog or digital form to optical signals The transmitter sends the optical signal by modulating the output of a light source usually an LED or a laser by varying its intensity The optical signal is transmitted over the optical fiber to a receiver At the receiver the optical signal is con verted back into an electrical signal by a photodiode The performance of a fiberoptic link depends on the numerical aperture NA atten uation and dispersion characteristics of the fiber As signals propagate through the fiber they become distorted due to attenuation and dispersion Numerical Aperture This is the most important parameter of an optical fiber The value of NA is dictated by the refractive indices of the core and cladding By definition the refractive index n of a medium is defined as speed of light in a vacuum speed of light in the medium 1 c um 1418 Since fim x0 in most practical cases n 1419 indicating that the refractive index is essentially the square root of the dielectric constant Keep in mind that er can be complex as discussed in Chapter 10 For common materials n 1 for air n 133 for water and n 15 for glass As a light ray propagates from medium 1 to medium 2 Snells law must be satisfied n2 sin 1420 Electrical data input Electrical to optical converter Optical fiber cable Transmission medium Light source Figure 148 A typical fiberoptic system Optical to electrical converter Light detector Electrical data output 144 OPTICAL FIBER 653 where d is the incident angle in medium 1 and 92 is the transmission angle in medium 2 The total reflection occurs when 82 90 resulting in 6C sin1 1421 where 9C is the critical angle for total internal reflection Note that eq 1421 is valid only if n n2 since the value of sin 9C must be less than or equal to 1 Another way of looking at the lightguiding capability of a fiber is to measure the ac ceptance angle da which is the maximum angle over which light rays entering the fiber will be trapped in its core We know that the maximum angle occurs when 6C is the critical angle thereby satisfying the condition for total internal reflection Thus for a stepindex fiber NA sin 6a n sin 6C 1422 where is the refractive index of the core and n2 is the refractive index of the cladding as shown in Figure 149 Since most fiber cores are made of silica ny 148 Typical values of NA range between 019 and 025 The larger the value of NA the more optical power the fiber can capture from a source Due to the numerous modes a fiber may support it is called a multimode stepindex fiber The mode volume V is given by V 1423 where d is the fiber core diameter and A is the wavelength of the optical source From eq 1423 the number N of modes propagating in a stepindex fiber can be estimated as N 1424 Cladding X N O 1 U Core Figure 149 Numerical aperture and acceptance angle L 654 Modern Topics Attenuation As discussed in Chapter 10 attentuation is the reduction in the power of the optical signal Power attenuation or fiber loss in an optical fiber is governed by dz ccP 1425 where a is the attenuation and P is the optical power In eq 1425 it is assumed that a wave propagates along z By solving eq 1425 the power P0 at the input of the fiber and the power P of the light after are related as P P0e 1426 It is customary to express attenuation a in dBkm and length I of the fiber in km In this case eq 1426 becomes at 101og10 P0 P 1427 Thus the power of the light reduces by a decibels per kilometer as it propagates through the fiber Equation 1427 may be written as P3 P0 10 crf10 For I 100 km P0 f 10100 for coaxial cable lO 2 for fiber indicating that much more power is lost in the coaxial cable than in fiber 1428 1429 Dispersion m This is the spreading of pulses of light as they propagate down a fiber As the pulses repre senting 0s spread they overlap epochs that represent Is If dispersion is beyond a certain limit it may confuse the receiver The dispersive effects in singlemode fibers are much smaller than in multimode fibers EXAMPLE 142 A stepindex fiber has a core diameter of 80 im a core refractive index of 162 and a nu merical aperture of 021 Calculate a the acceptance angle b the refractive index that the fiber can propagate at a wavelength of 08 fim c the number of modes that the fiber can propagate at a wavelength of 08 xm 144 OPTICAL FIBER 655 Solution a Since sin da NA 021 then sin1 021 1212 b From NA vn n we obtain c Hence n2 Vn NA2 Vl62 2 0212 1606 irdNA X A X TT80 X 106 X 021 08 X 10 6 65973 V1 N 2176 modes PRACTICE EXERCISE 142 A silica fiber has a refractive index of 148 It is surrounded by a cladding material with a refractive index of 1465 Find a the critical angle above which total inter nal reflection occurs b the numerical aperture of the fiber Answer a 8183 b 021 EXAMPLE 143 Light pulses propagate through a fiber cable with an attenuation of 025 dBkm Determine the distance through which the power of pulses is reduced by 40 Solution If the power is reduced by 40 it means that P P0 Hence 1 04 06 1 0 10 L 025 l 0 g l 06 8874 km I 656 M Modern Topics PRACTICE EXERCISE 143 A 10km fiber with an attenuation of 02 dBkm serves as an optical link between two cities How much of input power is received Answer 631 SUMMARY 1 Microwaves are EM waves of very short wavelengths They propagate along a straight line like light rays and can therefore be focused easily in one direction by antennas They are used in radar guidance navigation and heating 2 Electromagnetic compatibility EMC is the capability of electrical and electronic devices to operate in their intended electromagnetic environment without suffering or causing unacceptable degradation as a result of EMI 3 Electromagnetic interference EMI is the lack of EMC It can be suppressed by grounding shielding and filtering 4 An optical fiber is a dielectric waveguiding structure operating at optical frequencies and it consists of a core region and a cladding region 5 Advantages of optical fiber over copper wire include 1 large bandwidth 2 low at tenuation 3 immunity to EMI and 4 low cost REVIEW QUESTIONS 141 Microwaves have long wavelengths a True b False 142 The wavelength in free space of a microwave signal whose frequency is 3 GHz is a 1 mm b 10 mm c 10 cm d l m 143 Which of the following is not a source of EMI a Optical fiber b Personal computer c Police radar d Aircraft e Fluorescent lamp REVIEW QUESTIONS 657 144 Optical fiber is a A transmission line b A waveguide c Both 145 Unlike coax and twisted pair optical fibers are immune to a Highfrequency transmission b Signal attenuation c Power loss d Electromagnetic interference 146 As a consultant you have been asked to design a network for an auditorium Speed and cost are no issues However interference with a nearby radio station is of concern Which of the following media could be appropriate to implement the network a Microwave b Coaxial cable c Fiber optic d Radio 147 Applications of optical fiber include a Undersea cable b Longdistance telecommunication c Highspeed data transmission d Medical instrumentation e All of the above 148 Light rays are confined within a simple optical fiber by means of a Total internal reflection at the outer edge of the cladding b Total internal reflection at the corecladding interface c Reflection at the fibers jacket d Refraction e Defraction 149 An optical fiber has a core with a refractive index of 145 and a cladding with a refrac tive index of 142 The numerical aperture of the fiber is a 012 b 018 c 029 d 038 658 Modern Topics PROBLEMS 1410 A 20km long fiberoptic cable has an output power of 002 mW If the fiber loss is 048 dBkm what is the input power to the fiber a 52piW b 19xW c 7xW d 2ftW Answers 141b 142c 143a 144b 145d 146c 147e 148b 149c 1410a 141 Discuss briefly some applications of microwaves other than those discussed in the text 142 A useful set of parameters known as the scattering transfer parameters is related to the incident and reflected waves as l fr Ti2 b2 T2I T22a2 a Express the Tparameters in terms of the Sparameters b Find T when 02 04 S 04 02 143 The Sparameters of a twoport network are Sn 033 jO16 Sl2 S21 056 S22 044 jO62 Find the input and output reflection coefficients when ZL Zo 50 Q and Zg 2ZO 144 Why cant regular lumped circuit components such as resistors inductors and capaci tors be used at microwave frequencies 145 In free space a microwave signal has a frequency of 84 GHz Calculate the wavelength of the signal 146 An electrostatic discharge ESD can be modeled as a capacitance of 125 pF charged to 1500 V and discharging through a 2km resistor Obtain the current waveform 147 The insertion loss of a filter circuit can be calculated in terms of its A B C and D pa rameters when terminated by Zg and ZL as shown in Figure 1410 Show that AZL B CZgZL DZg IL 20 log 148 A silver rod has rectangular cross section with height 08 cm and width 12 cm Find a The dc resistance per 1 km of the conductor b The ac resistance per 1 km of the conductor at 6 MHz 149 The speed of light in a given medium is measured as 21 X 108 ms Find its refractive index PROBLEMS 659 Figure 1410 For Problem 145 1410 How will optical fiber be useful in EMI isolation 1411 A glass fiber has a core diameter of 50 jum a core refractive index of 162 and a cladding with a refractive index of 1604 If light having a wavelength of 1300 nm is used find a The numerical aperture b The acceptance angle c the number of transmission modes 1412 An optical fiber with a radius of 25 fjm and a refractive index of 145 is surrounded by an air cladding If the fiber is illuminated by a ray of 13 xm light determine a V b NA c An estimate of how many modes can propagate 1413 An optical fiber with an attenuation of 04 dBkm is 5 km long The fiber has 153 n2 145 and a diameter of 50 pm Find a The maximum angle at which rays will enter the fiber and be trapped b The percentage of input power received 1414 A laser diode is capable of coupling 10 mW into a fiber with attenuation of 05 dBkm If the fiber is 850 m long calculate the power received at the end of the fiber 1415 Attenuation a10 in Chapter 10 is in Npm whereas attenuation al4 in this chapter is in dBkm What is the relationship between the two 1416 A lightwave system uses a 30km fiber link with a loss of 04 dBkm If the system re quires at least 02 mW at the receiver calculate the minimum power that must be launched into the fiber 1417 a Discuss the advantages derived from using a fiber optic cable b What is pulse dispersion Chapter 15 NUMERICAL METHODS The recipe for ignorance is be satisfied with your opinions and content with your knowledge ELBERT HUBBARD 151 INTRODUCTION In the preceding chapters we considered various analytic techniques for solving EM prob lems and obtaining solutions in closed form A closed form solution is one in the form of an explicit algebraic equation in which values of the problem parameters can be substituted Some of these analytic solutions were obtained assuming certain situations thereby making the solutions applicable to those idealized situations For example in deriving the formula for calculating the capacitance of a parallelplate capacitor we assumed that the fringing effect was negligible and that the separation distance was very small compared with the width and length of the plates Also our application of Laplaces equation in Chapter 6 was restricted to problems with boundaries coinciding with coordinate surfaces Analytic solutions have an inherent advantage of being exact They also make it easy to observe the behavior of the solution for variation in the problem parameters However an alytic solutions are available only for problems with simple configurations When the complexities of theoretical formulas make analytic solution intractable we resort to nonanalytic methods which include 1 graphical methods 2 experimental methods 3 analog methods and 4 numerical methods Graphical experimental and analog methods are applicable to solving relatively few problems Numerical methods have come into prominence and become more attractive with the advent of fast digital computers The three most commonly used simple numerical techniques in EM are 1 moment method 2 finite difference method and 3 finite element method Most EM problems involve either partial differential equations or integral equations Partial differ ential equations are usually solved using the finite difference method or the finite element method integral equations are solved conveniently using the moment method Although numerical methods give approximate solutions the solutions are sufficiently accurate for engineering purposes We should not get the impression that analytic techniques are out dated because of numerical methods rather they are complementary As will be observed later every numerical method involves an analytic simplification to the point where it is easy to apply the method The Matlab codes developed for computer implementation of the concepts developed in this chapter are simplified and selfexplanatory for instructional purposes The notations 660 152 FIELD PLOTTING 661 used in the programs are as close as possible to those used in the main text some are defined wherever necessary These programs are by no means unique there are several ways of writing a computer program Therefore users may decide to modify the programs to suit their objectives 152 FIELD PLOTTING In Section 49 we used field lines and equipotential surfaces for visualizing an electrosta tic field However the graphical representations in Figure 421 for electrostatic fields and in Figures 78b and 716 for magnetostatic fields are very simple trivial and qualitative Accurate pictures of more complicated charge distributions would be more helpful In this section a numerical technique that may be developed into an interactive computer program is presented It generates data points for electric field lines and equipotential lines for arbitrary configuration of point sources Electric field lines and equipotential lines can be plotted for coplanar point sources with simple programs Suppose we have N point charges located at position vectors rt r2 rN the electric field intensity E and potential V at position vector r are given re spectively by y Qkrrk ti Airs r rk3 151 and ift 4ire r rk If the charges are on the same plane z constant eqs 151 and 152 become N E 2 N k xkf y ykff2 Qk xkf To plot the electric field lines follow these steps ykf m 152 153 154 1 Choose a starting point on the field line 2 Calculate Ex and Ey at that point using eq 153 3 Take a small step along the field line to a new point in the plane As shown in Figure 151 a movement A along the field line corresponds to movements AJC and Ay along x and ydirections respectively From the figure it is evident that Ax Ex E E2 X 2U2 662 Numerical Methods field line new point point Figure 151 A small displacement on a field line or Ax 155 Similarly y Ey 156 Move along the field line from the old point x y to a new point x x Ax y y Ay 4 Go back to steps 2 and 3 and repeat the calculations Continue to generate new points until a line is completed within a given range of coordinates On completing the line go back to step 1 and choose another starting point Note that since there are an infi nite number of field lines any starting point is likely to be on a field line The points generated can be plotted by hand or by a plotter as illustrated in Figure 152 To plot the equipotential lines follow these steps 1 Choose a starting point 2 Calculate the electric field Ex Ey at that point using eq 153 Figure 152 Generated points on field lines shown thick and equipotential lines shown dotted 152 FIELD PLOTTING 663 3 Move a small step along the line perpendicular to field line at that point Utilize the fact that if a line has slope m a perpendicular line must have slope Mm Since an fifield line and an equipotential line meeting at a given point are mutually or thogonal there Ax 157 Ay 158 Move along the equipotential line from the old point x y to a new point x Ax y Ay As a way of checking the new point calculate the potential at the new and old points using eq 154 they must be equal because the points are on the same equipotential line 4 Go back to steps 2 and 3 and repeat the calculations Continue to generate new points until a line is completed within the given range of x and After completing the line go back to step 1 and choose another starting point Join the points gener ated by hand or by a plotter as illustrated in Figure 152 By following the same reasoning the magnetic field line due to various current distri butions can be plotted using BiotSavart law Programs for determining the magnetic field line due to line current a current loop a Helmholtz pair and a solenoid can be developed Programs for drawing the electric and magnetic field lines inside a rectangular waveguide or the power radiation pattern produced by a linear array of vertical halfwave electric dipole antennas can also be written EXAMPLE 151 Write a program to plot the electric field and equipotential lines due to a Two point charges Q and 4Q located at x y 1 0 and 10 respectively b Four point charges Q QQ and Q located at xy 1 11 1 1 1 and 1 1 respectively Take QIAire landA 01 Consider the range 5 x y 5 Solution Based on the steps given in Section 152 the program in Figure 153 was developed Enough comments are inserted to make the program as selfexplanatory as possible For example to use the program to generate the plot in Figure 154a load program plotit in your Matlab directory At the command prompt in Matlab type plotit 1 4 1 0 1 0 1 1 01 001 8 2 5 where the numbers have meanings provided in the program Further explanation of the program is provided in the following paragraphs Since the field lines emanate from positive charges and terminate on negative charges it seems reasonable to generate starting points xs ys for the field lines on small circles centered at charge locations xQ yQ that is xs xQ r cos 0 y Vn rsind 1511a 1511b 664 Numerical Methods function plotitchargeslocationckEFieldckEqDLEDLVNLENLVPTS figure hold on Program for plotting the electric field lines and equipotential lines due to coplanar point charges the plot is to be within the range 5xy5 This is the correct usage function plotitcharges locationckEFieldckEqDLEDLVNLENLVPTS where charges a vector containing the charges location a matrix where each row is a charge location ckEField Flag set to 1 plots the Efield lines ckEq Flag set to 1 plots the Equipotential lines DLE or DLV the increment along E V lines NLE No of EField lines per charge NLV No of Equipotential lines per charge PTS Plots every PTS point ie if PTS 5 then plot every 5th point note that constant Q4PieErR is set equal to 10 Determine the EField Lines For convenience the starting points distributed about charge locations Qcharges XQ location1 YQ location2 JJ1 NQ lengthcharges if ckEField for K1NQ for I 1NLE THETA 2pi11NLE XSYS are radially XSXQK YSYQK XEXS YEYS JJJJ1 if modJJ plot XE end while1 01cos 01sin PTS YE Find i ncrement EX0 EY0 THETA THETA and new point XY Figure 153 Computer program for Example 151 152 FIELD PLOTTING 665 for J1NQ R sqrtXEXQJ2 YE YQJ A 2 EX EX QJXEXQJ FT3 EY EY QJYEYQJRA3 end E sqrtEX2 EYA2 CHECK FOR A SINGULAR POINT if E 00005 break end DX DLEEXE DY DLEEYE FOR NEGATIVE CHARGE NEGATE DX DY SO THAT INCREMENT IS AWAY FROM THE CHARGE if QK 0 DX DX DY DY end XE XE DX YE YE DY CHECK WHETHER NEW POINT IS WITHIN THE GIVEN RANGE OR TOO CLOSE TO ANY OF THE POINT CHARGES TO AVOID SINGULAR POINT if absXE 5 absYE 5 break end if sumabsXEXQ 05 absYEYQ 05 0 break end JJJJ1 if modJJPTS plot XEYE end end while loop end I 1NLE end K 1NQ end if NEXT DETERMINE THE EQUIPOTENTIAL LINES FOR CONVENIENCE THE STARTING POINTS XSYS ARE CHOSEN LIKE THOSE FOR THE EFIELD LINES if ckEq JJ1 DELTA 2 ANGLE 45pi180 Figure 153 Continued 666 Numerical Methods for K 1NQ FACTOR 5 for KK 1NLV XS XQK FACTORcosANGLE YS YQK FACTORsinANGLE if absXS 5 absYS 5 break end DIR 1 XV XS YV YS JJJJ1 if modJJPTS plotXVYV end FIND INCREMENT AND NEW POINT XVYV Nl while 1 EX 0 EY 0 for J 1NQ R sqrtXVXQJ2 YVYQJ2 EX EX QJXVXQJRA3 EY EY QJYVYQJR3 end EsqrtEXA2 EY2 if E 00005 FACTOR 2 FACTOR break end DX DLVEYE DY DLVEVE XV XV DIRDX YV YV DIRDY CHECK IF THE EQUIPOTENTIAL LINE LOOPS BACK TO XYS R0 sqrtXV XS2 YV YS2 if R0 DELTA N 50 FACTOR 2 FACTOR break end CHECK WHETHER NEW POINT IS WITHIN THE GIVEN RANGE IF FOUND OUT OF RANGE GO BACK TO THE STARTING POINT SYSBUT INCREMENT IN THE OPPOSITE DIRECTION if absXV 5 absYV 5 DIR DIR 2 XV XS YV YS Figure 153 Continued 152 FIELD PLOTTING m 667 if absDlR 1 FACTOR 2 FACTOR break end else if sumfabsXVXQ 005 absYVYQ 005 0 break end end JJJJ1 if modJJPTS NN1 plotXVYV end end WHILE loop end KK end K end if Figure 153 Continued where r is the radius of the small circle eg r 01 or 005 and 6 is a prescribed angle chosen for each field line The starting points for the equipotential lines can be generated in different ways along the x and yaxes along line y x and so on However to make the program as general as possible the starting points should depend on the charge loca tions like those for the field lines They could be chosen using eq 1511 but with fixed 0eg 45 and variable r eg 05 10 20 The value of incremental length A is crucial for accurate plots Although the smaller the value of A the more accurate the plots we must keep in mind that the smaller the value of A the more points we generate and memory storage may be a problem For example a line may consist of more than 1000 generated points In view of the large number of points to be plotted the points are usually stored in a data file and a graphics routine is used to plot the data For both the Efield and equipotential lines different checks are inserted in the program in Figure 153 a Check for singular point E 0 b Check whether the point generated is too close to a charge location c Check whether the point is within the given range of 5 xy 5 d Check whether the equipotential line loops back to the starting point The plot of the points generated for the cases of two point charges and four point charges are shown in Figure 154a and b respectively 668 Numerical Methods Figure 154 For Example 151 plots of field lines and equipoten tial lines due to a two point charges and b four point charges a twodimensional quadrupole PRACTICE EXERCISE 151 Write a complete program for plotting the electric field lines and equipotential lines due to coplanar point charges Run the program for iV 3 that is there are three point charges Q Q and Q located at xy 10 0 1 and 10 re spectively Take Q4irs 1 A 01 or 001 for greater accuracy and limit your plot to 5 S j t y 5 Answer See Figure 155 153 THE FINITE DIFFERENCE METHOD 669 Figure 155 For Practice Exercise 151 153 THE FINITE DIFFERENCE METHOD The finite difference method1 FDM is a simple numerical technique used in solving prob lems like those solved analytically in Chapter 6 A problem is uniquely defined by three things 1 A partial differential equation such as Laplaces or Poissons equations 2 A solution region 3 Boundary andor initial conditions A finite difference solution to Poissons or Laplaces equation for example proceeds in three steps 1 dividing the solution region into a grid of nodes 2 approximating the dif ferential equation and boundary conditions by a set of linear algebraic equations called difference equations on grid points within the solution region and 3 solving this set of algebraic equations For an extensive treatment of the finite difference method see G D Smith Numerical Solution of Partial Differential Equations Finite Difference Methods 2nd edition Oxford Clarendon 1978 L 670 Numerical Methods Step 1 Suppose we intend to apply the finite difference method to determine the elec tric potential in a region shown in Figure 156a The solution region is divided into rec tangular meshes with grid points or nodes as in Figure 156a A node on the boundary of the region where the potential is specified is called a fixed node fixed by the problem and interior points in the region are called free points free in that the potential is unknown Step 2 Our objective is to obtain the finite difference approximation to Poissons equation and use this to determine the potentials at all the free points We recall that Poissons equation is given by 159a For twodimensional solution region such as in Figure 156a pv is replaced by ps T 0 so d2V dz d2v d2v dx2 dy2 From the definition of the derivative of Vx y at point x0 yo V dV dx Vxo Ax yo Vxo Ax yo 2Ax 2 Ax 159b 1510 yo A v 1 h K b Figure 156 Finite difference solution pattern a division of the solution into grid points b finite difference fivenode molecule 153 THE FINITE DIFFERENCE METHOD 671 where Ax is a sufficiently small increment along x For the second derivative which is the derivative of the first derivative V V dx2 dV Vxo Ax2vo Vxo Ax2yo dx Ax Vxo Axy0 2Vxoyo Vx0 Axy0 VilJ 2VU Ax1 7 iU Ax2 1511 Equations 1510 and 1511 are the finite difference approximations for the first and second partial derivatives of V with respect to x evaluated at x xo The approximation in eq 1510 is associated with an error of the order of the Ax while that of eq 1511 has an associated error on the order of Ax2 Similarly Vxoyo Ay 2Vxo yo Vxoyo Ay d2V dyz Ayf ViJl Ay2 1512 Substituting eqs 1511 and 1512 into eq 159b and letting Ax Ay h gives Vi lJ Vw ViJl or 1513 where h is called the mesh size Equation 1513 is the finite difference approximation to Poissons equation If the solution region is chargefree ps 0 eq 159 becomes Laplaces equation dx2 dy2 1514 The finite difference approximation to this equation is obtained from eq 1513 by setting ps 0 that is 1 4 ViJl Vui 1515 This equation is essentially a fivenode finite difference approximation for the potential at the central point of a square mesh Figure 156b illustrates what is called the finite differ 672 Numerical Methods ence fivenode molecule The molecule in Figure 156b is taken out of Figure 156a Thus eq 1515 applied to the molecule becomes V4 1516 This equation clearly shows the averagevalue property of Laplaces equation In other words Laplaces equation can be interpreted as a differential means of stating the fact that the potential at a specific point is the average of the potentials at the neighboring points Step 3 To apply eq 1516 or eq 1513 to a given problem one of the following two methods is commonly used A Iteration Method We start by setting initial values of the potentials at the free nodes equal to zero or to any reasonable guessed value Keeping the potentials at the fixed nodes unchanged at all times we apply eq 1516 to every free node in turn until the potentials at all free nodes are cal culated The potentials obtained at the end of this first iteration are not accurate but just ap proximate To increase the accuracy of the potentials we repeat the calculation at every free node using old values to determine new ones The iterative or repeated modification of the potential at each free node is continued until a prescribed degree of accuracy is achieved or until the old and the new values at each node are satisfactorily close B Band Matrix Method Equation 1516 applied to all free nodes results in a set of simultaneous equations of the form A V B 1517 where A is a sparse matrix ie one having many zero terms V consists of the unknown potentials at the free nodes and B is another column matrix formed by the known potentials at the fixed nodes Matrix A is also banded in that its nonzero terms appear clustered near the main diagonal because only nearest neighboring nodes affect the potential at each node The sparse band matrix is easily inverted to determine V Thus we obtain the potentials at the free nodes from matrix V as V B 1518 The finite difference method can be applied to solve timevarying problems For example consider the onedimensional wave equation of eq 101 namely d2p dt2 1519 153 THE FINITE DIFFERENCE METHOD 673 where u is the wave velocity and P is the E or field component of the EM wave The difference approximations of the derivatives at xo to or Oyth node shown in Figure 157 are dx dt2 Ax2 v At2 Inserting eqs 1520 and 1521 in eq 1520 and solving for 4iJi gives 21 a PtJ w where a u At Ax 1520 1521 1522 1523 It can be shown that for the solution in eq 1522 to be stable a 1 To start the finite difference algorithm in eq 1522 we use the initial conditions We assume that at t 0 dtifidt 0 and use central difference approximation see Review Question 152 to get i JL X JL ili 0 dt 2At or 1524 Substituting eq 1524 into eq 1522 and taking j 0 t 0 we obtain o io 21 ai0 PiA r At 1 A i i 1 2A 2 Figure 157 Finite difference solution pattern for wave equation 2 I 1 2 A 2Av vu Av xo x A x x 2Av 674 Numerical Methods or i io iio 21 aj0 1525 With eq 1525 as the starting formula the value of 4 at any point on the grid can be obtained directly from eq 1522 Note that the three methods discussed for solving eq 1516 do not apply to eq 1522 because eq 1522 can be used directly with eq 1525 as the starting formula In other words we do not have a set of simultaneous equations eq 1522 is an explicit formula The concept of FDM can be extended to Poissons Laplaces or wave equations in other coordinate systems The accuracy of the method depends on the fineness of the grid and the amount of time spent in refining the potentials We can reduce computer time and increase the accuracy and convergence rate by the method of successive overrelaxation by making reasonable guesses at initial values by taking advantage of symmetry if possible by making the mesh size as small as possible and by using more complex finite difference molecules see Figure 1541 One limitation of the finite difference method is that inter polation of some kind must be used to determine solutions at points not on the grid One obvious way to overcome this is to use a finer grid but this would require a greater number of computations and a larger amount of computer storage EXAMPLE 152 Solve the onedimensional boundaryvalue problem P x2 0 x 1 subject to P0 0 Pl Use the finite difference method Solution First we obtain the finite difference approximation to the differential equation P x2 which is Poissons equation in one dimension Next we divide the entire domain 0 x 1 into N equal segments each of length h IA as in Figure 158a so that there are N 1 nodes d 2i dx 2 Pxo 2Pxo Pxo h h Figure 158 For Example 152 x 1 a l b 153 THE FINITE DIFFERENCE METHOD 675 or Thus or r 2 A 2Pj h 2P x2h2 l 2 2 2 Using this finite difference scheme we obtain an approximate solution for various values of N The Matlab code is shown in Figure 159 The number of iterations NI depends on the degree of accuracy desired For a onedimensional problem such as this NI 50 may suffice For two or threedimensional problems larger values of NI would be required see Table 151 It should be noted that the values of P at end points fixed nodes are held fixed The solutions for N 4 and 10 are shown in Figure 1510 We may compare this with the exact solution obtained as follows Given that d2Pdx2 x2 integrating twice gives P Ax B 12 ONEDIMENSIONAL PROBLEM OF EXAMPLE 152 SOLVED USING FINITE DIFFERENCE METHOD h MESH SIZE ni NO OF ITERATIONS DESIRED P n2 0 ni1000 110 h 1n phizerosn11 xh0n xlx2n for klni phi 2 n phi 3 nl phi lnl xlN2h2 2 end CALCULATE THE EXACT VALUE ALSO phiexx10xA3120 diary atestout lnl phi phiex diary off Figure 159 Computer program for Example 152 L 676 Numerical Methods 05 04 1 0 3 02 01 1 1 1 1 f 1 1 1 1 I i i i 1 2 3 4 5 6 7 9 1 Figure 1510 For Example 152 plot of Px Continuous curve is for iV 10 dashed curve is for N 4 where A and B are integration constants From the boundary conditions 05 0 1 1 or A 01 00 A Hence the exact solution is P xl x312 which is calculated in Figure 159 and found to be very close to case N 10 PRACTICE EXERCISE 152 Solve the differential equation d2yldx2 y 0 with the boundary conditions y0 0 yl 1 using the finite difference method Take Ax 14 Answer Compare your result with the exact solution yx sinc sinl EXAMPLE 153 Determine the potential at the free nodes in the potential system of Figure 1511 using the finite difference method Solution This problem will be solved using the iteration method and band matrix method 25 m 20 V ov 1 2 3 4 5 6 7 8 153 THE FINITE DIFFERENCE METHOD Figurel5il For Example 153 677 30 V Method 1 Iteration Method We first set the initial values of the potential at the free nodes equal to zero We apply eq 1516 to each free node using the newest surrounding potentials each time the potential at that node is calculated For the first iteration Vl 140 20 0 0 5 V2 145 0 0 0 125 V3 145 20 0 0 625 V4 14125 625 0 0 1875 and so on To avoid confusion each time a new value at a free node is calculated we cross out the old value as shown in Figure 1512 After V8 is calculated we start the second iter ation at node 1 Vx 140 20 125 625 6875 V2 146875 0 0 1875 2187 and so on If this process is continued we obtain the uncrossed values shown in Figure 1512 after five iterations After 10 iterations not shown in Figure 1512 we obtain V 1004 V2 4956 V3 1522 V4 9786 V5 2105 V6 1897 V7 1506 V8 1126 Method 2 Band Matrix Method This method reveals the sparse structure of the problem We apply eq 1516 to each free node and keep the known terms prescribed po tentials at the fixed nodes on the right side while the unknown terms potentials at free nodes are on the left side of the resulting system of simultaneous equations which will be expressed in matrix form as A V B For node 1 4V V2 V3 20 0 678 H Numerical Methods O s Ox Ox 9659 9rh5T fTT 6r875 fr 4705 e 20 0 1485 4T9 e 9545 fr 20 2087 e 1884 fctr fr 1502 T4 OO 1125 409 e 20 30 30 30 30 15 Figure 1512 For Example 153 the values not crossed out are the solutions after five iterations For node 2 For node 3 For node 4 For node 5 V 4V V4 0 0 j 4V3 V4 V5 2 0 v3 4V4 y6 o V3 4V5 Vb 2 0 30 153 THE FINITE DIFFERENCE METHOD 679 For node 6 For node 7 For node 8 V4 V5 4V6 V7 30 V6 4V7 Vs 30 0 V7 4V8 0 0 30 Note that we have five terms at each node since we are using a fivenode molecule The eight equations obtained are put in matrix form as 4 1 1 6 0 0 0 0 I 4 0 1 0 0 0 0 1 0 4 1 1 y 0 0 0 i I 4 0 1 6 0 0 0 r 0 4 i o 0 0 0 0 1 1 4 1 0 0 0 0 P o 1 4 1 0 0 0 0 0 6 r 4 v2 v3 v4 v5 V6 v7 vs 2 0 0 2 0 0 5 0 3 0 3 0 3 0 or A V B where A is the band sparse matrix V is the column matrix consisting of the unknown potentials at the free nodes and 6 is the column matrix formed by the potential at the fixed nodes The band nature of A is shown by the dotted loop Notice that matrix A could have been obtained directly from Figure 1511 without writing down eq 1516 at each free node To do this we simply set the diagonal or self terms A 4 and set Atj lifi and 7 nodes are connected or Atj 0 if i andj nodes are not directly connected For example A23 A32 0 because nodes 2 and 3 are not con nected whereas A46 A64 1 because nodes 4 and 6 are connected Similarly matrix B is obtained directly from Figure 1511 by setting Bt equal to minus the sum of the potentials at fixed nodes connected to node i For example B5 20 30 because node 5 is con nected to two fixed nodes with potentials 20 V and 30 V If node i is not connected to any fixed node B 0 By inverting matrix A using Matlab we obtain M AV1 B or V 1004 V2 4958 V3 1522 V4 9788 V5 2105 V6 1897 V7 1506 V8 1126 which compares well with the result obtained using the iteration method 680 I I Numerical Methods Figure 1513 For Practice Exercise 153 50 V PRACTICE EXERCISE 153 Use the iteration method to find the finite difference approximation to the potentials at points a and b of the system in Figure 1513 Answer Va 1001 V Vb 283 V EXAMPLE 154 Obtain the solution of Laplaces equation for an infinitely long trough whose rectangular cross section is shown in Figure 1514 Let V 10 V V2 100 V V3 40 V and V4 0 V Solution We shall solve this problem using the iteration method In this case the solution region has a regular boundary We can easily write a program to determine the potentials at the grid points within the trough We divide the region into square meshes If we decide to use a 15 X 10 grid the number of grid points along x is 15 1 16 and the number of grid points along y is 10 1 11 The mesh size h 1515 01 mThe 15 X 10 grid is il Figure 1514 For Example 154 10m V 0 V K 100 V Vl 10 V 15 m 0 8 10 1 153 THE FINITE DIFFERENCE METHOD 681 Figure 1515 For Example 154 a 15 X 10 t5 10 grid 80 150 lustrated in Figure 1515 The grid points are numbered starting from the lower left hand corner of the trough Applying eq 1515 and using the iteration method the com puter program in Figure 1516 was developed to determine the potential at the free nodes At points xy 05 05 08 08 10 05 and 08 02 corresponding to ij 5 5 8 8 10 5 and 8 2 respectively the potentials after 50 100 and 200 iterations are shown in Table 151 The exact values see Problem 618c obtained using the method of separation of variables and a program similar to that of Figure 611 are also shown It should be noted that the degree of accuracy depends on the mesh size h It is always desirable to make h as small as possible Also note that the potentials at the fixed nodes are held constant throughout the calculations USING FINITE DIFFERENCE ITERATION METHOD THIS PROGRAM SOLVES THE TWODIMENSIONAL BOUNDARYVALUE PROBLEM LAPLACES EQUATION SHOWN IN FIG 1514 ni NO OF ITERATIONS nx NO OF X GRID POINTS ny NO OF Y GRID POINTS vij POTENTIAL AT GRID POINT ij OR xy WITH NODE NUMBERING STARTING FROM THE LOWER LEFTHAND CORNER OF THE TROUGH vl 100 v2 1000 v3 400 v4 00 ni 2 00 nx 16 ny 11 SET INITIAL VALUES EQUAL TO ZEROES v zerosnxny FIX POTENTIALS ARE FIXED NODES Figure 1516 Computer Program for Example 154 682 Numerical Methods for i2nxl v i1 vl v i ny v3 end for j2nyl v1j v4 vnxj v2 end vl1 05vl v 4 vnx1 05vl v 2 vlny 05v3 v 4 vnxny 05v2 v 3 NOW FIND vij USING EQ 1515 AFTER ni ITERATIONS for klni for i 2nxl for j2nyl vij 025 vilj end end end diary atestlout v66 v99 vll6 v93 lnx lny vij diary off Figure 1516 Continued TABLE 151 Solution of Example 154 Iteration Method at Selected Points Coordinates y 05 05 08 08 1005 08 02 Number of Iterations 50 2091 377 4183 1987 100 2244 3856 4318 2094 200 2249 3859 432 2097 Exact Value 2244 3855 4322 2089 PRACTICE EXERCISE 154 Consider the trough of Figure 1517 Use a fivenode finite difference scheme to find the potential at the center of the trough using a a 4 X 8 grid and b a 12 X 24 grid Answer a 238 V b 2389 V 60 V 50 V 154 THE MOMENT METHOD Figure 1517 For Practice Exercise 154 683 154 THE MOMENT METHOD Like the finite difference method the moment method2 or the method of moments MOM has the advantage of being conceptually simple While the finite difference method is used in solving differential equations the moment method is commonly used in solving integral equations For example suppose we want to apply the moment method to solve Poissons equa tion in eq 159a It can be shown that an integral solution to Poissons equation is V Alter 1526 We recall from Chapter 4 that eq 1526 can be derived from Coulombs law We also recall that given the charge distribution pvx y z we can always find the potential Vx y z the electric field Ex y z and the total charge Q If on the other hand the po tential V is known and the charge distribution is unknown how do we determine pv from eq 1526 In that situation eq 1526 becomes what is called an integral equation An integral equation is one involving ihc unknown function under the inloiiral sign It has the general form of Vx Kx t pt dt 1527 where the functions Kx i and Vt and the limits a and b are known The unknown func tion pt is to be determined the function Kx t is called the kernel of the equation The 2The term moment method was first used in Western literature by Harrington For further exposi tion on the method see R F Harrington Field Computation by Moment Methods Malabar FL Krieger 1968 684 Numerical Methods moment method is a common numerical technique used in solving integral equations such as in eq 1527 The method is probably best explained with an example Consider a thin conducting wire of radius a length LL S a located in free space as shown in Figure 1518 Let the wire be maintained at a potential of Vo Our goal is to de termine the charge density pL along the wire using the moment method Once we deter mine pL related field quantities can be found At any point on the wire eq 1526 reduces to an integral equation of the form Vn L PLdl 4reor 1528 Since eq 1528 applies for observation points everywhere on the wire at a fixed point yk known as the match point L PLJy dy J0 yky 1529 We recall from calculus that integration is essentially finding the area under a curve If Ay is small the integration of fly over 0 y L is given by fly dyflylAy fy2 Ay fyN Ay 1530 fly 0 kl where the interval L has been divided into N units of each length Ay With the wire divided into N segments of equal length A as shown in Figure 1519 eq 1529 becomes A 1 Pi A 4irsoVo ykyu 1531 where A LIN Ay The assumption in eq 1531 is that the unknown charge density pk on the kth segment is constant Thus in eq 1531 we have unknown constants pu la Figure 1518 Thin conducting wire held at a constant potential 154 THE MOMENT METHOD 685 pN Figure 1519 Division of the wire into N segments Pi Pk p2 pN Since eq 1531 must hold at all points on the wire we obtain N similar equations by choosing N match points at yu y2 y ys o n t n e w i r e Thus we obtain P A 4TT OV O 4ireoVo yi y Pi 7 P2A 7 1532a 1532b 4TTSOVO Pi A Pi A yNy yN 1532c The idea of matching the lefthand side of eq 1529 with the righthand side of the equa tion at the match points is similar to the concept of taking moments in mechanics Here lies the reason this technique is called moment method Notice from Figure 1519 that the match points yu y2 yN are placed at the center of each segment Equation 1532 can be put in matrix form as B A p 1533 where B 4TTSOVO 1534 686 Numerical Methods A An A12 A21 A 2 2 IN A2N 1535a ym y Pi PN m n 1535b 1536 In eq 1533 p is the matrix whose elements are unknown We can determine p from eq 1533 using Cramers rule matrix inversion or Gaussian elimination technique Using matrix inversion p A1 B 1537 where A is the inverse of matrix A In evaluating the diagonal elements or self terms of matrix A in eq 1532 or 1535 caution must be exercised Since the wire is con ducting a surface charge density ps is expected over the wire surface Hence at the center of each segment V center 2x A2 psa dp ay 2iraps 4TTSO In A2 A22 a2m A2 A22 a22 Assuming A a 4Treo 1538 where pL 2TT aps Thus the self terms m n are 21nAa 1539 154 THE MOMENT METHOD 687 Equation 1533 now becomes 2 In yi y 2 l n Pi P2 PV 4ITP V 1 1 1 1540 Using eq 1537 with eq 1540 and letting Vo 1 V L 1 m a 1 mm and N 10 A LIN a Matlab code such as in Figure 1520 can be developed The program in Figure 1520 is selfexplanatory It inverts matrix A and plots pL against y The plot is shown in Figure 1521 The program also determines the total charge on the wire using Q which can be written in discrete form as 1541 1542 k With the chosen parameters the value of the total charge was found to be Q 8536 pC If desired the electric field at any point can be calculated using E which can be written as 1543 1544 where R and R r rk x xkax y z zkaz r x y z is the position vector of the observation point and rk xk yk zk is that of the source point Notice that to obtain the charge distribution in Figure 1521 we have taken N 10 It should be expected that a smaller value of N would give a less accurate result and a larger value of N would yield a more accurate result However if A7 is too large we may have the computation problem of inverting the square matrix A The capacity of the computing fa cilities at our disposal can limit the accuracy of the numerical experiment L 688 Numerical Methods THIS PROGRAM DETERMINES THE CHARGEDISTRIBUTION ON A CONDUCTING THIN WIRE OF RADIUS AA AND LENGTH L MAINTAINED AT VO VOLT THE WIRE IS LOCATED AT 0 Y L ALL DIMENSIONS ARE IN SI UNITS MOMENT METHOD IS USED N IS THE NO OF SEGMENTS INTO WHICH THE WIRE IS DIVIDED RHO IS THE LINE CHARGE DENSITY RHO INVAB FIRST SPECIFY PROBLEM PARAMETERS ER 10 EO 88541e12 VO 10 AA 0001 L 10 N 20 DELTA LN SECOND CALCULATE THE ELEMENTS OF THE COEFFICIENT MATRIX A YDELTAIO5 for ilN for jlN ifi j Ai jDELTAabsYiYj else Aij20logDELTAAA end end end NOW DETERMINE THE MATRIX OF CONSTANT VECTOR B AND FIND Q B 40piEOERVOonesN1 C invA RHO CB SUM 00 for I1N SUM SUM RHOI end QSUMDELTA diary aexaml45aout EOQ 1N Y RHO diary off FINALLY PLOT RHO AGAINST Y plotYRHO xlabely cm ylabelrhoL pCm Figure 1520 Matlab code for calculating the charge distribution on the wire in Figure 1518 154 THE MOMENT METHOD 689 105 0 01 02 03 04 05 06 07 08 09 1 ycm Figure 1521 Plot of pL against y EXAMPLE 155 Use the moment method to find the capacitance of the parallelplane capacitor of Figure 1522 Take a 1 m b 1 m d 1 m and sr 10 Solution Let the potential difference between the plates be Vo 2 V so that the top plate Px is main tained at 1 V while the bottom plate P2 is at 1 V We would like to determine the surface charge density ps on the plates so that the total charge on each plate can be found as Q PsdS Figure 1522 Parallelplate ca pacitor for Example 155 I 690 U Numerical Methods Once Q is known we can calculate the capacitance as C Q Q V 2 To determine ps using the moment method we divide P into n subsections AS AS2 ASn and P2 into n subsections A5n1 ASn2 AS2n The potential V at the center of a typical subsection AS is 1 y pi 47TO PjdS 2n ASt It has been assumed that there is uniform charge distribution on each subsection The last equation can be written as where Thus dS In 1L 71 Vn In 71 In 7 1 PjAnj 154 THE MOMENT METHOD 691 yielding a set of In simultaneous equations with 2n unknown charge densities pj In matrix form An A21 A12 A22 P Pi Pin 1 1 1 1 or A p B Hence p B where B is the column matrix defining the potentials and A is a square matrix contain ing elements A To determine A consider the two subsections i and j shown in Figure 1523 where the subsections could be on different plates or on the same plate 1 dxdy where Ru xj Xi2 yj yd2 zj 2il2 For the sake of convenience if we assume that the subsections are squares x2 x A y2 y it can be shown that AS A2 Figure 1523 Subsections i and j for Example 155 692 11 Numerical Methods and An lnl V2 08814 7TSO 7TEO With these formulas the Matlab code in Figure 1524 was developed With n 9 C 2651 pF with n 16 C 2727 pF and with n 25 C 2774 pF USING THE METHOD OF MOMENT THIS PROGRAM DETERMINES THE CAPACITANCE OF A PARALLELPLATE CAPACITOR CONSISTING OF TWO CONDUCTING PLATES EACH OF DIMENSION AA x BB SEPARATED BY A DISTANCE D AND MAINTAINED AT 1 VOLT AND 1 VOLT ONE PLATE IS LOCATED ON THE Z 0 PLANE WHILE THE OTHER IS LOCATED ON THE ZD PLANE ALL DIMENSIONS ARE IN SI UNITS N IS THE NUMBER IS SUBSECTIONS INTO WHICH EACH PLATE IS DIVIDED FIRST SPECIFY THE PARAMETERS ER 10 EO 88541e12 AA 10 BB 10 D 10 N 9 NT 2N M sqrtN DX AAM DY BBM DL DX SECOND CALCULATE THE ELEMENTS OF THE COEFFICIENT MATRIX A K 0 for Kll2 for K21M for K31M K K 1 XK DXK2 05 YK DYK3 05 end end end Figure 1524 Matlab program for Example 155 154 THE MOMENT METHOD 693 for KllN ZK1 00 ZK1N D end for 11NT for J1NT ifIJ AIJ DL08814piEO else R sqrt XIXJA2 YIYJ A2 Z IZ J A2 AIJ DLA24piEOR end end end NOW DETERMINE THE MATRIX OF CONSTANT VECTOR B for K1N BK 10 BKN 10 end INVERT A AND CALCULATE RHO CONSISTING THE UNKNOWN ELEMENTS ALSO CALCULATE THE TOTAL CHARGE Q AND CAPACITANCE C F invA RHO FB SUM 00 for I1N SUM SUM RHOI end Q SUMDLA2 VO 20 C absQVO diary aexaml45bout C 1NT X Y Z RHO diary off Figure 1524 Continued PRACTICE EXERCISE 155 Using the moment method write a program to determine the capacitance of two iden tical parallel conducting wires separated at a distance yo and displaced by xo as shown in Figure 1525 If each wire is of length L and radius a find the capacitance tor cases xo 00204 10mTakeyo 05 mL 1 ma 1 mmer 1 Answer For N 10 number of segments per wire see Table 152 L 694 Numerical Methods Figure 1525 Parallel conducting wires of Practice Ex ercise 155 x L TABLE 152 for Practice x0 m 00 02 04 06 08 10 Capacitance Exercise 155 CpF 491 4891 4853 4789 471 4643 155 THE FINITE ELEMENT METHOD The finite element method FEM has its origin in the field of structural analysis The method was not applied to EM problems until 19683 Like the finite difference method the finite element method is useful in solving differential equations As noticed in Section 153 the finite difference method represents the solution region by an array of grid points its application becomes difficult with problems having irregularly shaped boundaries Such problems can be handled more easily using the finite element method The finite element analysis of any problem involves basically four steps a discretiz ing the solution region into a finite number of subregions or elements b deriving govern ing equations for a typical element c assembling of all elements in the solution region and d solving the system of equations obtained A Finite Element Discretization We divide the solution region into a number of finite elements as illustrated in Figure 1526 where the region is subdivided into four nonoverlapping elements two triangular and two quadrilateral and seven nodes We seek an approximation for the potential Ve within an 3See P P Silvester and R L Ferrari Finite Elements for Electrical Engineers Cambridge England Cambridge Univ Press 1983 155 THE FINITE ELEMENT METHOD 695 Figure 1526 A typical finite element subdi vision of an irregular domain node no element no 7 Actual boundary Approximate boundary element e and then interrelate the potential distributions in various elements such that the potential is continuous across interelement boundaries The approximate solution for the whole region is Vx y X vex y 1545 where N is the number of triangular elements into which the solution region is divided The most common form of approximation for Ve within an element is polynomial ap proximation namely for a triangular element and Vex y a bx cy Vex y a bx cy dxy 1546 1547 for a quadrilateral element The potential Ve in general is nonzero within element e but zero outside e It is difficult to approximate the boundary of the solution region with quadrilat eral elements such elements are useful for problems whose boundaries are sufficiently regular In view of this we prefer to use triangular elements throughout our analysis in this section Notice that our assumption of linear variation of potential within the triangular element as in eq 1546 is the same as assuming that the electric field is uniform within the element that is Ee VVe bax cay 1548 B Element Governing Equations Consider a typical triangular element shown in Figure 1527 The potential VeU Ve2 and Ve3 at nodes 1 2 and 3 respectively are obtained using eq 1546 that is 1 xx yx 1 x 2 y 2 1 X VQ 1549 696 Numerical Methods The coefficients a b and c are determined from eq 1449 as 1 1 1 Xl x2 x3 yi y2 yi vei ve2 ve3 1550 Substituting this into eq 1546 gives Ve 1 x v 2A x2y3 y2 x3 x3y2 y3 x 2 X3J Xxy3 y3 y Xi X3 W 2 Xj y2 Xi or 1551 where 1 1 012 U 1 a 3 2 A and A is the area of the element e that is y2 1552a 1552b 1552c 2A 1 x3 xy2 xxy3 x3y2 or A 12 x2 xYy3 yi x3 Xy2 1553 The value of A is positive if the nodes are numbered counterclockwise starting from any node as shown by the arrow in Figure 1527 Note that eq 1551 gives the potential at any point x y within the element provided that the potentials at the vertices are known This is unlike the situation in finite difference analysis where the potential is known at the grid points only Also note that a are linear interpolation functions They are called the element shape functions and they have the following properties 1554a 155 THE FINITE ELEMENT METHOD 697 Figure 1527 Typical triangular element the local node numbering 123 must be counterclockwise as indicated by the arrow 20 i 1554b The shape functions a and ai for example are illustrated in Figure 1528 The energy per unit length associated with the element e is given by eq 496 that is 1555 eVVe2dS where a twodimensional solution region free of charge ps 0 is assumed But from eq 1551 1556 Substituting eq 1556 into eq 1555 gives 1 ii j If we define the term in brackets as f Vej Cf I Va Vaj dS 1557 1558 3 Figure 1528 Shape functions a and a2 for a triangular element 698 U Numerical Methods we may write eq 1557 in matrix form as We e VeT Ce Ve where the superscript T denotes the transpose of the matrix Ve2 ve3 1559 1560a and eh Ce Mee C 2 C 33 1560b The matrix Ce is usually called the element coefficient matrix The matrix element Cf of the coefficient matrix may be regarded as the coupling between nodes andy its value is obtained from eqs 1552 and 1558 For example n Va dS Similarly Also 77i Kyi y AA y2 el 7 Ky2 yx i 44 44 1 1 L x3 2x x3 I dS 1561a x3 1561b 1561c 1561d 1561e 1561f C 3 Me Wei L32 L23 1561g 155 THE FINITE ELEMENT METHOD 699 However our calculations will be easier if we define Pi yi y3 Pi y3 i Cv yd Qi x3 x2 Q2 xi x3 Q3 x2 xi 1562a With Pt and Qt i 1 2 3 are the local node numbers each term in the element coeffi cient matrix is found as 1562b where A P2Q3 P3Q2 1562c Note that Px P2 0 Ql Q2 Q3 and hence may be used in checking our calculations 0 2 ct This 71 C Assembling of All Elements Having considered a typical element the next step is to assemble all such elements in the solution region The energy associated with the assemblage of all elements in the mesh is el sVTCV 1563 where VI v v2 1564 n is the number of nodes N is the number of elements and C is called the overall or global coefficient matrix which is the assemblage of individual element coefficient matri ces The major problem now is obtaining C from Cw The process by which individual element coefficient matrices are assembled to obtain the global coefficient matrix is best illustrated with an example Consider the finite element mesh consisting of three finite elements as shown in Figure 1529 Observe the numberings of the nodes The numbering of nodes as 1 2 3 4 and 5 is called global numbering The numbering ijk is called local numbering and it corresponds with 1 23 of the element in L 700 Numerical Methods Figure 1529 Assembly of three elements ijk cor responds to local numbering 1 23 of the element in Figure 1527 Figure 1527 For example for element 3 in Figure 1529 the global numbering 354 cor responds with local numbering 123 of the element in Figure 1527 Note that the local numbering must be in counterclockwise sequence starting from any node of the element For element 3 for example we could choose 435 or 543 instead of 354 to correspond with 123 of the element in Figure 1527 Thus the numbering in Figure 1529 is not unique However we obtain the same C whichever numbering is used Assuming the par ticular numbering in Figure 1529 the global coefficient matrix is expected to have the form C c2l c3i c4i c5 C2 C22 C32 C42 c52 C3 C23 C33 c43 c53 C34 C 34 C44 C 25 35 45 55 1565 which is a 5 X 5 matrix since five nodes n 5 are involved Again Cy is the coupling between nodes and j We obtain Cy by utilizing the fact that the potential distribution must be continuous across interelement boundaries The contribution to the ij position in C comes from all elements containing nodes a n d To find C u for example we observe from Figure 1529 that global node 1 belongs to elements 1 and 2 and it is local node 1 in both hence Cn Cft Cft 1566a For C22 global node 2 belongs to element 1 only and is the same as local node 3 hence C22 C 1566b For C44 global node 4 is the same as local nodes 2 3 and 3 in elements 1 2 and 3 re spectively hence a C44 C m 33 1566c For C14 global link 14 is the same as the local links 12 and 13 in elements 1 and 2 respec tively hence 4 T 2 Since there is no coupling or direct link between nodes 2 and 3 c23 c32 0 1566d 1566e 155 THE FINITE ELEMENT METHOD 701 Continuing in this manner we obtain all the terms in the global coefficient matrix by in spection of Figure 1529 as 2 31 0 0 C22 r Ln 0 CXi C 2 C3 r0 0 0 1567 Note that element coefficient matrices overlap at nodes shared by elements and that there are 27 terms nine for each of the three elements in the global coefficient matrix C Also note the following properties of the matrix C 1 It is symmetric Cy C just as the element coefficient matrix 2 Since Cy 0 if no coupling exists between nodes and j it is evident that for a large number of elements C becomes sparse and banded 3 It is singular Although this is not so obvious it can be shown using the element co efficient matrix of eq 1560b D Solving the Resulting Equations From variational calculus it is known that Laplaces or Poissons equation is satisfied when the total energy in the solution region is minimum Thus we require that the partial derivatives of W with respect to each nodal value of the potential be zero that is dW dW dW dV2 0 or dW 0 k 12 n 1568 For example to get d Wd V 0 for the finite element mesh of Figure 1529 we substitute eq 1565 into eq 1563 and take the partial derivative of W with respect to V We obtain dW 0 2VlCu V2CU V3Cl3 V4Cl4 V5Cl5 v2c2l v3c31 v4c4l v5c51 or 0 ViCn V2C l2 v4ci4 v5c5 1569 L 702 Numerical Methods In general 3WdVk 0 leads to 0 V Cik 1570 where n is the number of nodes in the mesh By writing eq 1570 for all nodes k 12 n we obtain a set of simultaneous equations from which the solution of VT Vu V2 Vn can be found This can be done in two ways similar to those used in solving finite difference equations obtained from Laplaces or Poissons equation Iteration Method This approach is similar to that used in finite difference method Let us assume that node 1 in Figure 1529 for example is a free node The potential at node 1 can be obtained from eq 1569 as v y vc Lll i2 In general the potential at a free node k is obtained from eq 1570 as 1571 1572 This is applied iteratively to all the free nodes in the mesh with n nodes Since Cki 0 if node k is not directly connected to node only nodes that are directly linked to node k con tribute to Vk in eq 1572 Thus if the potentials at nodes connected to node k are known we can determine Vk using eq 1572 The iteration process begins by setting the potentials at the free nodes equal to zero or to the average potential Vave 12 Vn x 1573 where Vmin and Vmax are the minimum and maximum values of the prescribed potentials at the fixed nodes With those initial values the potentials at the free nodes are calculated using eq 1572 At the end of the first iteration when the new values have been calcu lated for all the free nodes the values become the old values for the second iteration The procedure is repeated until the change between subsequent iterations becomes negligible Band Matrix Method If all free nodes are numbered first and the fixed nodes last eq 1563 can be written such that W eVf VI Cff Cfp fp Vf 1574 155 THE FINITE ELEMENT METHOD 703 where subscriptsand p respectively refer to nodes with free and fixed or prescribed po tentials Since Vp is constant it consists of known fixed values we only differentiate with respect to Vso that applying eq 1568 to eq 1574 yields cf o or This equation can be written as Cff VA Cfc Vp A V B 1575 1576a or V B 1576b where V Vf A Cff and B C y Vp Since A is in general non singular the potential at the free nodes can be found using eq 1575 We can solve for V in eq 1576a using Gaussian elimination technique We can also solve for V in eq 1576b using matrix inversion if the size of the matrix to be inverted is not large Notice that as from eq 1555 onward our solution has been restricted to a two dimensional problem involving Laplaces equation V2V 0 The basic concepts devel oped in this section can be extended to finite element analysis of problems involving Poissons equation V2V pve V2A fij or wave equation V2 y2j 0 A major problem associated with finite element analysis is the relatively large amount of computer memory required in storing the matrix elements and the associated computa tional time However several algorithms have been developed to alleviate the problem to some degree The finite element method FEM has a number of advantages over the finite differ ence method FDM and the method of moments MOM First the FEM can easily handle complex solution region Second the generality of FEM makes it possible to construct a generalpurpose program for solving a wide range of problems A single program can be used to solve different problems described by the same partial differential equations with different solution regions and different boundary conditions only the input data to the problem need be changed However FEM has its own drawbacks It is harder to under stand and program than FDM and MOM It also requires preparing input data a process that could be tedious EXAMPLE 156 Consider the twoelement mesh shown in Figure 1530a Using the finite element method determine the potentials within the mesh 704 Numerical Methods Node 1 2 3 V0 4 xy 08 18 14 14 2121 12 27 Figure 1530 For Example 156 a twoelement mesh b local and global numbering of the ele ments K0 a b Solution The element coefficient matrices can be calculated using eq 1562 For element 1 consisting of nodes 124 corresponding to the local numbering 123 as in Figure 1530b P 13 P2 09 P3 04 Q 02 Q2 04 Q3 06 A 12054 016 035 Substituting all these into eq 1562b gives 1236 07786 04571 07786 06929 00857 04571 00857 03714 1561 Similarly for element 2 consisting of nodes 234 corresponding to local numbering 1 23 as in Figure 1530b P 06 P2 13 P3 07 G 09 Q2 02 07 A 12091 014 0525 155 THE FINITE ELEMENT METHOD 705 Hence 05571 04571 01 04571 08238 03667 01 03667 04667 Applying eq 1575 gives v2 c42 c44j v4 C21 C23 C41 C43 This can be written in a more convenient form as 1 0 0 0 0 C22 0 C42 0 0 1 0 0 c24 0 C44 v2 v4 1 C21 0 C41 0 C 2 3 1 C43 or QV B The terms of the global coefficient matrix are obtained as follows C22 C22 Cff 06929 05571 125 C42 C24 C Cfl 00857 01 00143 C44 C3f 03714 04667 08381 C2i C2 07786 1562 1563 1564a 1564b C23 C2 04571 C4 C3V 04571 Q3 C3f 03667 Note that we follow local numbering for the element coefficient matrix and global num bering for the global coefficient matrix Thus the square matrix C is obtained as 1 0 0 0 0 125 0 00143 0 0 1 0 0 00143 0 08381 1565 706 Numerical Methods and the matrix B on the righthand side of eq 1564a is obtained as 0 4571 100 3667 1566 By inverting matrix C in eq 1565 we obtain V 0 3708 100 4438 Thus V 0 V2 3708 V3 10 and V4 4438 Once the values of the potentials at the nodes are known the potential at any point within the mesh can be determined using eq 1551 PRACTICE EXERCISE 156 Calculate the global coefficient matrix for the twoelement mesh shown in Figure 1531 when a node 1 is linked with node 3 and the local numbering j k is as indicated in Figure 1531a b node 2 is linked with node 4 with local numbering as in Figure 1531b Answer a b 09964 005 02464 08 1333 00777 00 1056 005 07 075 00 07777 08192 098 02386 02464 075 15964 06 00 098 204 106 08 00 06 i4 1056 02386 106 1877 N o d e l 2 1 Node 3 2 2 4 Node 2 3 25 Node 4 15 16 Figure 1531 For Practice Exercise 156 155 THE FINITE ELEMENT METHOD 707 EXAMPLE 157 Write a program to solve Laplaces equation using the finite element method Apply the program to the twodimensional problem shown in Figure 1532a Solution The solution region is divided into 25 threenode triangular elements with the total number of nodes being 21 as shown in Figure 1532b This is a necessary step in order to have input data defining the geometry of the problem Based on our discussions in Section 155 a general Matlab program for solving problems involving Laplaces equa tion using threenode triangular elements was developed as in Figure 1533 The devel 10 Figure 1532 For Example 157 a twodimensional electrostatic problem b solution region di vided into 25 triangular elements 100 V 10 a 708 Numerical Methods FINITE ELEMENT SOLUTION OF LAPLACES EQUATION FOR TWODIMENSIONAL PROBLEMS TRIANGULAR ELEMENTS ARE USED ND NO OF NODES NE NO OF ELEMENTS NP NO OF FIXED NODES WHERE POTENTIAL IS PRESCRIBED NDPI NODE NO OF PRESCRIBED POTENTIAL I12NP VALI VALUE OF PRESCRIBED POTENTIAL AT NODE NDPI NLIJ LIST OF NODES FOR EACH ELEMENT I WHERE Jl23 REFERS TO THE LOCAL NODE NUMBER CEIJ ELEMENT COEFFICIENT MATRIX CIJ GLOBAL COEFFICIENT MATRIX BI RIGHTHAND SIDE MATRIX IN THE SYSTEM OF SIMULTANEOUS EQUATIONS SEE EQ 1564 XI YI GLOBAL COORDINATES OF NODE I XLJ YLJ LOCAL COORDINATES OF NODE Jl23 VI POTENTIAL AT NODE I MATRICES PI AND Q I ARE DEFINED IN EQ 1562a FIRST STEP INPUT DATA DEFINING GEOMETRY AND BOUNDARY CONDITIONS clear input Name of input data file SECOND STEP EVALUATE COEFFICIENT MATRIX FOR EACH ELEMENT AND ASSEMBLE GLOBALLY B zerosND1 C zerosNDND for 11NE FIND LOCAL COORDINATES XLJ YLJ FOR ELEMENT I K NLI 1 3 XL XK YL YK Pzeros3 Qzeros3 Pl P2 P3 QD Q2 Q3 AREA 1 1 YL2 YL3 YL1 XL3 XL 1 XL2 05abs YL3 YL1 YL2 XL 2 XL 3 XL1 P2Q3 Q2P3 Figure 1533 Computer program for Example 157 155 THE FINITE ELEMENT METHOD 709 DETERMINE COEFFICIENT MATRIX FOR ELEMENT I CEPPQQ40AREA ASSEMBLE GLOBALLY FIND CIJ AND BI for Jl3 IR NLIJ IFLAG1O CHECK IF ROW CORRESPONDS TO A FIXED NODE for K 1NP if IR NDPK CIRIR 10 BIR VALK IFLAG11 end end end for K 1NP ifIFLAGl 0 for L 13 IC NLIL IFLAG20 CHECK IF COLUMN CORRESPONDS TO A FIXED NODE for K1NP if IC NDPK BIR BIR CEJLVALK IFLAG21 end end end for K1NP ifIFLAG2 0 CIRIC CIRIC CEJL end end end for Ll3 end end ififiagl 0 end end for Jl3 end end for 11NE THIRD STEP SOLVE THE SYSTEM OF EQUATIONS V invCB VV FOURTH STEP OUTPUT THE RESULTS 6 diary examl47out ND NE NP 1ND X Y V diary off Figure 1533 Continued I 710 Numerical Methods opment of the program basically involves four steps indicated in the program and ex plained as follows Step 1 This involves inputting the necessary data defining the problem This is the only step that depends on the geometry of the problem at hand Through a data file we input the number of elements the number of nodes the number of fixed nodes the prescribed values of the potentials at the free nodes the x and y coordinates of all nodes and a list identifying the nodes belonging to each element in the order of the local numbering 123 For the problem in Figure 1532 the three sets of data for coordinates elementnode relationship and prescribed potentials at fixed nodes are shown in Tables 153 154 and 155 respec tively TABLE of the 153 Finite of Figure 15 Node 1 2 3 4 5 6 7 8 9 10 11 X 00 02 04 06 08 10 00 02 04 06 08 Nodal Coordinates Element Mesh 32 Y 00 00 00 00 00 00 02 02 02 02 02 Node 12 13 14 15 16 17 18 19 20 21 X 00 02 04 06 00 02 04 00 02 00 y 04 04 04 04 06 06 06 08 08 10 TABLE 154 ElementNode Identification Local Node No Local Node No Element No 1 2 3 4 5 6 7 8 9 10 11 12 13 1 1 2 2 3 3 4 4 5 5 7 8 8 9 2 2 8 3 9 4 10 5 11 6 8 13 9 14 3 7 7 8 8 9 9 10 10 11 12 12 13 13 Element No 14 15 16 17 18 19 20 21 22 23 24 25 1 9 10 10 12 13 13 14 14 16 17 17 19 2 10 15 11 13 17 14 18 15 17 20 18 20 3 14 14 15 16 16 17 17 18 19 19 20 21 155 THE FINITE ELEMENT METHOD 711 TABLE 155 Prescribed at Fixed Nodes Node No 1 2 3 4 5 6 11 15 Prescribed Potential 00 00 00 00 00 500 1000 1000 Potentials Node No 18 20 21 19 16 12 7 Prescribed Potential 1000 1000 500 00 00 00 00 TABLE 156 Input Data for the Finite Element Program in Figure 1533 NE ND NP NL 25 21 15 2 8 2 3 8 X Y NDP VAL 1 7 8 3 3 4 4 5 5 7 2 9 4 7 8 9 10 9 5 10 11 10 6 8 13 8 9 9 10 10 12 13 13 14 14 16 17 17 19 9 11 12 12 13 14 13 10 14 0 0 0 0 0 0 15 11 13 17 14 18 15 17 20 18 20 0 2 6 0 2 4 i o 5 5 14 15 16 16 17 17 18 19 19 20 21 0 0 0 0 0 0 2 0 0 0 0 0 2 4 0 0 2 6 3 0 0 0 0 0 0 4 5 0 0 100 0 0 4 6 2 0 2 6 6 0 0 0 0 0 0 0 0 11 0 6 8 4 0 2 6 15 0 100 0 0 0 0 0 0 0 0 0 0 8 0 0 0 4 8 18 0 0 1 0 0 0 0 0 20 0 100 0 0 2 2 0 4 8 21 0 0 0 0 0 0 0 1 IS 0 4 0 2 4 0 16 12 7 1000 0 712 Numerical Methods Step 2 This step entails finding the element coefficient matrix C for each element and the global coefficient matrix C The procedure explained in the previous example is applied Equation 1564 can be written in general form as 1 0 0 C ff Vp or C V B Both global matrix C and matrix B are calculated at this stage Step 3 The global matrix obtained in the previous step is inverted The values of the po tentials at all nodes are obtained by matrix multiplication as in eq 1576b Instead of in verting the global matrix it is also possible to solve for the potentials at the nodes using Gaussian elimination technique Step 4 This involves outputting the result of the computation The input and output data are presented in Tables 156 and 157 respectively TABLEi57 the Program Node 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 X 000 020 040 060 080 100 000 020 040 060 080 000 020 040 060 000 020 040 000 020 000 Output 1Data of in Figure 1533 Y 000 000 000 000 000 000 020 020 020 020 020 040 040 040 040 060 060 060 080 080 100 Potential 0000 0000 0000 0000 0000 50000 0000 18182 36364 59091 100000 0000 36364 68182 100000 0000 59091 100000 0000 100000 50000 y 25 SUMMARY 713 22 23 24 25 26 Figure 1534 For Practice Exercise 157 PRACTICE EXERCISE 157 Rework Example 153 using the finite element method Divide the solution region into triangular elements as shown in Figure 1534 Compare the solution with that obtained in Example 153 using the finite difference method Answer See Example 153 SUMMARY 1 Electric field lines and equipotential lines due to coplanar point sources can be plotted using the numerical technique presented in this chapter The basic concept can be ex tended to plotting magnetic field lines 2 An EM problem in the form of a partial differential equation can be solved using the finite difference method The finite difference equation that approximates the differen tial equation is applied at grid points spaced in an ordered manner over the whole solu tion region The field quantity at the free points is determined using a suitable method 3 An EM problem in the form of an integral equation is conveniently solved using the moment method The unknown quantity under the integral sign is determined by match ing both sides of the integral equation at a finite number of points in the domain of the quantity 4 While the finite difference method is restricted to problems with regularly shaped solu tion regions the finite element method can handle problems with complex geometries This method involves dividing the solution region into finite elements deriving equa tions for a typical element assembling all elements in the region and solving the re sulting system of equations 714 Numerical Methods Typical examples on how to apply each method to some practical problems have been shown Computer programs for solving the problems are provided wherever needed 151 At the point 1 2 0 in an electric field due to coplanar point charges E 03 ax 04 ay Vm A differential displacement of 005 m on an equipotential line at that point will lead to point a 1042030 b 096 1970 c 1041970 d 096 203 0 152 Which of the following is not a correct finite difference approximation to dVdx at xo if h Ax a b c d e Vxo h Vxo Vxo Vxo Vxo Vxo h Vxo h f h h f h 2h f h2 h Vxo Vx0 Vxo h h h2 153 The triangular element of Figure 1535 is in free space The approximate value of the po tential at the center of the triangle is a 10 V b 75 V c 5 V d 0 V 154 For finite difference analysis a rectangular plate measuring 10 by 20 cm is divided into eight subregions by lines 5 cm apart parallel to the edges of the plates How many free nodes are there if the edges are connected to some source a 15 b 12 c 9 REVIEW QUESTIONS 715 23 Figure 1535 For Review Questions 153 and 1510 20 V 30 d 6 e 3 155 Using the difference equation Vn Vnl Vn1 with Vo Vs 1 and starting with initial values Vn 0 for 1 4 the value of V2 after the third iteration is a 1 b 3 c 9 d 15 e 25 156 The coefficient matrix A obtained in the moment method does not have one of these properties a It is dense ie has many nonzero terms b It is banded c It is square and symmetric d It depends on the geometry of the given problem 157 A major difference between the finite difference and the finite element methods is that a Using one a sparse matrix results in the solution b In one the solution is known at all points in the domain c One applies to solving partial differential equation d One is limited to timeinvariant problems 158 If the plate of Review Question 144 is to be discretized for finite element analysis such that we have the same number of grid points how many triangular elements are there a 32 b 16 c 12 d 9 716 Numerical Methods PROBLEMS 159 Which of these statements is not true about shape functions a They are interpolatory in nature b They must be continuous across the elements c Their sum is identically equal to unity at every point within the element d The shape function associated with a given node vanishes at any other node e The shape function associated with a node is zero at that node 1510 The area of the element in Figure 1535 is a 14 b 8 c 7 d 4 Answers 151a 152c4 153a 154e 155c 156b 157a 158b 159e 1510d 151 Using the program developed in Example 151 or your own equivalent code plot the electric field lines and equipotential lines for the following cases a Three point charges 1 2 and 1 C placed at 1 0 02 and 10 respec tively b Five identical point charges 1 C located at 1 1 1 1 1 1 1 1 and 0 0 respectively 152 Given the onedimensional differential equation dx2 0 x subject to y0 0 yl 10 use the finite difference iterative method to find y025 You may take A 025 and perform 5 iterations dV d2V 153 a From the table below obtain and dx dx2 atx 015 X V 01 10017 015 15056 02 20134 025 25261 03 30452 b The data in the table above are obtained from V 10 sinh x Compare your result in part a with the exact values 4The formula in a is known as a forwarddifference formula that in b as a backwarddifference formula and that in d or e as a centraldifference formula PROBLEMS 717 m 0 Figure 1536 Finite difference grid in cylindri cal coordinates for Problem 155 o 154 Show that the finite difference equation for Laplaces equation in cylindrical coordi nates V Vp z is Vp0 Zo I Vpo Z Vpo A where h Az Ap V P o fc Z o 155 Using the finite difference representation in cylindrical coordinates p 4 at a grid point P shown in Figure 1536 let p m Ap and j n Af so that Vp 4p VmAp nAt Vm Show that 1 w Atf Vl 2 V V 156 A square conducting trough has its four sides held at potentials 10 0 30 and 60 V Determine the potential at the center of the trough 157 Use FDM to calculate the potentials at nodes 1 and 2 in the potential system shown in Figure 1537 30 V Figure 1537 For Problem 157 20 V 1 2 20 V 0V I 718 Numerical Methods 100 V Figure 1538 For Problem 159 1 3 2 4 OV 100 158 Rework Problem 157 if ps nCm h 0lm and e eo where h is the mesh size 159 Consider the potential system shown in Figure 1538 a Set the initial values at the free nodes equal to zero and calculate the potential at the free nodes for five iterations b Solve the problem by the band matrix method and compare result with part a 1510 Apply the band matrix technique to set up a system of simultaneous difference equations for each of the problems in Figure 1539 Obtain matrices A and B 1511 a How would you modify matrices A and B of Example 153 if the solution region had charge density ps b Write a program to solve for the potentials at the grid points shown in Figure 1540 assuming a charge density ps xy 1 nCm Use the iterative finite difference method and take er 10 oovi v a d V 100 V b e OV c f 1 4 7 2 5 8 3 6 A 1 5 V a b Figure 1539 For Problem 1510 10V 20 a d g V b e h c f i ov PROBLEMS Figure 1540 For Problem 1511 719 10 1512 The twodimensional wave equation is given by c2 dt2 By letting m n denote the finite difference approximation of Pxm z tj show that the finite difference scheme for the wave equation is 2 a Pj 2 P J 2 P J where Ax Az and a cAth2 1513 Write a program that uses the finite difference scheme to solve the onedimensional wave equation d2V d2V dxz 0 x 1 0 given boundary conditions V0 i 0 Vl t 0 t 0 and the initial condition dVdt x 0 0 Vx 0 sin 7TJC 0 JC 1 Take Ax At 01 Compare your solution with the exact solution Vx t sin 7rx COS TT for 0 t 4 1514 a Show that the finite difference representation of Laplaces equation using the nine node molecule of Figure 1541 is Vo 18 V V2 V3 V4 V5 V6 V1 Vs b Using this scheme rework Example 154 1515 A transmission line consists of two identical wires of radius a separated by distance d as shown in Figure 1542 Maintain one wire at 1 V and the other at 1 V and use MOM to find the capacitance per unit length Compare your result with exact formula for C in Table 111 Take a 5 mm d cm 5 m and e eo 1516 Determine the potential and electric field at point 14 5 due to the thin conducting wire of Figure 1519 Take Vo 1 V L 1 m a 1 mm L 720 Numerical Methods 4 3 0 2 Figure 1541 Ninenode molecule of Problem 1514 1517 Two conducting wires of equal length L and radius a are separated by a small gap and in clined at an angle 6 as shown in Figure 1543 Find the capacitance between the wires using the method of moments for cases 6 10 20 180 Take the gap as 2 mm a 1 mm L 2 m er 1 1518 Given an infinitely long thin strip transmission line shown in Figure 1544a we want to determine the characteristic impedance of the line using the moment method We divide each strip into N subareas as in Figure 1544b so that on subarea i V where Ry is the distance between the rth and jth subareas and V 1 or 1 depending on whether the rth subarea is on strip 1 or 2 respectively Write a program to find the char acteristic impedance of the line using the fact that Zn C Figure 1542 For Problem 1515 PROBLEMS 721 i For Problem 1517 gap 2a where C is the capacitance per unit length and and Vd 2 V is the potential difference between strips Take H 2 m W 5 m and iV 20 1519 Consider the coaxial line of arbitrary cross section shown in Figure 1545a Using the moment method to find the capacitance C per length involves dividing each conductor into N strips so that the potential on theyth strip is given by 2N where A In A I At z In 15 2xe r ij J a I igure 1544 Analysis of strip transmission line using moment method for Problem 1518 722 Numerical Methods AS b Figure 1545 For Problem 1519 coaxial line of a arbitrary cross section and b elliptical cylindrical cross section and Vj 1 or 1 depending on whether A lies on the inner or outer conductor respec tively Write a Matlab program to determine the total charge per length on a coaxial cable of elliptical cylindrical cross section shown in Figure 1545b using and the capacitance per unit length using C Q2 a As a way of checking your program take A B 2 cm and a b 1 cm coaxial line with circular cross section and compare your result with the exact value of C 2ire lnAa b Take A 2 cm B 4 cm a 1 cm and b 2 cm Hint For the inner ellipse of Figure 1545b for example r V sin2 v2cos2 where v alb di r dj Take ro 1 cm 1520 The conducting bar of rectangular cross section is shown in Figure 1546 By dividing the bar into Nequal segments we obtain the potential at theyth segment as where 4ireoRy 2eo j PROBLEMS 723 Figure 1546 For Problem 1520 and A is the length of the segment If we maintain the bar at 10 V we obtain 107 where 1 1 1 if and q pvthA a Write a program to find the charge distribution pv on the bar and take 2 m h 2 cm t 1 cm and N 20 b Compute the capacitance of the isolated conductor using C QV q2 qN10 1521 Another way of defining the shape functions at an arbitrary point x y in a finite element is using the areas AXA2 and A3 shown in Figure 1547 Show that a 4 4 1 2 3 A where A A A2 A3 is the total area of the triangular element 1522 For each of the triangular elements of Figure 1548 a Calculate the shape functions b Determine the coefficient matrix 1523 The nodal potential values for the triangular element of Figure 1549 are VY 100 V V2 50 V and V3 30 V a Determine where the 80 V equipotential line intersects the boundaries of the element b Calculate the potential of 2 1 Figure 1547 For Problem 1521 724 Numerical Methods 051 00 a Figure 1548 Triangular elements of Problem 1522 12 252 2 04 b 1524 The triangular element shown in Figure 1550 is part of a finite element mesh If Vx 8 V V2 12 V and V3 10 V find the potential at a 12 and b the center of the element 1525 Determine the global coefficient matrix for the twoelement region shown in Figure 1551 1526 Find the global coefficient matrix of the twoelement mesh of Figure 1552 1527 For the twoelement mesh of Figure 1552 let Vx 10 V and V3 30 V Find V2 and V4 j j 2 22 3 1 Figure 15W For Problem 1523 00 14 Fimire 1550 For Problem 1524 2 1 PROBLEMS Fijiure 1551 For Problem 1525 725 00 80 01 22 Figure 1552 For Problem 1526 and 1527 30 1528 The mesh in Figure 1553 is part of a large mesh The shading region is conducting and has no elements Find C5S and C51 1529 Use the program in Figure 1533 to solve Laplaces equation in the problem shown in Figure 1554 where Vo 100 V Compare the finite element solution to the exact solu tion in Example 65 that is i 4 V ST sin WTT x sinh nwy Vx y V 7T o n sinh nit n 2k 1 4 cm 7 Finre 1553 For Problem 1528 L 726 Numerical Methods 3f 25 19 13 32 33 34 35 36 dr V 30 24 18 12 5 6 x10 Figure 1554 For Problem 1529 1530 Repeat the preceding problem for Vo 100 sin irx Compare the finite element solution with the theoretical solution similar to Example 66a that is Vx y 100 sin it x sinh ir y sinh 7T 1531 Show that when a square mesh is used in FDM we obtain the same result in FEM when the squares are cut into triangles INDEX Acceptance angle 653 Ac resistance 427 Amperes law 262 273 290 applications of 274278 Amperian path 274 Amplitude 412 Angle of incidence 451 Angular frequency 412 Antenna pattern See Radiation pattern Antenna arrays 612618 binomial type 621 broadside type 615 endfire type 615 Antennas 588618 types of 589 Array factor 613 Attenuation 649 654 Attenuation constant 421 479 Azimuthal angle 30 Baccab rule 16 Band matrix method 672 702 Bandwidth 638 649 Bessel differential equation 223 BiotSavarts law 262266 290 307 Bonding 647 Bonding effectiveness 648 Bounce diagram 514 Boundary conditions 182187 330332 385 Boundaryvalue problems 199 Brewster angle 455 Capacitance 224230 Capacitor 224230 Cartesian coordinates 29 53 Characteristic admittance 480 Characteristic impedance 479 525 Charge conservation 180 Charged ring 118 Charged sphere 128 Circular cylindrical coordinates 29 55 Circulation 60 Closed form solution 660 Coaxial capacitor 227 Coaxial line 276 Colatitude 33 Complex permittivity 422 Complex variables 728729 Components of a vector 6 Conductivity 162 164 values of 737 Conductors 161 164167 Conservation of magnetic flux 283 Conservative field 87 Constantcoordinates surfaces 4114 Constitutive relations 385 Continuity equation 180 385 Coulombs law 104107 305 Critical angle 653 Cross product 13 Curie temperature 328 Curl 7580 definition of 76 properties of 78 Current 162164 conduction type 164 convection type 163 definition of 162 displacement type 382 Current density definition of 163 Current reflection coefficient 487 Cutoff 549 Cutoff frequency 542 550 Cutoff wavelength 550 Dc resistance 427 647 Definite integrals 734 Degenerate modes 576 Del operator 63 Derivatives 731 Diamagnetism 327 Dielectric breakdown 175 Dielectric constant 175 values of 738 Dielectric strength 175 Dielectrics 161 Difference equations 669 Differential displacement 53 55 56 89 Differential normal area 54 55 57 89 Differential solid angle 606 Differential volume 54 55 57 89 Dipole antenna 589 Dipole moment 143 Directional derivative 67 Directive gain 606 Directivity 606 Dispersion 654 Displacement current density 381 Distance vector 8 Distortionless line 481 Divergence 6973 definition of 69 properties of 72 Divergence theorem 72 125 Dominant mode 554 578 Dot product 12 Effective area 621 Effective relative permittivity 525 Electric dipole 142 Electric displacement 123 Electric field intensity 106 Electric flux 123 Electric flux density 122 123 Electric flux lines 144 Electric susceptibility 174 Electrical length 486 Electrohydrodynamic pump 203 Electromagnetic compatibility EMC 644 Electromagnetic interference EMI 644 Electromagnetics EM 3 Electrometer 179 Electromotive force emf 370 Electrostatic field 103 592 Electrostatic shielding 186 Electrostatics 103 Element coefficient matrix 698 763 764 Index Emf See Electromotive force Energy 146 339341 Equipotential line 144 Equipotential surface 144 Evanescent mode 549 Exponential identities 730 External inductance 338 Far field 592 Faradays law 370 Ferromagnetism 328 Field classification of 8688 definition of 5 harmonic type 84 timeharmonic type 389 uniform type 8 Field pattern 604 Field plotting 661663 Filtering 649 Finite difference method 669674 Finite element method 694703 Finite elements 694 Fivenode molecule 672 Fixed node 670 Force 104 304308 349 Flux 60 Flux linkage 336 Free node 670 Frequency 412 Fresnels equations 455 457 Friss transmission formula 623 Gausss law 124 125 283 applications of 126130 224228 Gaussian surface 126 Global coefficient matrix 699 Gradient 6567 Gradient operator 63 Grounding 647 Group pattern 613 Group velocity 563 Halfwave dipole antenna 594 Helmholtzs equation 419 Helmholtzs theorem 88 Hertzian dipole 590 Homogeneous material 175 Hyperbolic functions 729 Hysteresis 329 Indefinite integrals 732734 Inductance 336 Inductive field 592 Inductor 336 Infinite line charge 114 127 Infinite line current 274 Infinite sheet of charge 115 128 Infinite sheet of current 275 Input impedance 484486 Insertion loss 649 Insulators 161162 See also Dielectrics Integral equation 683 Internal inductance 338 Intrinsic impedance 420 of free space 424 Irrotational field 87 Isolated sphere 228 Isotropic antenna 606 Isotropic material 175 Iteration method 672 702 Joules law 167 Kernel 683 Kirchhoffs current law 180 348478 Kirchhoff s voltage law 477 Laplaces equation 84 199202 285 671703 Laplacian 8385 Lattice diagram See Bounce diagram Lenzslaw 371 374 Line charge 112 242 Line integral 60 Linear material 175 Logarithmic identities 730 Lorentz condition 388 Lorentz force 305 314 Lorentz force equation 305 384 Loss angle 422 Loss tangent 422 Lossless line 480 Lossy dielectric 418 Lossy line 480 Magnetic dipole 318 Magnetic dipole moment 317 Magnetic field intensity 263 281 Magnetic flux density 281 Magnetic potentials 284287 Magnetic susceptibility 326 Magnetization 324 Magnetization 329 Magnetization volume current density 325 Magnetization surface current density 325 Magnetomotive force 347 Magnetostatic field 261 Magnitude 5 Matched line 489 Maxwells equations 4 125 139 182 273 283 369389 418438 451 543 545 Medium velocity 563 Mesh size 671 Method of images 240 Microstrip lines 524526 Microwave components 639 Microwaves 638 Mode 546 548 Moment method 683687 Monopole 143 Motional emf 373 Mutual inductance 337 Newtons laws 308 Noise susceptibility 649 Nonmagnetic materials 327 Normalized impedance 493 Numerical aperture 652 Ohms law 164 166 181348 Ohmic resistance 608 Opencircuited line 489 Optical fiber 649 definition of 650 Orthogonal system 28 Parallelplate capacitor 225 226 Paramagnetism 327 Pattern multiplication 613 Penetration depth See Skin depth Period 412 Permeability 326 of free space 281 Permeance 348 Permittivity 175 of free space 104 Phase 412 Phase constant 412421479 Phase matching conditions 452 Phase velocity 563 Phasor 389 Plane of incidence 451 Point charge 104 126241 Poissons equation 199202 291 670 671683 Polarization 171174425 Polarization surface charge density 173 Polarization volume charge density 173 Position vector 7 106 135 451 Potential 134 Potential difference 133 Power 167435138 Power gain 607 608 Power pattern 604 Poyntings theorem 436 437 Poynting vector 436 Pressure 350 Propagation constant 419 479 Propagation vector 451 Quality factor 578 579 Quarterwave monopole antenna 598 Quarterwave transformer 505 Radar 641625 types of 627 Radar cross section 626 Radar range equation 627 Radar transmission equation 627 Radiated emissions 644 Radiation 588 Radiation efficiency 608 Radiation field 592 Radiation intensity 605 Radiation pattern 604 Radiation resistance 593 Reactance circle 495 Reflection coefficient 442 642 643 Refraction law 185332 Refractive index 453 Relative permeability 326 values of 739 Relative permittivity 175 See also Dielectric constant Relaxation time 181 229 Reluctance 348 Resistance 166 223 Resistance circle 494 Resistivity 167 Resonant frequency 577 Resultant pattern 613 Retarded potentials 389 Righthand rule 14 80 263 372 Righthand screw rule 80 263 Satellite 639 Scalar 5 Scalar component 16 Scalar product 1215 Scattering cross section 626 Scattering parameters 641 Selfinductance 336 Semiconductor 162 Separation constant 212 221 Separation of variables 212 221 Shape functions 696 Shielding effectiveness 648 Shorted line 489 Singlestub tuner 506 Skin depth 426 Skin effect 427 Skin resistance 428 Slotted line 507 Small loop antenna 599 Smith chart 492498 Snells law 453 Solenoid 271 Solenoidal field 87 Spectrum 415 Spherical capacitor 227 Spherical coordinates 3256 Standing wave 442 Standing wave ratio SWR 444 Stokess theorem 79 Superconductors 162 Superposition 106 135 216 Surface charge 114 Surface integral 60 Tensor 176 Timeharmonic field 389 Torque 316 Total reflection 653 Transformation of point 34 of vector 35 Transformer emf 373 Transient 512 Transmission coefficient 442 Transmission line equaoccs 1 Transmission line parameters I Transmission lines 473526 Transverse electric TE mode 54 552556 Transverse electromagnetic TEM wave 425 546 Transverse magnetic TM mode 547551 Trigonometric identities 727728 Uniform plane wave 425 Unit vector 5 6 Uniqueness theorem 201202 Vector addition of 6 definition of 5 multiplication of 11 subtraction of 6 Vector component 16 Vector identities 735 Vector product 1315 Voltage reflection coefficient 486 487 Volume charge 115 Volume integral 62 Wave 410 definition of 411 Wave equation 388 411 419 479 Wave number 412 Wave velocity 411 Waveguide resonator 575 Waveguide wavelength 563 Wavelength 412 Work done 133 Xerographic copying machine 204 Appendix A MATHEMATICAL FORMULAS A1 TRIGONOMETRIC IDENTITIES tan A sec A sin A cos A 1 cos A cot A 1 esc A tan A 1 sin A sin2 A cos2 A 1 1 tan2 A sec2 A 1 cot2 A esc2 A sin A B sin A cos B cos A sin B cos A B cos A cos B sin A sin B 2 sin A sin B cos A B cos A B 2 sin A cos B sin A B sin A B 2 cos A cos B cos A B cos A B sin A sin B 2 sin B A B cos A B A B sin A sin B 2 cos sin A B cos A cos B 2 cos A B cos A n A B A B cos A cos B 2 sin sin cos A 90 sinA sin A 90 cos A tan A 90 cot A cos A 180 cos A sin A 180 sin A 727 728 Appendix A tan A 180 tan A sin 2A 2 sin A cos A cos 2A cos2 A sin2 A 2 cos2 A 1 1 2 sin2 A tan A B tan A B tan 2A 1 tan A tan B 2 tan A 1 tan2 A sin A ejA eiA cos A 2 2 ejA cos A y sin A Eulers identity TT 31416 1 rad 57296 2 COMPUX VARIABLES A complex number may be represented as z x jy rl reje r cos 0 j sin where x Re z r cos 0 y Im z r sin 0 7 l T y The complex conjugate of z z x jy r 0 re je r cos 0 j sin 0 ej9 ejn6 cos 0 j sin 0 de Moivres theorem If Z x jyx and z2 2 i1 then z z2 only if x1 JC2 and j y2 Zi Z2 xi x2 jyi y2 or nr2o APPENDIX A 729 i j y or Z2 Vz VxTjy Trem Vr fl2 2n x y r enfl rn nd n integer z X yj r 1n e r Vn din 27rfcn t 0 1 2 n In re In r In e7 In r jO jlkir k integer A3 HYPERBOLIC FUNCTIONS sinhx tanh x u ex ex 2 sinh x cosh x 1 coshx COttlJt ex 1 sechx tanhx 1 coshx cosjx coshx coshyx cosx sinhx sinyx j sinhx sinhyx j sinx sinh x y sinh x cosh y cosh x sinh y cosh x y cosh x cosh y sinh x sinh y sinh x jy sinh x cos y j cosh x sin y cosh x jy cosh x cos y j sinh x sin y sinh 2x sin 2y tanh x jy cosh 2x cos 2y cosh 2x cos 2y cosh2 x sinh2 x 1 sech2 x tanh2 x 1 sin x yy sin x cosh y j cos x sinh y cos x yy cos x cosh y j sin x sinh y L 730 Appendix A A4 LOGARITHMIC IDENTITIES If log xy log x log y X log log x log y log x n log x log10 x log x common logarithm loge x In x natural logarithm llnl x x A5 EXPONENTIAL IDENTITIES ex where e 27182 X f e e In x2 2 4 V 1 x3 3 exy X x4 4 A6 APPROXIMATIONS FOR SMALL QUANTITIES If x Z 1 1 xn 1 ra 1 x In 1 x x sinx x or hm sinx 1 0 X COS 1 tanx x APPENDIX A K 731 A7 DERIVATIVES If U Ux V Vx and a constant dx dx dx dx dx dU U dx dx V2 aUn naUni dx dx U dx d In U 1 dU dx U dx d v t dU a d In a dx dx dx dx dx dx sin U cos U dx dx d dU cos U sin U dx dx d dU tan U sec dx dx d dU sinh U cosh dx dx cosh t sinh dx dx d dU tanh sech2t ix dx dx 732 Appendix A A8 INDEFINITE INTEGRALS lfU Ux V Vx and a constant a dx ax C UdVUV VdU integration by parts Unl Un dU C n 1 n 1 dU U In U C au dU C a 0 a In a eudU eu C eaxdx eax C a xeax dx rax 1 C x eaxdx a 2x2 lax 2 C a In x dx x In x x C sin ax cfcc cos ax C a cos ax ax sin ax C tan ax etc In sec ax C In cos ax C a a sec ax ax In sec ax tan ax C a APPENDIX A 733 2 x sin 2ax sin axdx 1 C 2 4a x sin 2ax 2 x cos ax dx I 22 4a C sin ax dx sin ax ax cos ax C x cos ax dx x cos ax ax sin ax C eax sin bx dx r a sin bx b cos to C a ft eajc cos bx dx a cos ftx ft sin x C a b sin a ftx sin a bx 2 2 sin ax sin ox ax TT a b sin ax cos bx dx la b la cos a bx cos a bx cos ax cos bx dx a ft 2a ft sin a ftx sin a ftx 2a ft 2a b C a1 C a2 b2 sinh flitfa cosh ax C a cosh c a sinh ax C a tanh axdx In cosh ax C a ax 1 x r r tan C x 2 a2 a a X X l 2 2 2 2 x a I C x2 dx x r x a tan C x2 a 734 Appendix A dx x a x2a2 x2a2 1 a x 2 2 T In h C x a 2a a x dx 2 Vx xdx x2 2 a x sin C In x Vx 2 a2 C a2 C dx xaz x2 a232 C xdx x2 a232 x2 a2 x2dx x2 a2f2 In a2 x a a V a2 C dx xz az 1 x 1 rf i j tan l C la x a a a A9 DEFINITE INTEGRALS o sin mx sin nx dx cos mx cos nx dx m n ir2 m n i w m n even sin mx cos nx dx I o i r m n odd m sin mx sin nx dx sin mx sin nx dx J m F n w m n sin ax ir2 a 0 dx 0 a 0 ir2 a 0 sin 2x APPENDIX A 735 f sin ax x xneaxdx 1 dx w 1 Iv 2 V a aax2bxc x J e M cos bx dx e1 sin bxdx a2 b2 A10 VECTOR IDENTITIES If A and B are vector fields while U and V are scalar fields then V U V VU VV V tV U VV V Vt VVL0 V V n V 1 VV integer V A B A V B B V A A X V X B B X V X A V A X B B V X A A V X B V VA V V A A W V VV V2V V V X A 0 VXA B V X A V X B V X A X B A V B B V A B VA A VB V x VA VV X A VV X A 736 Appendix A V x VV 0 V X V X A VV A V2A A d I V X A d S Vd I VV X dS A dS V A dv K VdS Wdv A X J S Appendix D MATERIAL CONSTANTS TABLE B1 Approximate Conductivity of Some Common Materials at 20C Material Conductors Silver Copper standard annealed Gold Aluminum Tungsten Zinc Brass Iron pure Lead Mercury Carbon Water sea Semiconductors Germanium pure Silicon pure Insulators Water distilled Earth dry Bakelite Paper Glass Porcelain Mica Paraffin Rubber hard Quartz fused Wax Conductivity siemensmeter 61 X 10 58 X 10 41 X 10 35 X 10 18 x 10 17 x 10 11 x 10 10 5 X 106 106 3 X 104 4 22 44 X 104 io4 io5 io io lO 1 2 io2 io5 lO 1 5 io5 io 10 The values vary from one published source to another due to the fact that there are many varieties of most materials and that conductivity is sensitive to temperature moisture content impurities and the like 737 738 Appendix B TABLE B2 Approximate Dielectric Constant or Relative Permittivity er and Strength of Some Common Materials Material Barium titanate Water sea Water distilled Nylon Paper Glass Mica Porcelain Bakelite Quartz fused Rubber hard Wood Polystyrene Polypropylene Paraffin Petroleum oil Air 1 atm Dielectric Constant er Dimensionless 1200 80 81 8 7 510 6 6 5 5 31 2580 255 225 22 21 1 Dielectric Strength RVm 75 x 106 12 X 10 35 x 106 70 X 106 20 X 106 30 X 106 25 X 106 30 X 106 12 X 106 3 X 106 The values given here are only typical they vary from one published source to another due to different varieties of most materials and the dependence of er on temperature humidity and the like APPENDIX B 739 TABLE B3 Relative Permeability of Some Materials Material Diamagnetic Bismuth Mercury Silver Lead Copper Water Hydrogen stp Paramagnetic Oxygen stp Air Aluminum Tungsten Platinum Manganese Ferromagnetic Cobalt Nickel Soft iron Siliconiron Vr 0999833 0999968 09999736 09999831 09999906 09999912 10 0999998 100000037 1000021 100008 10003 1001 250 600 5000 7000 The values given here are only typical they vary from one published source to another due to different varieties of most materials Appendix C ANSWERS TO ODDNUMBERED PROBLEMS CHAPTER 1 11 13 08703aJC03483a03482a a 5a 4a 6s b 5 3 3s 23a c 0439a 011a03293az d 11667a 070843 07084az 17 Proof 19 a 28577 b 02857a 08571a c 6591 111 7236 5966 14391 113 a B AA A AB b A BA X A A 115 2572 117 a 7681 b c 13743C d 11022 e 17309 119 a Proof b cos 0 cos 02 sin i 0 i 04286a AA X B 2a 5a7 sin 02 cos 0i cos 02 sin 0 sin 02 c sin 121 a 103 b 2175ax 1631a 4893a c 0175ax 0631ay 1893a 740 APPENDIX C 741 CHAPTER 2 21 a P05 0866 2 b g0 1 4 c 1837 10612121 d 7346420 23 a pz cos 0 p2 sin 0 cos 0 pz sin 0 b r2l sin2 8 sin2 0 cos 8 25 a 2 4sin0 pap 4az I sin 8 H ar sin 0 cos i V x 2 y2 z 29 Proof 211 a xl yz yz 3 b rsin2 0 cos 0 r cos3 0 sin 0 ar r sin 0 cos 0 cos 0 r cos 0 sin 0 a 3 213 a r sin 0 sin 0 cos 0 r sin 0 cos 0 ar sin 0 r cos2 0 sin 0 cos 0 ag 3 cos 0 a 5a 2121a0 p z a A 4472ap 2236az 215 a An infinite line parallel to the zaxis b Point 2 1 10 c A circle of radius r sin 9 5 ie the intersection of a cone and a sphere d An infinite line parallel to the zaxis e A semiinfinite line parallel to the xy plane f A semicircle of radius 5 in the xy plane 217 a a ay 7az b 14326 c 8789 219 a ae b 0693lae c a e O6931a0 d 06931a 221 a 3a0 25a 156ar lOa0 b 2071ap 1354a0 04141a c 05365ar 01073a9 08371a 223 sin 8 cos3 0 3 cos 9 sin2 0 ar cos 8 cos3 0 2 tan 8 cos 6 sin2 0 sin 6 sin2 0 ae sin 0 cos 0 sin 0 cos 0 a0 742 If Appendix C CHAPTER 3 31 33 35 37 39 311 313 315 317 319 321 323 325 327 329 331 333 335 a b c a b c 2356 05236 4189 6 110 4538 06667 a b 4a a b a b c 50 395 1333az 2 0 62 2a 2 At 05578ax 0 25ap 25a0 a r 0866a Along 2a 2a a b a b 2z y2ax 2zay p2 3z2a0 COt 7 COS p Proof 2xyz z y 2 y Proof a b c d 6yzax 3xy2ay Ayzax 3xy2a3 6xyz 3xy3 2x2 y2 z2 Proof a b c a b c 6xy2 2x2 x 3zcos 4 sin 4 er sin 6 cos 7 6 7 6 Yes 50265 a b c Proof both sides Proof both sides Proof both sides 53 ms 8367ay 3047a 1732az az x 0 4p2az 0 1 c o s t r COS r V sin 6 3x2yzaz f 4x2yzaz x2yz 5y2exz 2446 81961 A L j 08277 equal 1667 equal 13157 equal 13623 6 a 0 APPENDIX C 743 CHAPTER 4 337 a 4TT 2 b 1K 339 0 341 Proof 343 Proof 345 a 1 0 7 1 41 5746a 1642a 4104a mN 43 a 3463 nC b 187 nC 45 a 05 C b 1206 nC c 1579 nC MVm 413 415 417 419 421 423 a Proof b 04 mC 3161aiVm 0591ax018azN Derivation a 884xyax 884x2a pCm2 b 884pCm3 5357 kJ Proof 0 p 1 p 2 28 P 425 1050 J 427 a 1250J b 3750 nJ c 0J d 8750 nJ 429 a 2xax Ayay 8zaz b xax yay zaz cos x2 y2 z2m c 2pz 1 sin 4 ap pz 1 cos j a0 p2 sin t az d er sin 6 cos 20 ar cos 6 cos 20 ae H sin 20 431 433 a b 72ax 30 Proof 95 27a PC 36a Vm 744 Appendix C 435 a b c CHAPTER 5 2po 2p0 I5eor2 n I5eor 1 Psdr 5 J r eoV20 6 2po 1 poa 15eo 60sn 15 d Proof 437 a 1136 akVm b a 02a X 107 ms 439 Proof 441 2 sin 0 sin 0 ar cos 0 sin t ae cos 0 a Vm 443 6612 nJ 51 6283 A 53 5026 A 55 a 16ryz eo b 1131 mA 57 a 35 X 107 Sm aluminum b 566 X 106Am2 59 a 027 mil b 503 A copper 97 A steel c 0322 mfi 511 1000182 513 a 1273zaznCm2 1273 nCm3 b 7427zaz nCm2 7472 nCm3 515 a 4rr2 b 0 o e 1 Q 4Kb2 517 2472a 3295ay 9886a Vm 519 a Proof b 521 a 0442a 0442ay 01768aznCm2 b 02653a 05305ay 07958a 523 a 4623 A b 4598 uCm3 525 a 182 b 2058 c 1923 APPENDIX C 745 527 a 1061a 1768a 1547az nCm2 b 07958a 1326a 1161aznCm2 c 3979 529 a 3878ap 4524a 6786azVm 12a 14a0 21aznCm2 b 4a 2a 3az nCm2 0 c 1262 mJm3 for region 1 and 9839 mJm3 for region 2 531 a 7059 Vm 0 glass 6000 Vm 0 air b 19405 Vm 846 glass 24786 Vm 512 air 533 a 38197 nCm2 0955a 2 b 5nCm r c 1296 pi CHAPTER 6 61 120a 1203 12az 5 3 0 5 21 PvX3 PoX2 fV0 pod py PaX Vo pod pod s0V0 s0V0 pod b 3 d d 6 65 15708 9425y2 30374 kV 67 Proof 69 Proof 611 25z kV 25az kVm 332az nCm2 332az nCm2 613 952 V 18205ap Vm 0161a nCm2 615 117 V 1786aeVm 617 Derivation 619 621 623 625 a HA 4V L x x Proof Proof Proof 00 V CO n odd Ddd sin i sin 1 m I niry a 1 b n sinh nira b sinh nirx a n sinh nwb a niry b n h n 7 r n b sinh 746 Appendix C 627 629 631 633 635 637 639 641 643 645 647 649 05655 cm2 Proof a 100 V b 995 nCm2 99 a 25 pF b 63662 nCm2 4x 1 1 1 1 c d be 2185 pF 6931 s Proof Proof 07078 mF a lnC b 525 nN 01891 a av 5 nCm2 1 1 a b a7N 651 a 13824a x 18432a Vm b 1018 nCm2 CHAPTER 7 71 b 02753ax 0382ay H 73 09549azAm 75 a 2847 ay mAm b 13a 13a mAm c 51a 17ay mAn d 51ax 17a mAm 77 a 06792az Am b 01989azmAm c 01989ax 79 a 1964azAm b 178azAm c d 711 a b c 713 a 01404a7Am 01989a Am 01178a Am 03457a 03165ay 01798azAm Proof 178 Am 1125 Am Proof 136a7Am b 0884azAm 715 a 6963 Am b 3677 Am APPENDIX C 747 717 b 0 p2a2 2KP b2 a2 I p a a pb pb 719 721 723 725 727 729 731 a 2a Am2 b Proof both sides equal 30 A a 8Oa0nWbm2 b 1756i Wb a 31433 Am b 1279ax 63663 Am 137 nWb a magnetic field b magnetic field c magnetic field 14a 42a0 X 104 Am 1011 Wb IoP a 2ra2 733 735 737 739 Am2 28x a 50 A b 250 A Proof 8Xo CHAPTER 8 81 44ax 13a 114a kVm 83 a 2 1933 3156 b 1177 J 85 a Proof 87 864azpN 89 1559 mJ 811 1949axmNm 813 2133a 02667ay Wbm2 815 a 1852azmWbm2 b 4amWbm2 c Ilia 786amWbm2 I 748 Appendix C 817 819 821 823 825 827 829 831 833 835 837 839 841 843 a b c d 55 8168ax 204 220az Am 95 mJm2 47668 kAm 2 a a b 26 a b c a b 11 25ap 15a0 6665 Jm3 57 833 30ay 5a Am 6 35ay Am 2ay 3267az jtWbm2 50az mWbm2 7 Jm3 3396a Am 283a jtWbm2 110ayWbm2 5ay Am 6283ay iWbm2 1674 6181 kJm3 58 mm 5103 turns Proof 1908 A t 19080 88 a b 5 mWbm2 666 mN 1885 mN Proof Am CHAPTER 9 91 04738 sin 377 93 54 V 95 a 04 V b 2 2 97 9888 JUV point A is at higher potential 99 097 mV 911 6A counterclockwise 913 2778 Am2 7778 A 915 36 GHz 917 a V Es pje VH s 0 V x E 5 BDX dDy BDZ b ox dy oz dBx dBv dBz dx dy dz dz dEy dBx dy dz dt V X H a Pv 0 APPENDIX C 749 dEx dz dEy dx dHz dy dHx dz dHy dx 919 Proof dEz dx dEx dy dHy dz dH1 dx dHx dy j Jx Jy Jz dB dt dBk dt I BDX dt dDy dt dDz dt 921 03z 2sinl0 4rmCm 3 923 0833 radm 1005 sin j3x sin at ay Vm 925 a Yes b Yes c No d No 927 3 cos j cos 4 X 106ra Am2 8482 cos j sin 4 X 106faz kVm l03p r t 7 4TT 929 2 pl tepaz Wbm2 931 a 6392424 b 0227220214 c 13871768 d 0034968 933 a 5 cos at Bx 3637a3 20 b cos at 2zap 2236 c j cos at j 6343 sin 0 a0 935 Proof CHAPTER 10 101 a along ax b 1 us 1047 m 1047 X 106 ms c see Figure C 1 103 a 54105 y6129m b 1025 m c 5125 X 107ms d 101414144 0 e 59A6eJ4h44 e I 750 Appendix C 25 I Figured For Problem 101 25 25 25 25 2 5 I t 778 2 t 774 t Til 105 107 109 1011 a 1732 b 1234 c 1091 jl89 X 10nFin d 00164 Npm a 5 X 105 ms b 5 m c 0796 m d 140545 U a 005 j2 m b 3142 m c 108ms d 20 m a along xdirection b 7162 X 1010Fm c 1074 sin 2 X 108 6xazVm APPENDIX C B 751 1013 a lossless b 1283 radm 049 m c 2566 rad d 4617 11 1015 Proof 1017 576 02546 sin109r 8xay 03183 cos 109r 8xa Am 1019 a No b No c Yes 1021 2183 m 3927 X 107 ms 1023 01203 mm 0126 n 1025 294 X 106m 1027 a 1316 a b 01184 cos2 2ir X 108r 6xaxWm2 c 03535 W 1029 a 2828 X 108 rads 0 225 sin cor 2za Am 9 b sin2 cor 2zaz Wm2 P c 1146 W 1031 a2 b 10 cos cor zax Vm 2653 cos cor zay mAm 1033 26038 X 106 Hm 1035 a 05 X 108 radm b 2 c 2653 cos 05 X 108r zax mAm d 1061a Wm2 1037 a 6283 m 3 X 108 rads 732 cos cor zay Vm b 00265 cos cor zax Am c 02680732 d Et 10 cos cor zay 268 cos ut zay Vm E2 732 cos cor zay Vm Pave 01231a Wm2 P2me 01231a Wm2 1039 See Figure C2 1041 Proof Hs ky sin k sin kyyax kx cos jfc cos kyyay C0Xo 1043 a 3687 b 79583 1061a mWm2 c 1518ay 2024a sin cor Ay 3z Vm 1877a 5968av sin cor 9539y 3z Vm 1045 a 15 X 108 rads b 8a 6a 5az sin 15 X 108r 3x Ay Vm 752 Appendix C i 0 Figure C2 For Problem 1039 curve n corre sponds to n778 n 0 1 2 A4 CHAPTER 11 111 00104 nm 5026 nHm 221 pFm 0 Sm 113 Proof 115 a 133436240 2148 X 107ms b 1606 m 117 Proof y 119 sin at j8z A 1111 1113 1115 1117 1119 1121 1123 1125 1127 1129 1131 1133 a b 79S Proof 2 n 1 ii 2 iii 0 iv 1 S3 radm 3542 X 107 ms Proof a b 02 a b 041122397 34634O650 Q 40A 4687 0 4839 V Proof io a b 2 7I38 a 07222154 62 7300 n 15 7075 U 035 yO24 a b c a b 125 MHz 72 772 n 0444120 35 734 a 0375X APPENDIX C 753 1135 a 245 0 b 5533 Cl 611A 1 1137 1025 W 1139 20 yl5 mS 7IO mS 6408 j5189 mS 20 J15 mSJIO mS 2461 j5691 mS 1141 a 342 7414 0 b 038X 0473X c 265 1143 4 06900 276 y528 Q 1145 211 1764 GHz 03574450 70 j40 0 1147 See Figure C3 1149 See Figure C4 1151 a 7777 1 18 b 0223 dBm 4974 dBm c 3848 m 1153 9112 Q Z O 21030 V0t 144 V 12 V Figure C3 For Problem 1147 24 V 228 V 10 t us 150 mA 1425 mA 10 754 II Appendix C Vht 80 V 7467 V 75026 V 0 10 mA t us 5333 0 4978 50017 0 1 2 3 Figure C4 For Problem 1149 t us CHAPTER 12 121 123 125 127 129 1211 1213 Proof a b c a b 43C 375 a b c a b See Table CI i7TEn 57383 Q rTM15 3058 fi 3096 X 107ms No Yes Ins AQ 08347 W TE23 y4007m 9853 0 Proof 406 X 108 ms 2023 cm 5669 X 108 ms 2834 cm APPENDIX C U 755 1215 a 1193 b 08381 1217 4917 TABLE C1 Mode TEo TE10 TE02 TEnTM TE I 2TM I 2 TE03 TE l 3 TM l 3 TEM TE 1 4TM 1 4 TE05 TE23 TM 2 3 TE l 5 TM15 fc GHz 08333 1667 1863 2357 25 3 3333 3727 4167 4488 4ir i b 1221 004637 Npm 4811 m 1223 a 2165 X 102Npm b 4818 X 103Npm 1225 Proof 1227 Proof j 1229 a TEo b TM110 c TE101 1231 See Table C2 r mzx niry Ho sin cos cos V a J b J piK c TABLE Mode Oil 110 101 102 120 022 C2 fr GHz 19 3535 3333 38 4472 38 1233 a 6629 GHz b 6387 1235 25 sin 30TTX COS 30X3 cos 30irx sin 3070 sin 6 X 109 756 M Appendix C CHAPTER 13 131 CfD sin w 3rsin Aa cos 6 cos tae Vm Am sin oit 0rsin j6 cos 8 cos fir 133 9425 mVm jO25 mAm 135 1974 fl 137 2847 A 139 a fe jnhei0r sinfl t f fi OTT b 15 1311 a 09071 xA b 25 nW 1313 See Figure C5 1315 See Figure C6 1317 8 sin 6 cos t 8 1319 a 15 sin 0 b 15 Figure C5 For Problem 1313 1 3X2 1X 1 5x8 APPENDIX C Figure C6 For Problem 1315 c 15A2sin20 1321 1323 1325 d 3084 fl 9997 a 15 sin2 9 5 b 6 sin2 0 cos2 j 6 c 6605 cos2 0 sin2 j2 6605 1 sin 6 cos 2irr 1327 See Figure C7 1329 See Figure C8 1331 02686 1333 a Proof b 128 1335 2128 pW 1337 19 dB 13d cos 6 Figure C7 For Problem 1327 758 Appendix C Figure C8 For Problem 1329 Nl N4 1339 a 1708 Vm b 1136tiVm c 3095 mW d 191 pW 1341 7752 W CHAPTER 14 141 143 145 147 149 1411 1413 1415 1417 Discussion 033 yO 15 05571 0626 3571 Proof 1428 a 02271 b 1313 c 376 a 2923 b 631 aw 8686a14 Discussion APPENDIX C 759 CHAPTER 15 151 See Figure C9 153 a 10117 156 b 101131506 155 Proof 157 6 V 14 V 159 V V2 375 V3 V4 125 Figure C9 For Problem 151 760 Appendix C 1511 a Matrix A remains the same but h2pss must be added to each term of matrix B b Va 4276 Vb 9577 Vc 11126 Vd 2013 Ve 2919 Vf 6069 Vg 3424 Vh 0109 V 2909 1513 Numerical result agrees completely with the exact solution eg for t 0 V0 0 0 V01 0 03090 V02 0 05878 V03 0 0809 V04 0 09511 V05 0 10 V06 0 09511 etc 1515 1277 pFm numerical 1212 pFm exact 1517 See Table C3 TABLE C3 6 degrees 10 20 30 40 170 180 CpF 85483 90677 8893 8606 1132 86278 1519 a Exact C 8026 pFm Zo 4156 fi for numerical solution see Table C4 TABLE C4 N C pFm Zo ft 10 82386 40486 20 80966 41197 40 80438 41467 100 80025 41562 b For numerical results see Table C5 TABLE C5 N C PFm Zo ft 10 10951 30458 20 10871 30681 40 10827 30807 100 10793 30905 APPENDIX C 761 1521 Proof 1523 a At 15 05 along 12 and 09286 09286 along 13 b 5667 V 0 06708 12 01248 1408 0208 1525 08788 208 0 06708 1527 18 V 20 V 1529 See Table C6 0208 1528 12 01248 0208 10036 TABLE C6 Node No 8 9 10 11 14 15 16 17 20 21 22 23 26 27 28 29 FEM 4546 7197 7197 4546 1098 1705 1705 1098 2235 3295 3295 2235 4545 5949 5949 4545 Exact 4366 7017 7017 4366 1066 1684 1684 1060 2178 3316 3316 2178 4563 6060 6060 4563 1531 Proof