Chapter 27 - Resolução Sears - Vol 3

MAGNETIC FIELD AND MAGNETIC FORCES\n27.1. IDENTIFY and SET UP: Apply Eq (27.2) to calculate \\( \\mathbf{F} \\). Use the cross products of unit vectors from Section 1.10. EXECUTE: \\( \\mathbf{E} = (4.19 \\times 10^{-6} \\, \\text{N s/m}^2) \\hat{j} - (-3.85 \\times 10^{-10} \\, \\text{N s/m}^2) \\hat{i} \\)\n(a) \\( \\mathbf{B} = (1.40 \\, \\text{T}) \\hat{j} \\)\n\\( \\mathbf{F} = q \\mathbf{v} \\times \\mathbf{B} = (-1.24 \\times 10^{-6} \\, \\text{C})(1.40 \\, \\text{T})(4.19 \\times 10^{-6} \\, \\text{m/s}) \\hat{i} \\times \\hat{j} \\hat{k} \\)\n\\( \\hat{i} = -\\hat{j} \\hat{k} \\)\n\\( \\hat{k} = -\\hat{j} \\hat{i} \\)\n\\( \\hat{j} = -\\hat{i} \\hat{k} \\)\n\\( \\mathbf{F} = (-2.72 \\times 10^{-10} \\hat{j}) + (6.68 \\times 10^{-10} \\hat{j}) \\)\n\\( \\mathbf{F} = -(7.27 \\times 10^{-10} \\hat{j}) \\)\nEVALUATE: The directions of \\( \\mathbf{v} \\) and \\( \\mathbf{B} \\) are shown in Figure 27.1a.\nThe right-hand rule gives that \\( \\mathbf{F} \\times \\mathbf{B} \\) is directed out of the paper (\\(-z\\)-direction). The charge is negative so \\( \\mathbf{F} \\) is opposite to \\( \\mathbf{v} \\times \\mathbf{B} \\).\n\\( \\mathbf{F} \) is in the \\(-z\\)-direction. This agrees with the direction calculated with unit vectors.\n\\( \\mathbf{F} = (1.40 \\, \\text{T})(-3.85 \\times 10^{-10} \\, \\text{N s/m}^2)(4.19 \\times 10^{-6} \\, \\text{m/s}) \\hat{j} \\hat{j} + ... \\rightarrow \\text{rest of calculations}\\)\nEVALUATE: Chapter 27\n27.2 IDENTIFY: The magnetic field could also have a component along the north-south direction, that would not contribute to the force, but then the field wouldn't have minimum magnitude.\nEXECUTE: The force \\( \\mathbf{F} \\) on the particle is in the direction of the deflection of the particle. Apply the right-hand rule to the directions of \\( \\mathbf{v} \\) and \\( \\mathbf{B} \\). Set if your thumb is in the direction of \\( \\mathbf{F} \\) or opposite to that direction. Use \\( \\mathbf{F} = |q|\\mathbf{v}|B|\\sin \\phi \\) with \\( \\phi = 90° \\) to calculate \\( \\mathbf{F} \\).\nSET UP: The directions of \\( \\mathbf{v} \\), \\mathbf{B} \\) and \\( \\mathbf{F} \\) are shown in Figure 27.3.\nEXECUTE: (a) When you apply the right-hand rule to \\( \\mathbf{v} \\) and \\( \\mathbf{B} \\), your thumb points east. \\( \\mathbf{F} \\) is in this direction, so the charge is positive.\n(b) \\( \\mathbf{F} = |q|\\mathbf{v}|B|\\sin \\phi = (8.50 \\times 10^{-6} \\, \\text{C})(4.75 \\times 10^{-10} \\, \\text{m/s})(0.25 \\sin 90^{18}) = 0.0505 \\)\nEVALUATE: If the particle had negative charge and \\( \\mathbf{v} \\) and \\( \\mathbf{B} \\) are unchanged, the particle would be deflected toward the west.\n\\( \\cdots \\rightarrow \\text{continuation}\\) 27.3 EXECUTE: (a) The smallest possible acceleration is zero, when the motion is parallel to the magnetic field. The greatest acceleration is when the velocity and magnetic field are at right angles:\na = \\frac{q\\mathbf{B}h}{m} = \\frac{(1.60 \\times 10^{-19} )\\cdots}{0.25}{7.40 \\times 10^{-10} \\, \text{m/s}^2} = 3.25 \\times 10^{-3} \\, \\text{m/s}^2.\\ \nIf \\( m = 9.11 \\times 10^{-31} \\text{kg} \\), then sin \\( \\delta = 0.25 \\) and \\( \\theta = 145° \\).\nEVALUATE: The force and acceleration decrease as the angle approaches zero.\n27.4 IDENTIFY: Apply Newton's second law, with the force being the magnetic force.\nSET UP: \\( \\mathbf{F} = |q|\\mathbf{v}|B|\\sin \\phi \\) and solve for \\( v \\).\nSET UP: An electron has: \\( q = -1.60 \\times 10^{-19} \, \\text{C} \\).\nEXECUTE: \\( \\mathbf{F} = 4.60 \\times 10^{-15} \\hat{N} \\);\nEVALUATE: Only the component \\( \\mathbf{B} \\sin \\phi \\) of the magnetic field perpendicular to the velocity contributes to the force.\n27.6 IDENTIFY: Apply Newton's second law and \\( |F| = |q|\\mathbf{v} \\sin \\phi \\).\nSET UP: \\( \\phi \\) is the angle between the direction of \\mathbf{F} and the direction of \\mathbf{B}. \\ 27-6 Chapter 27\n\nEXECUTE: (a) B = \\frac{mv}{qr} = \\frac{(1.67 \\times 10^{-27} kg)(1.41 \\times 10^{6} m/s)}{(1.60 \\times 10^{-19} C)(0.050 m)} = 0.294 T\nThe direction of the magnetic field is out of the page (the charge is positive), in order for F to be directed to the right at point A.\n\n(b) The time to complete half a circle is t = \\frac{R}{v} = 1.11 \\times 10^{-7} s.\n\n27.17. IDENTIFY and SET UP: Use conservation of energy to find the speed of the ball when it reaches the bottom of the shaft. The right-hand rule gives the direction of F and E_{2}(z) gives its magnitude. The number of excess electrons determines the charge of the ball:\n\nEXECUTE: q = (4.00 \\times 10^{-12} C) = -6.408 \\times 10^{-12} C\n\nspeed at bottom of shaft: v_{1} = mg_{s} + m_{g} = \sqrt{2gy} = 49.5 m/s\n\nv is downward and B is west, so v \\times B is north. Since \\alpha < 0, F is south.\n\nF = |q|vB\\sin \\theta = (4.608 \\times 10^{-10} C)(49.5 m/s)(0.250 T)(sin 90^{\\circ}) = 7.931 \\times 10^{-9} N\n\nEVALUATE: Both the charge and speed of the ball are relatively small to the magnetic force is small, much less than the gravity force of 1.5 N.\n\n27.18. IDENTIFY: Since the particle moves perpendicular to the uniform magnetic field, the radius of its path is\n\n|r| = \\frac{mv}{|q|B|}\n\nSet Up: The alpha particle has charge q = e^{2} = 2.320 \\times 10^{-19} C.\n\nEXECUTE: (a) r = (6.45 \\times 10^{-27} kg)(3.65 \\times 10^{6} m/s) = 6.73 \\times 10^{-14} m. The alpha particle moves in a circular arc of diameter 2R = 1.35 m.\n\n(b) For a very short time interval the displacement of the particle is in the direction of the velocity. The magnetic force is always perpendicular to the direction of motion, and so the work-energy theorem says that the kinetic energy of the particle, and hence speed, is constant.\n\nThe acceleration is a = F_{B}/m = \\frac{|q|vB \\sin \\beta}{m}\n\nThen also use r = \\frac{mv}{|q|B} and realize for part (a) to calculate 6.73 \\times 10^{-14} m = 1.88 \\times 10^{7} m/s. We can also use r = R^{*} and the result for part (a) to calculate 6.73 \\times 10^{-14} m = 1.88 \\times 10^{7} m/s.\n\nEVALUATE: (b) The electromagnetic force is perpendicular to both E and B.\n\nSet Up: In part (a), apply conservation of energy to the motion of the two nuclei. In part (b) apply |q|vB = m\\over R.\n\nEXECUTE: (a) K_{i} + U_{i} = K_{f} + U_{f}. U_{i} = K_{f} = U_{f} = 0, K_{f} = U_{i} = 0 and \\frac{1}{2}mv^{2} = k_{e}\\frac{2k}{r}.\n\nv = \\sqrt{\\frac{2k}{mr}} = (1.602 \\times 10^{-19} C)(3.34 \\times 10^{-10} m) = 1.2 \\times 10^{7} m/s.\n\n(b) \\sum F = ma gives \\sum qB = m\\frac{mv}{r}.\n\nNow \\frac{q}{r} = (1.602 \\times 10^{-19} C)(2.50 m) = 0.10 T.\n\n27.20. EXECUTE: (a) F = |q|vB\\sin \\beta - B -\\frac{F}{|q||B|}\\sin \\beta; F = \\frac{F}{|q|singd}\\frac{F}{|q||B|} = 50.0 T. If the angle > 90^{\\circ}, larger field is needed to produce the same force. The direction of the field must be toward the south so that v \\times B is downward.\n\nEXECUTE: (a) F = |q|vB\\sin \\beta - B -\\frac{F}{|q||B|}\\sin \\beta; F = \\frac{F}{|q|singd}\\frac{F}{|q||B|} = 50.0 T. 27.21. (a) IDENTIFY and SET UP: Apply Newton's 2nd law, with a = v^2 / R since the path of the particle is circular.\nEXECUTE: ΣF = ma says |q|vB = m(v^2/r)\n|q| / m = (1.602x10^-19 C)(2.50 T)(9.6x10^6 m/s) / (3.43x10^-3 kg) = 8.35x10^10 m/s²\n\n(b) IDENTIFY and SET UP: The speed is constant so t = distance/v.\nEXECUTE: t = r / |v| = (R)(6.96x10^10 m/s) = 2.62x10^5 s.\n\n(c) IDENTIFY and SET UP: Kinetic energy gained = electric potential energy lost.\nEXECUTE: 3m/v2 = |q|E|\n(3.43x10^2 kg)(8.35x10^10 m/s)2 / 2(1.602x10^-19 C)\nEVALUATE: The detrun has a much larger mass to charge ratio than an electron so a much larger B is required for the same v and R. The detrun has positive charge so gains kinetic energy when it goes from high potential to low potential.\n27.22. IDENTIFY: For motion in an arc of a circle, a = v^2 / R and the net force is radially inward, toward the center of the circle.\nEXECUTE: (a) If the direction of the force is shown in Figure 27.22, the mass of a proton is 1.67x10^-27 kg.\n\nF = F = |q|vBsinθ = r^-90° and gives:\n|q|vB = 3.61x10^-19 C(2.50 T)(0.475 m) = 1.21x10^-17 kg.\n\nw = mg = (1.02|7.67|10^-10 kg)(9.80 m/s²) = 1.96x10^-16 N. The magnetic force is much larger than the weight of the particle, so it is a very good approximation to neglect gravity.\n\nEVALUATE: (e) The magnetic force is always perpendicular to the path and does no work. The particles move with constant speed.\n\nFigure 27.22 27.23. IDENTIFY: Example 27.3 shows that B = mf / |q|, where f is the frequency, in Hz, of the electromagnetic waves that are produced.\nSET UP: An electron has charge q = -e and mass m = 9.11x10^-31 kg. A proton has charge q = +e and mass m = 1.67x10^-27 kg.\n\nEXECUTE: (a) B = mf / |q| = (9.11x10^-31 kg)(2π(3.00x10^7 Hz)) / (1.60x10^-19 C) = 10.7 T. This is about 2.4 times the greatest magnitude of magnetic field yet obtained on earth.\n(b) Protons have a greater mass than the electrons, so a greater magnetic field would be required to accelerate them with the same frequency and there would be no advantage in using them.\n\n27.24. IDENTIFY: The magnetic force on the beam bends it through a quarter circle.\nSET UP: The distance that particles on the beam travel is R = r0 and the radius of the quarter circle is R = mv / qB.\nEXECUTE: Solving for R gives R = (9.1x10^-31 kg)(2.00x10^7 m/s) / (1.60x10^-19 C)(0.250 T) = 1.1x10^-2 m.\n\nB = mv/qR = (1.67x10^-27 kg)(1.00x10^7 m/s)(1.60x10^-19 C) / (0.075 m) = 2.13x10^-5 T.\nEVALUATE: When a particle of charge -e is accelerated through a potential difference of magnitude V, its kinetic energy K.E. when it moves in a circular path of radius R is acceleration is R.\nSET UP: An electron has charge q = -e and mass 9.11x10^-31 kg.\nEXECUTE: ½mv^2 = eV and v = \\sqrt{2eV/m} = \\sqrt{ |q|E} / R = |q|E/mv = |q|vB / |q|\n\n|F| = |B|v|q| gives F = |q|v × B = |q|E.\n(b) Yes. The electric field exerts a force in the direction of the electric field, since the charge of the proton is positive, and there is a component of acceleration in this direction.\n(c) EXECUTE: Let the plane perpendicular to B (the xy-plane) the motion is circular. But there is a velocity component in the direction of B so the motion is helical. The electric field does not affect the circular motion but does affect the radius of the helix.\n\nwith the +i direction. This force produces an acceleration in the -j direction and this causes the pitch of the helix to vary. The force does not affect the circular motion in the xy-plane, so the electric field does not affect the radius of the helix. 27.28. IDENTIFY: For no deflection the magnetic and electric forces must be equal in magnitude and opposite in direction.\nSET UP: v = E / B for no deflection. With only the magnetic force, |q|vB = mv / R\nEXECUTE: (a) v = E / B = (1.56x10^3 V/m) / (4.62x10^-2 T) = 3.83x10^6 m/s.\n\n(b) The directions of the three vectors, E and B are sketched in Figure 27.28.\n|q|E / |B| = (91.1x10^-9 kg)(3.83x10^6 m/s) / (1.60x10^-19 C)(4.62x10^-2 T) = 4.17x10^-10 m.\n\nE = 2q / |q|\n|B| = 2πk / |q| = 2π(7.74x10^-5 s).\n\nEVALUATE: For the field directions shown in Figure 27.28, the electric force is toward the top of the page and the magnetic force is toward the bottom of the page.\n\nFigure 27.28\n\n27.29. IDENTIFY: For the alpha particles to emerge from the plates undeflected, the magnetic force on them must exactly cancel the electric force. The battery produces an electric field between the plates, which acts on the alpha particles.\nSET UP: First use energy conservation to find the speed of the alpha particles as they enter the plates: v = √(2E / m).\nThe electric field between the plates due to the battery is E = V / d. For the alpha particles to be undeflected, the magnetic force must cancel the electric force, so qE = qvB, giving B = E / v.\nEXECUTE: Solve for the speed of the alpha particles just as they enter the region between the plates. Their charge is 2e:\nv = (2E / m) = (2(4.60x10^-6 C)(750 V)) / (6.64x10^-3 kg) = 4.11x10^6 m/s\n\nThe electric field between the plates produced by the battery is,\nE = V / d = (15 V)(0.0082 m) = 18,300 V/m.\n\nThe magnetic force must cancel the electric force:\nB = E / v = (18,300 V/m) / (4.11x10^6 m/s) = 0.00445 T.\n\nThe magnetic field is perpendicular to the electric field. If the charges are moving to the right and the electric field points upward, the magnetic field is out of the page.\n\nEVALUATE: The sign of the charge of the alpha particle does not enter the problem, so negative charges of the same magnitude would also not be deflected.\n\n27.30. IDENTIFY: For no deflection the magnetic and electric forces must be equal in magnitude and opposite in direction.\nSET UP: v = E / B for no deflection.\nEXECUTE: To pass undeflected in both cases, E = vB = (5.85x10^6 m/s)(1.35 T) = 7988 N/C.\n\n(a) If q = 0.640x10^-19 C, the electric field direction is given by -(j) - (k) = j, since it must point in the opposite direction to the magnetic force.\n\n(b) If q = -0.320x10^-19 C, the electric field direction is given by -(j)-(k) = -i, since the electric force must point in the opposite direction as the magnetic force.\n\nEVALUATE: The same configuration of electric and magnetic fields works as a velocity selector for both positively and negatively charged particles.\n\n27.31. IDENTIFY AND SET UP: Use the fields in the velocity selector to find the speed of the particles that pass through. Apply Newton's 2nd law with a = v / R in the circular motion in the second region of the spectrometer. Solve for the mass m of the ion.\nEXECUTE: In the velocity selector |q|E = |q|vB:\nv = E / B = (1.12x10^6 V/m) / (0.540 T) = 2.074x10^6 m/s 27-10 Chapter 27\n\nIn the region of the circular path \\( \\sum F = ma \\) gives \\( | q | vB = m \\frac{v^2}{r} \\) so \\( m = \\frac{|q|}{B} \\)\n\nSingly charged ion, so \\( |q| = e = 1.602 \\times 10^{-19} \\) C\n\nm = (1.602 \\times 10^{-19} C)(3.10 \\times 10^{6} T) = 1.29 \\times 10^{-19} kg\n\n2.074 \\times 10^{-15} m/s\n\nMass number: 78 gives atomic mass units, so \\( 1.29 \\times 10^{-19} \) kg = 78.\n\n27.32. IDENTIFY and SET UF: For a velocity selector, \\( E = vB \\). For parallel plates with opposite charge, \\( V = Ed \\).\nEXECUTE: (a) \\( E = vB = (8.32 \\times 10^{-6} m/s)(0.650 T) = 1.81 \\times 10^{4} V/m. \\)\n(b) \\( V = Ed = (1.81 \\times 10^{4} V/m)(5.20 \\times 10^{-7} m) = 6.14 kV. \\)\n\nEVALUATE: Any charged particle with \\( v = 8.2 \\times 10^{6} m/s \\) will pass through undeflected, regardless of the sign and magnitude of its charge.\n\n27.33. IDENTIFY: The magnetic force is \\( F = I B sin \\theta \\). For the wire to be completely supported by the field requires that \\( F = mg \\) and that \\( B \\) and \\( I \\) are in opposite directions.\nSET UP: The magnetic force is maximum when \\( \\theta = 90^\\circ \\). The gravity force is downward.\nEXECUTE: (a) \\( F = I B = (0.150 kg)(9.80 m/s^2) = 1.34 \\times 10^{-1} N. \\) This is a very large current and ohmic heating due to the resistance of the wire would be severe; such a current isn't feasible.\n(b) The magnetic force must be upward. The directions of \\( I \\) and \\( B \\) are shown in Figure 27.33, where we assumed that \\( B \\) is south to north. To produce an upward magnetic force, the current must be eastward and perpendicular to the earth's magnetic field.\nEVALUATE: The magnetic force is perpendicular to both the direction of \\( I \\) and the direction of \\( B \\).\n\nFigure 27.33\n\n27.34. IDENTIFY: \\( F = I B sin \\theta. \\)\nSET UP: \\( l = 0.050 m \\) is the length of wire in the magnetic field. Since the wire is perpendicular to \\( B \\), \\( \\theta = 90^\\circ. \\)\nEXECUTE: \\( F = I B = (108 A)(0.0500 m)(0.550 T) = 0.297 N. \\)\nEVALUATE: The force per unit length of wire is proportional to both \\( B \\) and \\( I. \\)\nIDENTIFY: Apply \\( F = I B sin \\theta. \\)\nSET UP: Label the three segments in the field as \\( a, b, \\) and \\( c. \\) Let \\( b \\) be the length of segment a. Segment b has length 0.30 m and segment a has length 0.600 m -- x. Figure 27.35 shows the direction of the force on each segment. For each segment, \\( \\theta = 90^\\circ. \\) The total force on the wire is the vector sum of the forces on each segment.\nEXECUTE: \\( F_{c} = I B = -(4.50 A)(0.240 T), \\) \\( F_{a} = -(4.50 A)(0.600 m)(0.240 T). \\) Since \\( F_{b} \\) and \\( F_{c} \\) are in the same direction their vector sum has magnitude \\( F_{c} + F_{a} = -(4.50 A)(0.600 m)(0.240 T) = 0.648 N \\) directed toward the bottom of the page in Figure 27.35a. \\( F_{c} = (4.50 A)(0.300 m)(0.240 T) = 0.324 N \\) and is directed to the right. The vector addition diagram for \\( F_{c} \\) and \\( F_{b} \\) is given in Figure 27.35b.\nF = \\( \\sqrt{F_{c}^{2} + F_{b}^{2}} = \\sqrt{(0.648 N)^{2} + (0.324 N)^{2}} = 0.724 N. \\) tan \\( \theta = \\frac{F_{b}}{F_{c}} = \\frac{0.324 N}{0.648 N} \\) and \\( \\theta = 63.4^\\circ \\) in Figure 27.35b. The net force has magnitude 0.724 N and its direction is specified by \\( \\theta = 63.4^\\circ \\).