Chapter 28 - Resolução Sears - Vol 3

SOURCES OF MAGNETIC FIELD 28.1. IDENTIFY and SET UP: Use Eq.(28.2) to calculate B at each point. \[ \vec{B} = \frac{\mu_0 q \vec{v} \times \vec{r}}{4 \pi r^3} = \frac{\mu_0 q \vec{v} \times \vec{r}}{4 \pi r^3} \text{since} \= \frac{\vec{r}}{r}. \] \[ \vec{r} = (8.00 \times 10^4 \hat{m}) \hat{i} \; \text{and} \; \vec{r} \text{ is the vector from the charge to the point where the field is calculated.} \] EXECUTE: (a) \[ \vec{r} = (0.500 m) \hat{i}, r = 0.500 m \] \[ \vec{v} \times \vec{r} = v \hat{k} \times \hat{i} = -v \hat{j} \] \[ \vec{B} = \frac{\mu_0 q \vec{v} \times \vec{r}}{4 \pi r^3} = \frac{(1 \times 10^{-7} T \cdot m/A)(6.00 \times 10^{-6} C)(8.00 \times 10^4 m/s)}{(0.500 m)^3} \hat{k} \] \[ \vec{B} = -(1.92 \times 10^{-7} T) \hat{j} \] (b) \[ \vec{r} = (0.500 m) \hat{j}, r = 0.500 m \] \[ \vec{r} \times \vec{r} = v \hat{k} \times \hat{j} = 0 \text{ and } \vec{B} = 0. \] (c) \[ \vec{r} = (0.500 m) \hat{k}, r = 0.500 m \] \[ \vec{r} \times \vec{r} = v \hat{k} \times \vec{v} \hat{i} \] \[ \vec{B} = \frac{\mu_0 q \vec{v} \times \vec{r}}{4 \pi r^3} = \frac{(1 \times 10^{-7} T \cdot m/A)(6.00 \times 10^{-6} C)(8.00 \times 10^4 m/s)}{(0.500 m)^3} \hat{i} + \{1.92 \times 10^{-7} T\} \hat{j} \] (d) \[ \vec{r} = (0.500 m) \hat{i} - (0.500 m) \hat{k}, \; r = \sqrt{(0.500 m)^2 + (0.500 m)^2} = 0.701 m \] \[ \vec{r} \times \vec{r} = (v \hat{k} - v \hat{i}) \times \vec{v} \hat{j} = -(4.00 \times 10^4 m/s) \hat{i} \] \[ \vec{B} = \frac{(1 \times 10^{-7} T \cdot m/A)(6.00 \times 10^{-6} C)(4.00 \times 10^4 m/s)}{(0.707 m)^3} \hat{i} + \print{(6.79 \times 10^{-8} T) \hat{i} \]. \] EVALUATE: At each point B is perpendicular to both \vec{r} and r. \[ B = 0 \text{along the direction of} \; \vec{v}. \] 28.2. IDENTIFY: A moving charge creates a magnetic field as well as an electric field. SET UP: The magnetic field caused by a moving charge is \[ B = \frac{\mu_0 q \sin \theta}{4 \pi r^2}, \text{ and its electric field is } \frac{E = \frac{1}{4 \pi \epsilon_o} \frac{q}{r^2}. \] EXECUTE: Substitute the appropriate numbers into the above equations. EVALUATE: There are enormous fields within the atom! 28.3. IDENTIFY: A moving charge creates a magnetic field. SET UP: The magnetic field due to a moving charge is \[ B = \frac{\mu_0 q \sin \theta}{4 \pi r^2}. \] 28-2 Chapter 28 EXECUTE: Substituting numbers into the above equation gives \[ B = \frac{\mu_0 q \sin \theta}{4 \pi r^2}(1.6 \times 10^{-19} C)(3.0 \times 10^7 m/s) \sin{30^\circ} \] (a) \[ B = \frac{4 \pi 10^{-7} T \cdot m/A}{(2.00 \times 10^9 m)} \] \[ B = 4.00 \times 10^{-11} T, \text{ out of the page, and it is the same at point B.} \] (b) \[ B = (1.00 \times 10^{-7} T \cdot m/A)(1.60 \times 10^{-19} C)(3.0 \times 10^7 m/s)/(2.00 \times 10^9 m)^3 \] \[ B = 2.00 \times 10^{-11} T, \text{ out of the page.} \] (c) \[ B = 0 \] \text{since} \sin 180^\circ = 0. EVALUATE: Even at high speeds, these charges produce magnetic fields much less than the Earth's magnetic field. 28.4. IDENTIFY: Both moving charges produce magnetic fields, and the net field is the vector sum of the two fields. SET UP: Both fields point out of the page, so their magnitudes add, giving \[ B = B_{alpha} + B_{sat} = \frac{\mu_0 q}{4 \pi r^2}(\sin 40^\circ + 2 \sin 140^\circ) \] EXECUTE: Factoring out en and \vec{r} putting in the numbers gives \[ B = 4 \pi \times 10^{-7} T \cdot m/A \cdot (1.60 \times 10^{-19} C)(2.5 \times 10^7 m/s) \times \frac{1.75 \times 10^{-9} m}{4 \pi} \] \[ B = 2.52 \times 10^{-7} T = 2.52 mT, \text{ out of the page.} \] EVALUATE: At distances very close to the charges, the magnetic field is stronger enough to be important. 28.5. IDENTIFY: Apply \[ \vec{B} = \frac{\mu_0 q \vec{v} \times \vec{r}}{4 \pi r^3}. \] SET UP: Since the charge is at the origin, \vec{r} = x \hat{i} + y \hat{j} + z \hat{k}. EXECUTE: (a) \[ \vec{v} = v \hat{i}, \vec{r} = r \hat{i}, r = 0.500 m \] \[ \vec{r} \times \vec{v} = v \hat{i} \times v \hat{i} = 0. \] (b) \[ \vec{v} = v \hat{i}, \vec{r} = \vec{r}, \vec{v} \times \vec{r} = x \hat{i} \times y \hat{k} = \vec{v} \hat{j}. \] (c) \[ \vec{v} = v \hat{i}, \vec{r} = (0.500 m) \hat{i}, r = 0.500 m \] \[ \vec{B} = \frac{\mu_0 q v \left| \vec{r} \times \vec{v} \right|}{4 \pi r^3} \frac{(0.800 \cdot 10^7 C)(4.8 \cdot 10^4 m/s)}{(0.707 m)^3} \hat{i} = 1.31 \times 10^{-6} T. \] EVALUATE: In each case, \vec{B} is perpendicular to both \vec{r} and \vec{v}. 28.6. IDENTIFY: Apply \[ \vec{B} = \frac{\mu_0 q \vec{v} \vec{r} \sin \theta}{4 \pi r^2}, \] SET UP: In part (a), \vec{r} and \vec{v} is perpendicular to \vec{v} in each case, so \[ \vec{r} \frac{|\vec{r} \times \vec{v}|}{r^2/d^2}. \] EXECUTE: (a) \[ B_{\text{amp}} = B + B^r = \frac{\mu_0 q \vec{v} \vec{r} \sin \theta}{4 \pi r^2} \], \[ d^4 \] \[ B = \frac{\mu_0 q \vec{v} \left| \vec{r} \times \vec{v} \right|}{4 \pi r^3} \frac{(0.500 m) \cdot \hat{i}}{(0.707 m)^4} = 4.38 \times 10^{-4} T. \] \[ \text{The direction of} \; \vec{B} \; \text{is into the page.} \] 28-3 Sources of Magnetic Field (b) Following Example 28.1 we can find the magnetic force between the charges: \[ F_B = \frac{\mu_0 q q' v \vec{r} \sin \theta}{4 \pi r^2}(1.00 \times 10^{-7} N \cdot A)(3.00 \times 10^7 C)(3.00 \times 10^7 m/s)\] \[ F_B = 1.69 \times 10^{-17} N. \text{The force on the upper charge points up and the force on the lower charge points down.}\] Coulomb force between the charges is \[ F_C = \frac{k_C q_q'}{r^2}(8.99 \times 10^9 N \cdot m^2/C^2) \cdot (8.00 \times 10^{-8})(3.00 \times 10^7 N \cdot m^2/C)^2 = 3.75 N.\] The force on the upper charge points up and the force on the lower charge points down. The ratio of the Coulomb force to the magnetic force is \[ \frac{F_C}{F_M} = \frac{3.75 N}{1.69 \times 10^{-17} N} = 2.22 \times 10^{-10}: \text{the Coulomb force is much larger.} \] EVALUATE: When two charges have the same sign and move in opposite directions, the force between them is repulsive. When two charges have the same sign move in the same direction, the force between them is attractive. 28.7. IDENTIFY: Apply \[ \vec{F}_B = q \vec{v} \times \vec{B} \]. For the magnetic force on q, use \[ \vec{F}_B = q \vec{v} \vec{B}, \vec{B} = \frac{\mu_0 q \vec{v} \times \vec{r}}{4 \pi r^3}. \] \[ q \; \text{use} \; \text{q} \; q' \sin \theta \cdot d \] SET UP: In part (a), \vec{r}, and \vec{r} perpendicular to \vec{v} in each case, \[ = B \], \[ \frac{F_m}{4 \pi \epsilon r}(2.667). \] EXECUTE: (a) \[ \vec{F}_e = \vec{r} q \vec{v} \], \[ F_C = \frac{k_C q_q'}{r^2} \], \[ \vec{F}_e = m^2 \cdot \frac{1}{\epsilon_0}. \] \[ \vec{F}_C + s_{un} \text{where} \star = (av/ x). \] \[ F_e = \cfrac{k_C q_1}{4\epsilon_{04}\left(\frac{v^2}{r^2}\right)} = \frac{\mu_0 E}{4\pi \epsilon_{q_1} V_{q'}'} \] \[ \vec{i} = \frac{d}{F_e} = \frac{(14 \cdot \times10^{-3} N \cdot m^{-2})(0.07012 b \cdot C)}{(3.00 m/s)\} \sin(2 \cdot P)\ = 1.69 \times 10^{-13} N \text{from} \; \theta d F_{e]. \] EVALUATE: When charges of opposite sign move in opposite directions, and the fun is attractive. For the two charges move in the same direction, \text{ the force due to opposite sign is found as force }. Coulomb force between them. \text{In the electron force is attractive, the force less than decided to 2F\vec{e}}. 28.8. IDENTIFY: Both moving charges create magnetic fields, and the net field is the vector sum of the two. The magnetic force on a moving charge is \( F_{\text{m}} = q\vec{v}\vec{B} \) and the electrical force obeys Coulomb's law. SET UP: The magnetic field due to a moving charge is \( B = \frac{\mu_0 q\vec{v} \sin \theta}{4\pi r^2} \). EXECUTE: (a) Both fields are into the page, so their magnitudes add, giving \( B = B_r + B_a \frac{\mu_0 q\vec{v}(r/\vec{r})^2}{(5\times 10^{-7} m)^2 \cdot (4.0 \times 10^9 m)^3} \sin 90^\circ \frac{a}{i} \] \( B = \frac{4\mu_0 q (1.60\times 10^{-9} C)(845,000 m/s)}{(5.00 \times 10^{-7} m)^3} = 1.39 \times 10^{-19} T. \] (b) \text{Using} \( B = \frac{\mu_0 q \sin \theta}{4 \pi r}, \text{where} \; r = a= 90^\circ, \theta= 180\circ - \arctan(5/4) = 128.7^\circ, \;\text{we get} \] \( B = (4 \times10^{-7} T\cdot m/A)(1.60\times10^{-9} C)(\sec840,000 m/s)\sin 128.7^\circ, \] \( B = 2.58 \cdot 10^{-18} T, \text{into the page.} \] (c) \[ F_{\text{m}} = qvB\sin90^\circ = (1.60\cdot10^{-9} C)(845,000 m/s)(16.48\times10^{-27} T \] \( F_{\text{m}} = \text{Increased}{4} \times 10^{-3} \times (4\times10^{-7} Nm^2/C^3. \]