# solutions_chapter21 - 21 ELECTROMAGNETIC INDUCTION Answers...

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21-1 E LECTROMAGNETIC I NDUCTION 21 Answers to Multiple-Choice Problems 1. A 2. B 3. C 4. B 5. B 6. A 7. C 8. A 9. C 10. D 11. C 12. A 13. D 14. C 15. A Solutions to Problems 21.1. Set Up: The normal to the loop is in the z direction. The loop has area Solve: (a) (b) (c) Reflect: The flux is a maximum when the magnetic field is perpendicular to the plane of the loop and the flux is zero when the magnetic field is parallel to the plane of the loop. 21.2. Set Up: where is the angle between and the normal to the surface. Solve: (a) The normal to the surface is parallel to the x axis. and (b) The normal to the surface is parallel to the z axis and The flux is into the enclosed volume. (c) The normal to the surface and the direction of are shown in Figure 21.2. and The flux is out of the shaded volume. (d) The net flux is zero, since the magnitude of the flux into the volume equals the magnitude of the flux out of the volume. Figure 21.2 y B x normal 40.0cm 30.0cm f f F B 5 BA cos f 5 1 0.128 T 21 0.300 m 21 0.500 m 2 cos 53.1° 5 0.0115 Wb. f 5 53.1°. tan f 5 40.0 cm 30.0 cm B S F B 5 1 0.128 T 21 0.300 m 2 2 5 0.0115 Wb. f 5 0°. F B 5 0. f 5 90° B S f F B 5 BA cos f , F B 5 0. f 5 90°. F B 5 BA cos 53.1° 5 1 3.06 3 10 3 Wb 2 cos 53.1° 5 1.84 3 10 2 3 Wb. f 5 53.1°. F B 5 BA 5 1 0.230 T 21 1.33 3 10 2 2 m 2 2 5 3.06 3 10 2 3 Wb. f 5 0°. A 5 p r 2 5 p 1 6.50 3 10 2 2 m 2 2 5 1.33 3 10 2 2 m 2 . F B 5 BA cos f .
21.3. Set Up: Since the field is uniform over the plane of the coil, Solve: Half the flux means half the area, so with This gives 21.4. Set Up: A is constant and B is changing. Solve: (a) (b) 21.5. Set Up: A is constant and B is changing. Solve: 21.6. Set Up: is the flux through each turn of the coil. Solve: (a) The total flux through the coil is (b) 21.7. Set Up: is the angle between the normal to the loop and so and Solve: Reflect: The flux changes because the orientation of the coil relative to the magnetic field changes. 21.8. Set Up: and Solve: 21.9. Set Up: Since the field is uniform, Since the plane of the circuit is perpendicular to the field, and A is constant and B is changing, so and The current in the circuit is given by Ohm’s law, Solve: Reflect: The induced emf depends on the rate of change of the flux so it depends on the rate at which B is changing. It does not depend on the initial magnitude of the field. I 5 E R 5 0.045 V 15 V 5 3.0 mA. E 5 1 0.300 m 21 0.600 m 21 0.25 T / s 2 5 0.045 V. E 5 IR . P D B D t P 5 0.25 T / s. E 5 A P D B D t P E 5 P DF B D t P . cos f 5 1. f 5 F B 5 BA cos f . 5 30.2 m V E 5 2 m 0 nIA D t 5 2 1 4 p 3 10 2 7 T # m / A 21 40 3 10 2 turns / m 21 0.250 A 21 6.00 3 10 2 4 m 2 2 0.0500 s 5 3.02 3 10 2 5 V f 5 0°. B f 5 0. B i 5 m 0 nI F B 5 BA cos f . E 5 P DF B D t P . E 5 NBA 0 cos f f 2 cos f i 0 D t 5 1 80 21 1.10 T 21 0.250 m 21 0.400 m 2 0.0600 s 0 cos 2 cos 53.0° 0 5 58.4 V. f f 5 0°. f i 5 90.0° 2 37.0° 5 53.0° B S , f F B 5 BA cos f . E 5 P DF B D t P . E 5 P N F i 2 N F f D t P 5 1.44 3 10 2 5 Wb 0.040 s 5 3.6 3 10 2 4 V 5 0.36 mV. F B , f 5 BA cos 90° 5 0. N F B , i 5 1 200 21 7.2 3 10 2 8 Wb 2 5 1.44 3 10 2 5 Wb.