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Unformatted text preview: N 5 1 56.3 m 21 10 3 10 3 coils / m 2 5 5.63 3 10 5 turns. l 5 L m A a 2 5 8.89 H 1 4 p 3 10 2 7 T # m / A 2 p 1 0.0200 m 2 2 1 10 3 10 3 coils / m 2 2 5 56.3 m. a 5 10 coils / mm 5 10 3 10 3 coils / m. L 5 m A a 2 l . N 5 a l a L 5 m AN 2 l . L 5 2 U I 2 5 2 1 10.0 J 2 1 1.50 A 2 2 5 8.89 H L 5 m AN 2 l . U 5 1 2 LI 2 . I 5 Å 2 U L 5 Å 2 1 1.0 J 2 10.2 3 10 2 3 H 5 14 A. U 5 1 2 1 10.2 3 10 2 3 H 21 1.15 A 2 2 5 6.74 mJ U 5 1 2 LI 2 N 5 Å 4 p rU m AI 2 5 Å 4 p 1 0.150 m 21 0.390 J 2 1 4 p 3 10 2 7 Wb / m # A 21 5.00 3 10 2 4 m 2 21 12.0 A 2 2 5 2850. U 5 1 2 1 m N 2 A 2 p r 2 I 2 U 5 1 2 LI 2 . L 5 m N 2 A 2 p r . N 1 N 2 5 Å R 1 R 2 5 Å 12.8 3 10 3 V 8.00 V 5 40 1 V 1 V 2 2 2 5 R 1 R 2 N 1 N 2 5 V 1 V 2 . V 1 2 R 1 5 V 2 2 R 2 Electromagnetic Induction 2111 21.51. Set Up: The stored energy is The rate at which thermal energy is dissipated is Solve: (a) (b) (c) No. If I is constant then the stored energy U is constant. The energy being consumed by the resistance of the induc tor comes from the emf source that maintains the current; it does not come from the energy stored in the inductor. 21.52. Set Up: The time constant is The current as a function of time is The energy stored in the inductor is Solve: (a) (b) The maximum current is when and is equal to (c) (d) 21.53. Set Up: The loop rule applied to the circuit gives the voltage across the inductor. The current as a function of time is given by At at Solve: (a) At and (b) At and Initially, the voltage across the resistor is zero, and the full battery emf appears across the inductor. (c) The time constant is When (d) When Reflect: Initially and the full battery voltage is across the inductor. After a long time, the full battery voltage is across the resistor. 21.54. Set Up: After has been closed a long time, the current has reached a value When is opened and is closed, a current decay R L circuit is produced and Solve: (a) (b) (c) When 21.55. Set Up: At all times The voltage across the resistor depends on the current through it and the voltage across the inductor depends on the rate at which the current through it is changing. Immediately after closing the switch the current thorough the inductor is zero. After a long time the current is no longer changing. Solve: (a) so and The ammeter reading is (b) After a long time, and and The ammeter reading is (c) None of the answers in (a) and (b) depend on L so none of them would change. Reflect: The inductance L of the circuit affects the rate at which current reaches its final value. But after a long time the inductor doesn’t affect the circuit and the final current does not depend on L. A 5 1.67A. i 5 v 1 R 5 25.0 V 15.0 V 5 1.67 A. v 1 5 iR v 1 5 25.0 V. v 2 5 A 5 0. v 2 5 25.0 V. v 1 5 i 5 v 1 1 v 2 5 25.0 V. i S 0. t S ` , t 5 L R 5 300.0 3 10 2 3 H 500.0 V 5 6.00 3 10 2 4 s 5 0.600 ms I 5 0.500 A i 5 I e 2 1 R / L 2 t . S 2 S 1 250.0 V 500.0 V 5 0.500 A. S 1 i 5 i 5 E R 5 6.00 V 8.00 V 5 0.750 A. t S ` , i 5 E R 1 1 2 e 2 1 2 5 6.00 V 8.00 V 1 1 2 e 2 1 2 5 0.474 A....
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 Spring '09
 RODRIGUEZ
 Physics, Flux, Magnetic Field

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