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solutions_chapter21

# N f b 5 li f b e 1 5 0270 v e 1 5 m p d i 2 d t p e 2

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N F B 5 Li . F B E 1 5 0.270 V. E 1 5 M P D i 2 D t P , E 2 D i 1 D t E 2 5 1 3.25 3 10 2 4 H 21 830 A / s 2 5 0.270 V. E 2 5 M P D i 1 D t P . E 1 5 M P D i 2 D t P . E 2 5 P D i 1 D t P 5 1 2.88 3 10 2 7 H 21 37.5 A / s 2 5 1.08 3 10 2 5 V M 5 1 4 p 3 10 2 7 T # m / A 2 p 1 6.00 3 10 2 4 m 2 2 1 6750 21 15 2 0.500 m 5 2.88 3 10 2 7 H 5 0.288 m H m 0 5 4 p 3 10 2 7 T # m / A. M 5 m 0 AN 1 N 2 l , E 2 5 M P D i 1 D t P . i 1 E 2 Electromagnetic Induction 21-9

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21.38. Set Up: Example 21.10 shows that the inductance of a toroidal solenoid is Solve: 21.39. Set Up: Problem 21.37 shows that for a long solenoid Solve: 21.40. Set Up: A transformer transforms voltages according to The effective resistance of a secondary circuit of resistance R is and Solve: (a) (b) 21.41. Set Up: A transformer transforms voltages according to The effective resistance of a secondary circuit of resistance R is Resistance R is related to P and V by Conservation of energy requires so Solve: (a) and we want so use a step-down transformer with (b) so (c) The resistance R of the blower is The effective resistance of the blower is Reflect: Energy is provided to the primary at the same rate that it is con- sumed in the secondary. Step-down transformers step up resistance and the current in the primary is less than the cur- rent in the secondary. 21.42. Set Up: For an ideal transformer, Solve: (a) (b) (c) (d) so 21.43. Set Up: and Solve: (a) (b) (c) Reflect: Since the power supplied to the primary must equal the power delivered by the secondary, in a step-up transformer the current in the primary is greater than the current in the secondary. I 1 5 P V 1 5 110 W 120 V 5 0.917 A P 5 V 2 I 2 5 1 13,000 V 21 8.50 3 10 2 3 A 2 5 110 W N 2 N 1 5 V 2 V 1 5 13,000 V 120 V 5 108 N 1 N 2 5 V 1 V 2 . P 1 5 P 2 . P 5 VI V 2 5 13,000 V. V 1 5 120 V. 1 50.0 V 2 1 N 1 N 2 2 2 5 1 50.0 V 21 10 2 2 5 5000 V . R 5 V 2 P 5 1 120 V 2 2 2.88 W 5 5000 V . P 5 V 2 R P 5 VI 5 1 12.0 V 21 0.240 A 2 5 2.88 W I 2 5 V 2 R 5 12.0 V 50.0 V 5 0.240 A N 1 N 2 5 V 1 V 2 5 120 V 12.0 V 5 10 P 5 VI 5 V 2 R . V 2 V 1 5 N 2 N 1 . I 2 V 2 5 1 13.3 A 21 120 V 2 5 1600 W. R eff 5 9.00 V 1 1 / 2 2 2 5 36.0 V . R 5 V 2 P 5 1 120 V 2 2 1600 W 5 9.00 V . I 5 P V 5 1600 W 240 V 5 6.67 A. P 5 VI , N 2 / N 1 5 1 2 . V 2 5 120 V, V 1 5 240 V V 1 I 1 5 V 2 I 2 . P 1 5 P 2 P 5 V 2 R . R eff 5 R 1 N 2 / N 1 2 2 . V 2 V 1 5 N 2 N 1 . R eff 5 R 1 N 2 / N 1 2 2 5 125 V 1 834 / 275 2 2 5 13.6 V V 2 5 V 1 1 N 2 / N 1 2 5 1 25.0 V 21 834 / 275 2 5 75.8 V V 1 5 25.0 V. N 2 5 275 R eff 5 R 1 N 2 / N 1 2 2 . V 2 V 1 5 N 2 N 1 . L 5 1 4 p 3 10 2 7 T # m / A 2 p 1 7.50 3 10 2 4 m 2 2 1 50 2 2 5.00 3 10 2 2 m 5 1.11 3 10 2 7 H 5 0.111 m H L 5 m 0 N 2 A l . L 5 1 4 p 3 10 2 7 Wb / m # A 21 1800 2 2 1 0.480 3 10 2 4 m 2 2 2 p 1 0.120 m 2 5 2.59 3 10 2 4 H 5 0.259 mH L 5 m 0 N 2 A 2 p r . 21-10 Chapter 21
21.44. Set Up: and Solve: so 21.45. Set Up: Example 21.10 shows that the inductance of a toroidal solenoid is The energy stored in an inductor is Solve: and Reflect: The stored energy is proportional to the square of the number of turns even though the magnetic field within the solenoid is directly proportional to the number of turns. We will see in Section 21.10 that the magnetic field energy depends on the square of the magnetic field.

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