2146 set up solve a b this is not a reasonable

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21.46. Set Up: Solve: (a) (b) This is not a reasonable current in a circuit; it is too large. 21.47. Set Up: The self-inductance of a solenoid is found in Problem 21.37 to be The length l of the solenoid is the number of turns divided by the turns per unit length. Solve: (a) (b) If is the number of turns per unit length, then and For this coil This is not a practical length for laboratory use. Reflect: The number of turns is The length of wire in the solenoid is the circumference C of one turn times the number of turns. The length of wire is This length of wire will have a large resist- ance and electrical energy loses will be very large. 21.48. Set Up: The energy density is The energy U is u (volume). Solve: 21.49. Set Up: Solve: (a) (b) 21.50. Set Up: and Solve: I 2 5 I 1 1 U 2 U 1 2 1 / 2 5 I 1 9.0 mJ 3.0 mJ 2 1 / 2 5 I " 3 . U 1 I 1 2 5 U 2 I 2 2 . U I 2 5 1 2 L 5 constant U 5 1 2 LI 2 . L 5 2 U I 2 5 2 1 1.73 3 10 7 J 2 1 80.0 A 2 2 5 5.41 3 10 3 H Energy 5 1 200 W 21 24 h 21 3600 s / h 2 5 1.73 3 10 7 J U 5 1 2 LI 2 . Energy 5 Pt . U 5 B 2 2 m 0 1 volume 2 5 1 0.560 T 2 2 2 1 4 p 3 10 2 7 T # m / A 2 1 0.0290 m 3 2 5 3.62 3 10 3 J u 5 B 2 2 m 0 . I 2 R 1 0.126 m 21 5.63 3 10 5 2 5 7.1 3 10 4 m 5 71 km. C 5 p d 5 p 1 4.00 3 10 2 2 m 2 5 0.126 m. 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 0 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 0 A a 2 l . N 5 a l a L 5 m 0 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 0 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 0 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 0 N 2 A 2 p r 2 I 2 U 5 1 2 LI 2 . L 5 m 0 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 21-11
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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.
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