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Unformatted text preview: 1. (a) The period is T = 4(1.50 µ s) = 6.00 µ s. (b) The frequency is the reciprocal of the period: f T = = = × 1 1 6 00 167 10 5 . . µ s Hz. (c) The magnetic energy does not depend on the direction of the current (since U B ∝ i 2 ), so this will occur after onehalf of a period, or 3.00 µ s. 2. We find the capacitance from U Q C = 1 2 2 : C Q U = = × × = × − − − 2 6 2 6 9 2 160 10 2 140 10 914 10 . . C J F. c h c h 3. According to U L I Q C = = 1 2 2 1 2 2 , the current amplitude is I Q LC = = × × × = × − − − − 300 10 4 00 10 4 52 10 6 3 6 2 . . . C 1.10 10 H F A. c h c h ( ) ( ) ( ) ( ) ( ) ( ) 3 2 1 2 1 1 1 ( 1) 2 1 2 1 2.50 s , 2 2 2 2 2 10 Hz n n t T n T n T n f µ − − = + − = − = = = − × where n = 1, 2, 3, 4, … . The earliest time is ( n =1) 2.50 s. t µ = (c) At t T = 1 4 , the current and the magnetic field in the inductor reach maximum values for the first time (compare steps a and c in Fig. 311). Later this will repeat every half period (compare steps c and g in Fig. 311). Therefore, ( ) ( ) ( ) ( 1) 2 1 2 1 1.25 s , 4 2 4 L T n T T t n n µ − = + = − = − where n = 1, 2, 3, 4, … . The earliest time is ( n =1) 1.25 s. t µ = 4. (a) We recall the fact that the period is the reciprocal of the frequency. It is helpful to refer also to Fig. 311. The values of t when plate A will again have maximum positive charge are multiples of the period: t n T n f n n A = = = × = 2 00 10 500 3 . . , Hz s µ b g where n = 1, 2, 3, 4, … . The earliest time is ( n =1) 5.00 s. A t µ = (b) We note that it takes t T = 1 2 for the charge on the other plate to reach its maximum positive value for the first time (compare steps a and e in Fig. 311). This is when plate A acquires its most negative charge. From that time onward, this situation will repeat once every period. Consequently, 5. (a) All the energy in the circuit resides in the capacitor when it has its maximum charge. The current is then zero. If Q is the maximum charge on the capacitor, then the total energy is U Q C = = × × = × − − − 2 6 2 6 6 2 2 90 10 2 360 10 117 10 . . . C F J. c h c h (b) When the capacitor is fully discharged, the current is a maximum and all the energy resides in the inductor. If I is the maximum current, then U = LI 2 /2 leads to I U L = = × × = × − − − 2 2 1168 10 75 10 558 10 6 3 3 . . J H A. c h 6. (a) The angular frequency is ω = = = × = − k m F x m 8 0 050 89 13 . . . N 2.0 10 m kg rad s c h b g (b) The period is 1/ f and f = ω /2 π . Therefore, T = = = × − 2 2 7 0 10 2 π π 89 ω rad s s. . (c) From ω = ( LC ) –1/2 , we obtain C L = = = × − 1 1 89 50 2 5 10 2 2 5 ω rad s H F. b g b g . . 7. (a) The mass m corresponds to the inductance, so m = 1.25 kg....
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This note was uploaded on 05/11/2010 for the course PHYS 2101,2102 taught by Professor Giammanco during the Spring '10 term at LSU.
 Spring '10
 GIAMMANCO
 Physics, Current, Energy

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