PHY_solution-chapter29

PHY_solution-chapter29 - 29.1: ) . 37 cos 1 ( . 37 cos and...

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Unformatted text preview: 29.1: ) . 37 cos 1 ( . 37 cos and , - = = = NBA NBA NBA B B B i f V. 5 . 29 s 0600 . ) . 37 cos 1 )( m 25 . )( m 400 . )( T 10 . 1 )( 80 ( ) . 37 cos 1 ( = -- = -- = - = t NBA t B 29.2: Before: a) ) m 10 12 )( T 10 . 6 )( 200 ( 2 4 5-- = = NBA B : after ; m T 10 44 . 1 2 5 =- b) . V 10 6 . 3 s 040 . ) m 10 2 . 1 )( T 10 . 6 )( 200 ( 4 2 3 5--- = = = = t NBA t B 29.3: a) . R NBA Q NBA QR R t Q IR t NBA t B = = = = = = A credit b) card reader is a search coil. Data is c) stored in the charge measured so it is independent of time. 29.4: From Exercise (29.3), . C 10 16 . 2 . 12 80 . 6 ) m 10 20 . 2 ( T) 05 . 2 )( 90 ( 3 2 4-- = + = = R NBA Q 29.5: From Exercise (29.3), . T 0973 . ) m 10 20 . 3 )( 120 ( ) . 45 . 60 )( C 10 56 . 3 ( 2 4 5 = + = = =-- NA QR B R NBA Q 29.6: a) ( 29 4 4 5 ) s T 10 00 . 3 ( s) T 012 . ( ) ( t t dt d NA B dt d NA dt Nd B- + = = = ( 29 . ) s V 10 02 . 3 ( V 0302 . ) s T 10 2 . 1 ( ) s T 012 . ( 3 3 4 3 4 4 t t NA-- + = + = At b) V 0680 . ) s 00 . 5 )( s V 10 02 . 3 ( V 0302 . s 00 . 5 3 2 4 + = + = =- t . A 10 13 . 1 600 V 0680 . 4- = = = R I 29.7: a) for 2 sin 2 2 cos 1 - = -- = - = T t T NAB T t NAB dt d dt d B otherwise. zero ; T t < < . b) 2 at T t = = c) . 4 3 and 4 at occurs 2 max T t T t T NAB = = = From d) B t T , 2 < < is getting larger and points in the z + direction. This gives a clockwise current looking down the z- axis. From B T t T , 2 < is getting smaller but still points in the z + direction. This gives a counterclockwise current. 056ddba76af2e97e015697860a997452149a9ecc.doc Page 1 of 26 29.8: a) ) ( 1 ind A B dt d dt d = = B ( 29 t s 057 . ind 1 ) T 4 . 1 ( 60 sin 60 sin-- = = e dt d A dt dB A t s 057 . 1 2 1 ) s 057 . )( T 4 . 1 )( 60 )(sin (--- = e r t s e 1 057 . 1 2 ) s 057 . )( T 4 . 1 )( 60 (sin ) m 75 . (--- = = t s e 1 057 . V 12 .-- b) ) V 12 . ( 10 1 10 1 = = t s e 1 057 . V 12 . ) V 12 . ( 10 1-- = s 4 . 40 s 057 . ) 10 1 ( ln 1 = - =- t t B is getting c) weaker, so the flux is decreasing. By Lenzs law, the induced current must cause an upward magnetic field to oppose the loss of flux. Therefore the induced current must flow counterclockwise as viewed from above. 29.9: a) 4 / so and 2 2 2 c A r A r c = = = 2 ) 4 ( c B BA B = = dt dc c B dt d B = = 2 At m 570 . ) s 120 . )( s . 9 ( m 650 . 1 , s . 9 =- = = c t mV 44 . 5 ) s m 120 . )( m 570 . )( 2 1 )( T 500 . ( = = b) Flux is decreasing so the flux of the induced current is ind and I is clockwise....
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PHY_solution-chapter29 - 29.1: ) . 37 cos 1 ( . 37 cos and...

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