problem29_08 - 29.8 a d ind dt B d dt B1 A ind 1 dB d A sin...

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29.8: a) ) ( 1 ind A B dt d dt d = = Φ B ε ( 29 t s 057 . 0 ind 1 ) T 4 . 1 ( 60 sin 60 sin - - ° = ° = e dt d A dt dB A ε t s 057 . 0 1 2 1 ) s 057 . 0 )( T 4 . 1 )( 60 )(sin ( - - - ° = e r π t s e 1 057 . 0 1 2 ) s 057 . 0 )( T 4 . 1 )( 60 (sin ) m 75 . 0 ( - - - ° = π = t s e 1 057 . 0 V 12 . 0 - - b) ) V 12 . 0 ( 10 1 10 1 0 = = ε ε t s e 1 057 . 0 V 12 . 0 ) V 12 . 0 ( 10 1 - - = s 4 . 40 s 057 . 0 ) 10 1 ( ln 1 = - = - t t c) B is getting weaker, so the flux is decreasing. By Lenz’s law, the induced current
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