problem29_60

# problem29_60 - 29.60 a b B BA B0r0(1 3(t t0)2 B0r0 2 2 2...

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29.60: a) ). ) ( 2 ) ( 3 1 ( 3 0 2 0 2 0 0 t t t t πr B BA B + - = = Φ b) ) ) ( 6 ) ( 6 ( ) ) ( 2 ) ( 3 1 ( 2 0 0 0 2 0 0 3 0 2 0 2 0 0 t t t t t πr B t t t t dt d πr B dt d ε B + - - = + - - = Φ - = ckwise. counterclo V, 5 066 . 0 s 010 . 0 s 10 0 . 5 s 010 . 0 s 10 0 . 5 s 010 . 0 ) m 0420 . 0 ( 6 , , s 10 0 . 5 at so 6 3 2 3 2 0 3 0 2 0 0 2 0 0 = × - × - = × = - = - = - - - π ε ε B t t t t t t πr B c) . 2 .. 10 12 A 10 . 3 0 V . 0 0655 3 total total = - × = = + = = - r i R r R R i ε ε d) Evaluating the emf at 2 10 21 . 1 - × = t s, using the equations of part (b): , V 6 067 . 0 - = ε and the current flows clockwise, from
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• Spring '08
• ROSS
• Magnetism, Resistor, Electrical resistance, Clockwise

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