878_Physics ProblemsTechnical Physics

878_Physics ProblemsTechnical Physics - 218 P31.10 Faradays...

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218 Faraday’s Law P31.10 Φ B nI A = µ 0 bg solenoid εµ π επ ε =− × × −− N d dt Nnr dI dt t t B Φ 0 73 1 2 15 0 4 10 1 00 10 0 020 0 600 120 14 2 120 solenoid 2 TmA m m As mV ej e j b g af .. . c o s .c o s P31.11 For a counterclockwise trip around the left-hand loop, with BA t = d dt At a I R I R PQ 20 5 0 2 1 cos °− = and for the right-hand loop, d dt Ata I R I R PQ 2 2 30 +− = where II I PQ 12 is the upward current in QP . Thus, 2 5 0 2 2 Aa R I I I R PQ PQ −+ = di FIG. P31.11 and Aa I R I R PQ 2 2 3 += a f 26 5 3 0 22 Aa RI Aa I R PQ PQ I Aa R PQ = 2 23 upward, and since R == 0 100 0 650 0 065 0 . m m ΩΩ a f I PQ = × = 100 10 0650 23 0 065 0 283 3 2 . Ts m A upward a f . P31.12 F H G I K J =+ ∆Φ B t N dB dt AN tA 00100 00800 At t = 500 . s , 30 0 0 410 0 040 0 61 8 2 . . m mV P31.13 Bn In e t µµ 00 1.60 30 0 1 . A a f Φ B t BdA n e dA zz 0 1.60
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