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Q.

If a rod PQ of length 1.25m is moving with velocity 10ms-1 on parallel tracks placed in uniform magnetic field B0 = 0.2T . If resistance of ABCDA is 100Ω , then current through loop at this instant is

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a

20 mA

b

1.25 mA

c

10 mA

d

40 mA

answer is C.

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Detailed Solution

ε=Bvl and i=εRv=10ms−1,l=1m,B=0.2Tε= induced emf ==BlV=0.2×10×1=2 volt induced current in loop i=εR=2100=201000=20mA

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