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Question:

A rectangular loop has a sliding connector PQ of length l and resistance RΩ and it is moving with a speed v as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents I1, I2 and I are:

I1=−I2=BlvR, I=2BlvR

I1=I2=Blv3R, I=2Blv3R

I1=I2=I=BlvR

I1=I2=Blv6R, I=Blv3R

Solution:

A moving conductor is equivalent to a battery of emf=vBℓ(motion emf) Equivalent circuitI=I1+I2Applying Kirchoff's lawI1R+IR−vBl=0... (1)I2R+IR−vBP=0... (2)Adding(1)and(2)2IR+IR=2vBlI=2vBl3RSolving, we getI1=I2=vBℓ3R