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Q.

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

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a

I1=-I2=BlvR, I=2BlvR

b

I1=I2=I=BlvR

c

I1=I2=Blv6R, I=Blv3R

d

I1=I2=Blv3R, I=2Blv3R

answer is C.

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Detailed Solution

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Given circuit can be reduced to

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I1+I2=BlvR+R2=2Blv3R So, I1=I2=Blv3R

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