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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

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A
I1=I2=BlvR, I=2BlvR
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B
I1=I2=Blv3R, I=2Blv3R
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C
I1=I2=I=BlvR
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D
I1=I2=Blv6R, I=Blv3R
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Solution

The correct option is B I1=I2=Blv3R, I=2Blv3R
A moving conductor is equivalent to a battery of emf =vB (motion emf) Equivalent circuit

I=I1+I2

Applying Kirchoff's law
I1R+IRvBl=0 ............... (1)
I2R+IRvBP=0 ............... (2)

Adding (1) and (2)
2IR+IR=2vBl
I=2vBl3R

Solving, we get
I1=I2=vB3R

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