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Question

Two conducting rings P and Q of radii r and 2r slide uniformly in opposite directions with centre of mass velocities 2v and v respectively on a conducting surface S. There is an uniform magnetic field of magnitude B perpendicular to the plane of the rings. The potential difference between the highest points of the two rings is

A
Zero
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B
4 Bvr
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C
8 Bvr
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D
16 Bvr
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Solution

The correct option is C 8 Bvr
The surface is conducting, hence it is equipotential.

We know that motional e.m.f. is velocity × B×l
So,
vAvP=2vB×2r=4vBr (i)
vQvB=v 4r B
vP=vQ
vAvB=4 rvB+4 rvB=8 vBr

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