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Question

Two conducting rings of radii r and 2r move in opposite directions with velocities 2v and v respectively on a conducting surfaces S. There is a 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?
1033909_1cbc12e1c1904200a8242b314da97d7a.jpg

A
zero
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B
2rvB
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C
4rvB
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D
8rvB
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Solution

The correct option is D 8rvB
The correct option is D.

Given,

Radii=r,2r

Velocities=v,2v

Conductive surfaces is S and the magnetic field is B

So for the potential difference we have to reduce emf in the ring by the cell,

e1=B(2v)(2r=4Bvr)=4Bvr$

e2=B(v)(4r)=4Bvr

Therefore the pd between the heighest point of the two ring will be:

v2v1=e1+e2

Δv=4Bvr+4Bvr=8Bvr


980247_1033909_ans_b724082c078e442aa7cf40f049906e27.PNG

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