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Question

Two conducting rings $P$ and $Q$ of radii $r$ and $2 r$ slide uniformly in opposite directions with centre of mass velocities $2 v$ 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 B v r

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C

8 B v r

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D

16 B v r

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Solution

The correct option is C $8 B v r$
The surface is conducting, hence it is equipotential.

We know that motional e.m.f. is velocity $\times B \times l$
So,
$v_{A} - v_{P} = 2 v B \times 2 r = 4 v B r \dots \dots (i)$
$v_{Q} - v_{B} = v 4 r B$
$v_{P} = v_{Q}$
$v_{A} - v_{B} = 4 r v B + 4 r v B = 8 v B r$

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