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Standard XII

Physics

Newton's Second Law: Rotatory Version

A conducting ...

Question

A conducting ring of radius r with a conducting spoke is in pure rolling on a horizontal surface in a region having a uniform magnetic filed B as shown, v being the velocity of the centre of the ring. Then the potential difference $V_{0} - V_{A}$ is

\frac{B v r}{2}

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\frac{3 B v r}{2}

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\frac{- B v r}{2}

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\frac{- 3 B v r}{2}

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Solution

The correct option is C $\frac{- B v r}{2}$

$F = q V B$
Considering pure rolling of OA about A.
The induced emf across OA will be
$| e | = \frac{B w {(x)}^{2}}{2}$
From lens law, O will be the negative end, while A will be the positive end.
$∴ V_{0} - V_{A} = \frac{- ({B w r}^{2})}{2} (∵ s i n c e V = r w)$
$= \frac{- (B V r^{2})}{2 r}$
$V_{0} - V_{B} = \frac{- B V r}{2}$

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