A total charge Q is uniformly distributed over a non-conducting disc of radius r. There is a time varying magnetic field perpendicular to its plane and changing at the uniform rate of $\frac{d B}{d t} .$ The magnitude of torque experienced by the disc is

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$\frac{1}{4} Q r^{2} (\frac{d B}{d t})$

$\frac{Q r^{2}}{2} (\frac{d B}{d t})$

$\frac{Q r^{3}}{3} (\frac{d B}{d t})$

$\frac{1}{2} Q r^{3} (\frac{d B}{d t})$

answer is C.

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Detailed Solution

$E 2 π r = \frac{d B}{d t} (π r^{2}),$

$E = \frac{r}{2} \frac{d B}{d t},$

$d τ = (2 π r d r) σ . E . r,$

$τ = \int d τ = \int \frac{r}{2} \frac{d B}{d t} (2 π r d r) σ . r .$

$= \frac{1}{4} Q r^{2} (\frac{d B}{d t})$

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