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A non-conducting ring of mass $m$ , radius R, having charge q uniformly distributed over its circumference is placed on a rough horizontal surface. A vertical time varying magnetic field $B = 8 t^{2}$ switched on at time $t = 0$ . After 3 s, the ring starts rotating. The coefficient of friction between the ring and the table is

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$\frac{24 q R}{m g}$

$\frac{q R}{16 m g}$

$\frac{16 q R}{m g}$

$\frac{1}{24} (\frac{q R}{m g})$

answer is D.

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Detailed Solution

Due to the time varying magnetic field, an induced electric field E is set up and $F = q E$
And F= friction $q E = μ m g$
Also $E (2 π R) = π R^{2} (\frac{d B}{d t})$
$E = 8 R t$
$q (8 R t) = μ m g$
At 3 sec $μ = \frac{24 q R}{m g}$

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