A non-conducting ring having charge $q$ uniformly distributed over its circumference is placed on a rough horizontal surface. A vertical time varying magnetic field $B = 4 t^{2}$ is switched on at time t = 0. Mass of the ring is m and radius is R. The ring starts rotating after 2s, the coefficient of friction between the ring and the table is

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$\frac{4 q m R}{g}$

$\frac{2 q m R}{g}$

$\frac{8 q R}{m g}$

$\frac{q R}{2 m g}$

answer is C.

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Detailed Solution

$E l = \frac{d ϕ}{d t}$

or, $E (2 π R) = π R^{2} \cdot \frac{d B}{d t} = π R^{2} (8 t)$

$\begin{array}{rcl} ∴ E & = & 4 R t \\ F & = & q E = 4 q R t (tangential) \\ τ_{F} & = & 4 q R^{2} t \\ τ_{f} & = & (μ m g R) \end{array}$

When $τ_{F} > τ_{f}$ , ring will start rotating.

At 2s,

$8 q R^{2} = μ m g R \Rightarrow μ = \frac{8 q R}{m g}$

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