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**Hint**Firstly, we will find motional emf produced due to rotation of the ring by the formula $e = - \dfrac{{d\phi }}{{dt}}$ . Then we will calculate the mechanical force as well as the electromagnetic force on the ring. And finally, we will equate the different forces calculated above to calculate the required coefficient of friction.

**Complete Step by step solution**

We know that the motion emf produced due to the rotation of the ring is,

$\Rightarrow$ $e = - \dfrac{{d\phi }}{{dt}}$

Where, $\phi $ is the magnetic flux and it is equal to $\phi = BA$

Therefore, we have

$e = - \dfrac{{d(BA)}}{{dt}}$

Where, A is area of the ring and is equal to $\pi {R^2}$

Given time varying magnetic field $B = 4{t^2}$

Hence, we have

$\Rightarrow$ $

e = - \pi {R^2}\dfrac{{d(4{t^2})}}{{dt}} \\

e = - 8\pi {R^2}t \\

$

For $t = 2\sec $ we have

$

e = - 8\pi {R^2}(2) \\

V = - 16\pi {R^2}......(1) \\

$

Where, $V$ is potential due to the ring.

Now force on the ring is given by,

$\Rightarrow$ $F = qE$

Where, $E$ is electric field and is equal to potential gradient

Hence, we have

$\Rightarrow$ $F = qE = \dfrac{{qV}}{{2\pi R}}$

Using equation (1),

$

F = \dfrac{{16\pi q{R^2}}}{{2\pi R}} \\

F = 8qR......(2) \\

$

Now the minimum force required to move ring is,

$F = \mu mg......(3)$

Using equation (2) and (3) is,

$\Rightarrow$ $

\mu mg = 8qR \\

\mu = \dfrac{{8qR}}{{mg}} \\

$

Hence the required value of friction is $\mu = \dfrac{{8qR}}{{mg}}$

**Option (C) is correct.**

**Note**We have used Faraday's law of magnetic induction to find out the emf due to ring i.e., whenever the flux of the magnetic field through the area bounded by a closed conducting loop changes, an emf is produced in the loop. The emf is given by

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