A non- conducting ring of mass m and radius R has a charge Q uniformly distributed over its circumference. The ring is placed on a rough horizontal  surface such that plane of the ring is parallel to the surface. A vertical magnetic field  $\mathrm{B}={\mathrm{B}}_{°}{\mathrm{t}}^2$ tesla is switched on. After 2 s from switching on the magnetic field the ring is just about to rotate about vertical axis through its centre.
Friction coefficient  µ  between the ring and the surface is $\frac{{\mathrm{nB}}_{°}\mathrm{RQ}}{\mathrm{m}},\mathrm{then}$   n =
 

Magnitude of induced  electric  field due to change in magnetic flux is given by
$\oint \mathrm{E}.\mathrm{d}1=\frac{\mathrm{d\varphi }}{\mathrm{dt}}=\mathrm{S}.\frac{\mathrm{dB}}{\mathrm{dt}}$
Or          $\mathrm{El}={\mathrm{\pi R}}^2\left(2{\mathrm{B}}_0\mathrm{t}\right)\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left(\frac{\mathrm{dB}}{\mathrm{dt}}=2{\mathrm{B}}_0\mathrm{t}\right)$
Here,  E =  induced electric field due to change in magnetic flux
Or          $\mathrm{E}\left(2\mathrm{\pi R}\right)=2{\mathrm{\pi R}}^2{\mathrm{B}}_0\mathrm{t}$
$\mathrm{Or} \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{E}={\mathrm{B}}_0\mathrm{Rt}$
Hence ,      $\mathrm{F}=\mathrm{QE}={\mathrm{B}}_0\mathrm{QRt}$
This force is tangential to ring. Ring starts
Rotating when torque of this force is greater
Than the  torque due to maximum friction
$\left({\mathrm{f}}_{\max }=\mathrm{\mu mg}\right)\mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{when}$
${\mathrm{\tau }}_{\mathrm{F}}\ge {\mathrm{\tau }}_{{\mathrm{f}}_{\max }}$
Taking the limiting case
${\mathrm{\tau }}_{\mathrm{F}}={\mathrm{\tau }}_{{\mathrm{f}}_{\max }}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{FR}=\left(\mathrm{\mu mg}\right)\mathrm{R}$
Or          $\mathrm{F}=\mathrm{\mu mg}$       
Or           ${\mathrm{B}}_0\mathrm{QRt}=\mathrm{\mu mg}$
It is given that ring starts rotating after 2. So,
Putting t =  2 , we get
 $\mathrm{\mu }=\frac{2{\mathrm{B}}_0\mathrm{RQ}}{\mathrm{mg}}$