A non-conducting ring of mass $m$ and radius $R$ has a charge $Q$ uniformly distributed over the circumference. The ring is placed on a rough horizontal surface such that plane of the ring parallel to the surface. A vertical magnetic field $B=B_0{t}^2$ tesla is switched on. After $2s$of switching on the magnetic field the ring is just about to rotate about vertical axis through its centre. 

(a) Find friction coefficient $\mu$ between the ring and the surface. 

(b) If magnetic field is switched off after $4s$, then find the angle rotated by the ring before coming to stop after switching off the magnetic field. 

(a) Magnitude of induced electric field due to change in magnetic flux is given by

$\oint E.dl=\frac{d\varphi }{dt}=S.\frac{dB}{dt}$

or $El={\mathrm{\pi R}}^2\left(2{\mathrm{B}}_0\mathrm{t}\right)$             $\left(\frac{dB}{dt}=2B_{0}t\right)$

Here, $E=$induced electric field dur to change in magnetic flux 

or $E\left(2\mathrm{\pi R}\right)=2{\mathrm{\pi R}}^2{\mathrm{B}}_0\mathrm{t}$

or $E=B_{0}Rt$

Hence, $F=QE=B_{0}QRt$

This force is tangential to ring. Ring starts rotating when torque of this force is greater than the torque due to maximum friction

$\left(f_{max}=\mu mg\right)$ or when

${\tau }_F\ge {\tau }_{f_{max}}$

Taking the limiting case

${\tau }_F={\tau }_{f_{max}}$

or  $FR=\left(\mu mg\right)R$

or $F=\mu mg$

or $B_{0}QRt=\mu mg$

It is given that ring starts rotating after $2$. 

So, putting $t=2$, we get

$\mu =\frac{2B_{0}RQ}{mg}$

(b) After $2$

${\tau }_F>{\tau }_{f_{max}}$

Therefore, net torque is 

$\tau ={\tau }_F-{\tan }_{f_{max}}=B_{0}Q{R}^{2}t-\mu mgR$

Substituting, $\mu =\frac{2B_{0}QR}{mg}$, we get

$\tau =B_{0}Q{R}^2\left(t-2\right)$

or $I\left(\frac{d\omega }{dt}\right)=B_{0}Q{R}^2\left(t-2\right)$

or $m{R}^2\left(\frac{d\omega }{dt}\right)=B_{0}Q{R}^2\left(t-2\right)$

or ${\int }_0^{\omega }d\omega =\frac{B_{0}Q}{m}{\int }_2^4\left(t-2\right)dt$

or $\omega =\frac{2B_{0}Q}{m}$

Now, magnetic field is switched off, i.e., only retarding torque is present due to friction. So, angular retardation will be

$\alpha =\frac{{\tau }_{f_{max}}}{I}=\frac{\mu mgR}{m{R}^2}=\frac{\mu g}{R}$

Therefore, applying

${\omega }^2={{\omega }_0}^2-2\alpha \theta$

or $0={\left(\frac{2B_{0}Q}{m}\right)}^2-2\left(\frac{\mu g}{R}\right)\theta$

or $\theta =\frac{2{B_0}^2{Q}^{2}R}{\mu m^{2}g}$

Substituting $\mu =\frac{2B_{0}RQ}{mg}$

We get  $\theta =\frac{B_{0}Q}{m}$