Question

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 the plane of the ring is parallel to the surface. A vertical magnetic field $B=B_0{t}^2$ $tesla$ is switched on. After $2$ second from switching on the magnetic field, the ring is just about to rotate about a vertical axis through its centre. Find friction coefficient $\mu$ between the ring and the surface. [Given that $mg=B_{0}QR]$ (All physical quantities are in $S.I.$ unit)

Answer 1

Answer: 2

The magnitude of induced electric field due to change in magnetic flux is given by
$\oint \vec{E}\cdot d\vec{l}=\frac{d\varphi }{dt}=S\frac{dB}{dt}$
or $El=\pi R^2\left(2B_{0}t\right)\mid \because \frac{dB}{dt}=2B_{0}t$
Here $E=$ induced electric field due to change in magnetic flux or
$E(2\pi R)=2\pi R^2{B}_{0}t$
or $E=B_{0}Rt$
Hence $F=QE=B_{0}QRt$
This force is tangential to ring. The ring starts rotating when the torque of this force is greater than the torque due to maximum friction.
$\left(f_{\max }=\mu mg\right)$ or when ${\tau }_F\ge {\tau }_{f_{\max }}$
$\Rightarrow {\tau }_{f_{\max }}=f_{\max }R=(\mu mg)R$, taking the limiting case. $f_{\max }=\mu mg$ $\Rightarrow B_{0}QRt\times R=\mu mgR$
It is given that the ring starts rotating after 2 second. So, putting $t=2$
second we get $\mu =\frac{2B_{0}RQ}{mg}=2$