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 $\left.m g=B_{0} Q R\right]$ (All physical quantities are in $S.I.$ unit)

Answer 1

The magnitude of induced electric field due to change in magnetic flux is given by
$\oint \vec{E} \cdot d \vec{l}=\frac{ d \phi}{ d t}=S \frac{ d B}{ d t}$
or $E l=\pi R^{2}\left(2 B_{0} t\right) \quad \mid \because \frac{ d B}{ d t}=2 B_{0} t$
Here $E=$ induced electric field due to change in magnetic flux or
$E(2 \pi R)=2 \pi R^{2} B_{0} t$
or $E=B_{0} R t$
Hence $F=Q E=B_{0} Q R t$
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.
$\left(f_{\max }=\mu m g\right)$ or when $\tau_{ F } \geq \tau_{f_{\max }}$
$\Rightarrow \tau_{f_{\max }}=f_{\max } R=(\mu m g) R$, taking the limiting case. $f_{\max }=\mu m g$ $\Rightarrow B_{0} Q R t \times R=\mu m g R$
It is given that the ring starts rotating after 2 second. So, putting $t=2$
second we get $\mu=\frac{2 B_{0} R Q}{m g}=2$