Question

A non-conducting ring of mass $m=4\,kg$ and radius $R=10\,cm$ has a charge $Q=2C$ 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 $B=4t^{3}T$ is switched on at $t=0$ . At $t=5s$ ring starts to rotate about the vertical axis through the centre. The coefficient of friction between the ring and the surface is found to be $\frac{k}{24}$ . Then the value of $k$ is

Answer 1

The induced electric field at the periphery of the ring is
$E=\frac{R}{2}\frac{d B}{d t}$
$\Rightarrow E=\frac{R}{2}\left(12 t^{2}\right)=6Rt^{2}$
When the ring is about to slip,
$qER=\mu mgR$
$\Rightarrow 6qRt^{2}=\mu mg$
$\Rightarrow \mu =\frac{6 q R t^{2}}{m g}=\frac{3}{4}=\frac{18}{24}$
$k=18$