Question

A non conducting ring of mass m, radius R, having charge q uniformly distributed over its circumference is placed on a rough horizontal surface. A vertical time varying magnetic field $B=8t^2$ is switched on at time t = 0. After 3s, the ring starts rotating. The coefficient of friction between the ring and the table is

• A. $\frac{24qR}{mg}$
• B. $\frac{16qR}{mg}$
• C. $\frac{qR}{16mg}$
  
• D. $\frac{1}{24}\left(\frac{qR}{mg}\right)$

Answer 1

The correct option is A

$\frac{24qR}{mg}$

Due to the time varying magnetic field, an induced electric field E is set up, so the force on the ring is $F=qE$
If f be the frictional force between the ring and the table, then the ring starts rotating when
$F=f$

$\Rightarrow qE=\mu mg\dots \left(1\right)$

Also $\oint \overrightarrow{E}.d\overrightarrow{l}=\left|\frac{d\varphi }{dt}\right|$

$\Rightarrow E\left(2\pi R\right)=A\left(\frac{dB}{dt}\right)$

$\Rightarrow E\left(2\pi R\right)=\left(\pi R^2\right)\left(16t\right)\{\because \frac{dB}{dt}=16t\}$

$\Rightarrow E=8Rt\dots \left(2\right)$

Substituting (2) in (1), we get

$q\left(8Rt\right)=\mu mg$

$\Rightarrow \mu =\left(\frac{8qR}{mg}\right)t$

$\Rightarrow \mu {|}_{t=3}=\frac{24qR}{mg}$