Question

A ring of mass $m$, radius $r$ having charge $q$ uniformly distributed over it and free to rotate about its own axis is placed in a region having a magnetic field $B$ parallel to its axis. If the magnetic field is suddenly switched off, the angular velocity acquired by the ring is

• A. $\frac{qB}{m}$
• B. $\frac{2qB}{m}$
• C. $\frac{qB}{2m}$
• D. None of these

Answer 1

The correct option is C $\frac{qB}{2m}$

Due to change in magnetic flux the electric field $E$ produced can be found by$\int E.d\ell =\frac{-d\varphi }{dt}=-\pi r^2\frac{dB}{dt}$
$\tau =r\int \lambda Ed\ell =-r\lambda \pi r^2\frac{dB}{dt}$
$\int \tau dt=-r\lambda \pi r^2\int dB$
We know that change in angular momentum is equal to net external torque
$⟹\tau =\frac{dL}{dt}$ ($L$ $=$ angular momentum)

$⟹R\lambda \pi r^{2}B$ = $I$$\omega$ (where $I$ is the moment of inertia of ring)
$\Rightarrow \pi r^{2}B\frac{q}{2\pi r}=m{r}^2\omega \Rightarrow \omega =\frac{qB}{2m}$