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

Answer 1

$\int Edl=-\frac{d\varphi }{dt}=\pi r^2\frac{dB}{dt}$

Let $\lambda =\frac{q}{2\pi R}$be the charge per unit length of the ring

the torque is given by $N=\int \lambda Edl$

Angular momentum $=I\omega =\int Ndt$

On solving the above set of sequential equations

$R\lambda {\pi }^{2}B=I\omega$

$\frac{R\pi R^{2}Bq}{2\pi R}=m{R}^2\omega$

one gets the value of $\omega$ as $\frac{qB}{2m}$