$\begin {array} {1 1} (a)\;\large\frac{qB}{m} & \quad (b)\;\large\frac{2qB}{m} \\ (c)\;\large\frac{qB}{2m} & \quad (d)\;Zero \end {array}$

$ \int E\: d l=-\large\frac{d\phi}{dt}$$ = \pi r^2 \large\frac{dB}{dt}$

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

the torque is given by

$ N = \int \lambda\: E \:d l$

Angular momentum=$I\omega = \int N\: dt$

On solving the above set of sequential equations

$R\lambda \pi ^2 B= I\omega$,

$R\pi R^2 B q/2\pi R =mR^2 \omega$

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

Ans : (c)

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