Charge Q is uniformly distributed on a thin insulating ring of mass m which is initially at rest. Magnetic field B, perpendicular to the plane of ring is switched on. Then,

The changing magnetic field induces an electric field in the ring. Let us imagine the ring divided into small sections each of length  $\Delta s$ and denote the tangential component of the induced electric field by Et (in the general case Et can vary from point to point). The charge on a small section of the ring is
$\Delta Q=Q\frac{\Delta s}{2\pi r},$                               
Where r is the radius of the ring. The force exerted on it is   $\Delta F_t=\Delta Q{E}_t$                     
And the resultant torque is  $\Delta \tau =r\Delta F_t$
The total torque experienced by the ring is thus,
$\tau =\sum \Delta \tau =\sum rQ\frac{\Delta s}{2\pi r}{E}_t=\frac{Q}{2\pi }\sum E_t\Delta s$                      
Identifying the expression  $\sum E_t\Delta s$ as the induced electromotive force along the ring, which is directly proportional to the rate of change in the magnetic flux, we have
             $\sum E_t\Delta s=-\frac{\Delta \varphi }{\Delta t}=-\pi r^2\frac{\Delta B}{\Delta t}$                    
As a result of the torque, the ring, which has a moment of inertia $I=m{r}^2$, starts to spin with angular acceleration $\alpha$. During a time interval $\Delta t$ its angular velocity changes by
                 $\Delta \omega =\alpha \Delta t=\frac{\tau }{I}\Delta t=\frac{Q}{2\pi }(-\pi r^2\frac{\Delta B}{\Delta t})\frac{1}{m{r}^2}\Delta t=-\frac{Q}{2m}\Delta B$
Since the magnetic field strength increases from zero to B, the final angular velocity of the ring will be
                                      $\omega =-\frac{QB}{2m}$