$A$ magnetic field $B$ is confined to a region $r \le a$ and points out of the paper (the $z$-axis),$r = 0$ being the centre of the circular region. $A$ charged ring (charge $= Q$) of radius $b$,$b > a$ and mass $m$ lies in the $xy$-plane with its centre at the origin. The ring is free to rotate and is at rest. The magnetic field is brought to zero in time $\Delta t$. Find the angular velocity $\omega$ of the ring after the field vanishes.

(D) The change in magnetic flux $\Delta \phi$ through the ring is $\pi a^2 B$. According to Faraday's law,the induced electromotive force (emf) is $\varepsilon = \frac{\Delta \phi}{\Delta t} = \frac{B \pi a^2}{\Delta t}$.
This induced emf creates an induced electric field $E$ along the ring such that $\varepsilon = E(2 \pi b)$. Thus,$E = \frac{B a^2}{2 b \Delta t}$.
The force on the charge $Q$ is $F = QE = \frac{Q B a^2}{2 b \Delta t}$.
The torque $\tau$ acting on the ring is $\tau = F \cdot b = \frac{Q B a^2}{2 \Delta t}$.
Using the impulse-momentum theorem for rotation,$\tau \Delta t = \Delta L = I \omega$,where $I = m b^2$ is the moment of inertia of the ring.
Substituting the values: $\left( \frac{Q B a^2}{2 \Delta t} \right) \Delta t = m b^2 \omega$.
Therefore,$\omega = \frac{Q B a^2}{2 m b^2}$.