$A$ line charge $\lambda$ per unit length is lodged uniformly onto the rim of a wheel of mass $M$ and radius $R$. The wheel has light non-conducting spokes and is free to rotate without friction about its axis. $A$ uniform magnetic field extends over a circular region within the rim. It is given by,
$B = -B_{0} \hat{k}$ for $r \leq a$ (where $a < R$)
$B = 0$ otherwise.
What is the angular velocity of the wheel after the field is suddenly switched off?

(D) When the magnetic field is switched off,a changing magnetic flux induces an electric field. According to Faraday's law,the induced electromotive force $(EMF)$ is given by $\oint \vec{E} \cdot d\vec{l} = -\frac{d\Phi_B}{dt}$.
For a circular path of radius $r \leq a$,the magnetic flux is $\Phi_B = B \cdot \pi r^2$.
The induced electric field $E$ at radius $r$ is $E(2\pi r) = \frac{d}{dt}(B_0 \pi r^2) = \pi r^2 \frac{dB_0}{dt}$.
Thus,$E = \frac{r}{2} \frac{dB_0}{dt}$.
The torque $\tau$ acting on the rim (where charge $Q = \lambda (2\pi R)$ is located) is $\tau = r_f \times F = R(qE)$.
Since the charge is on the rim,$r=R$ for the force calculation,but the induced field is inside. The force on the rim charge $dq$ is $dF = dq E = (\lambda dl) E$.
The total torque is $\tau = \int R E dq = R E (\lambda 2\pi R) = 2\pi R^2 \lambda E$.
Substituting $E$ at $r=a$ (the boundary of the field),the impulse is $\int \tau dt = \Delta L = I\omega$.
$I = MR^2$.
$\int (2\pi R^2 \lambda) \frac{a}{2} \frac{dB_0}{dt} dt = MR^2 \omega$.
$\pi R^2 \lambda a B_0 = MR^2 \omega$.
$\omega = \frac{\pi a^2 \lambda B_0}{MR}$.