Question

A ring of radius $r$ carries a charge of uniform linear charge density $\lambda$. A uniform magnetic field $B_0$ exists in the space as shown. Find the angular impulse acting on the wheel when the field is switched off.

Answer 1

$\pi \lambda r^3 B_0$

Answer 2

When the magnetic field changes, an induced electric field is generated. According to Faraday's Law of Induction, $\oint \vec{E} \cdot d\vec{l} = -\frac{d\Phi_B}{dt}$. For a uniform magnetic field $B$ changing with time, the induced electric field at the radius $r$ of the ring is $E = -\frac{r}{2} \frac{dB}{dt}$. The force on an infinitesimal charge element $dq = \lambda dl$ is $dF = dq E$. The torque $d\tau$ on this element is $d\tau = r dF = r (dq) E$. Substituting $dq = \lambda r d\theta$ and $E = -\frac{r}{2} \frac{dB}{dt}$, we get $d\tau = -\frac{\lambda r^3}{2} \frac{dB}{dt} d\theta$. Integrating over the ring gives the total torque $\tau = -\pi \lambda r^3 \frac{dB}{dt}$. The angular impulse is $J_\theta = \int \tau dt = \int \left(-\pi \lambda r^3 \frac{dB}{dt}\right) dt = -\pi \lambda r^3 \int dB$. Since the field is switched off, the change in magnetic field is $\Delta B = B_{final} - B_{initial} = 0 - B_0 = -B_0$. Thus, $J_\theta = -\pi \lambda r^3 (-B_0) = \pi \lambda r^3 B_0$.