Question

A nonconducting ring of radius $r$ has charge $Q$. A magnetic field perpendicular to the ring changes at the rate of $\frac{dB}{dt}$. The torque experienced by the ring is

• A. zero
• B. $Q^2\frac{dB}{dt}$
• C. $\frac{1}{2}Q{r}^2\frac{dB}{dt}$
• D. $\pi r^{2}Q\frac{dB}{dt}$

Answer 1

The correct option is C $\frac{1}{2}Q{r}^2\frac{dB}{dt}$

The same emf is induced in a conducting and a nonconducting ring, and it is equal to $\pi r^2\frac{dB}{dt}$.

At any point on ring, $E=\frac{1}{2}r\frac{dB}{dt}$.

For any charge dQ on the ring, force $=dF=EdQ$, tangential to the ring

Torque about the centre due to $dF=dx=rdF=rEdQ$.

Total torque $=\sum rEdQ=r\left(\frac{1}{2}r\frac{dB}{dt}\right)Q$.