Question

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

• A. Zero
• B. $\frac{Q{r}^2}{2}\frac{dB}{dt}$
• 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 non-conducting ring, and it is equal to $\pi r^2\frac{dB}{dt}$. At any charge dQ on the ring, force dF = EdQ, tangential to the ring.
Torque about the centre due to $dF=d\tau =rdF=rEdQ$
Total torque$=\sum rEdQ=r\left(\frac{1}{2}r\frac{dB}{dt}\right)Q=\frac{Q{r}^2}{2}\frac{dB}{dt}$