Question

A non-conducting insulating ring of mass m and radius $R$, having charge $Q$ uniformly distributed over its circumference is hanging by a insulated thread with the help of a small smooth ring (not rigidly fixed with bigger ring). A time varying magnetic field $B=B_{0}sin\omega t$ is switched on at $t=0$ and the ring is released at the same time. Work done by all forces on ring in $0$ to $\left(\frac{\pi }{\omega }\right)$ time interval is :

• A. $0$
• B. $\frac{1}{8}\frac{(B_{0}QR{)}^2}{m}$
• C. $\frac{1}{4}\frac{(B_{0}QR{)}^2}{m}$
• D. $\frac{1}{2}\frac{(B_{0}QR{)}^2}{m}$

Answer 1

The correct option is A $0$

Work done = $\frac{1}{2}I{\omega }^{{}^{\prime }2}$
${\omega }^{\prime }=\frac{Q{B}_0}{2m}sin\omega t$,
At $t=\frac{\pi }{\omega }$,
${\omega }^{\prime }=0$
Work done $=0$