Question

A uniformly charged ring of mass $m$, charge $q$, and radius $r$ is rotating at angular velocity $\omega$ in this plane as shown in the figure. Suddenly, a magnetic field $B$ is switched on which is perpendicular to the plane of the ring. Then,

• A. angular velocity of the ring may increase
• B. angular velocity of the ring will remain same
• C. change in angular velocity of the ring will be $\frac{Bq}{2m}$
• D. work done by the magnetic force will be zero

Answer 1

Answer: (A), (C), (D)

The correct options are
A angular velocity of the ring may increase
C change in angular velocity of the ring will be $\frac{Bq}{2m}$
D work done by the magnetic force will be zero


$\oint E.dl=\frac{d\varphi }{dt}$
$E\times 2\pi r=\frac{B\pi r^2}{\Delta t}$
$⟹E=\frac{Br}{2\Delta t}$
Due to this electric field, the torque on the ring is
$\tau =Eqr=\frac{B{r}^{2}q}{2\Delta t}$
By $\tau =\frac{\Delta L}{\Delta t}$,
$\Delta L=\tau \Delta t$
$\Delta L=I\Delta \omega =\frac{B{r}^{2}q}{2}$
$m{r}^2\Delta \omega =\frac{B{r}^{2}q}{2}$
$\Delta \omega =\frac{Bq}{2m}$ (may be positive or negative depending on direction of magnetic field)
Work done by the magnetic force is always zero.