A thin non conducting ring of mass `m`, radius a carrying a charge `q` can rotate freely about its own axis which is vertical. At the initial moment, the ring was at rest in horizontal position and no magnetic field was present. At instant `t=0`, a uniform magnetic field is switched on which is vertically downward and increases with time according to the law `B=B_0t`. Neglecting magnetism induced due to rotational motion of ring.
Angular acceleration of ring is

Options

A. `(qB_0)/(2m)`
B. `(qB_0)/(4m)`
C. `(qB_0)/m`
D. `(2qB_0)/m`

Answer

A

Explanation

`F=qE=1/2qaB_0`
`tau=Fa=1/2qa^2B_0`
`alpha=tau/I=((1/2qa^2B_0))/(ma^2)=(qB_0)/(2m)`