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A thin non conducting ring of mass m, ra...

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.
The magnitude of induced emf of the closed surface of ring will be

A

`pia^2B_0`

B

`2a^2B_0`

C

zero

D

`1/2pia^2B_0`

Text Solution

Verified by Experts

The correct Answer is:
A
`|e|=|(dphi)/(dt)|=S(dB)/(dt)=(pia^2) B_0`
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