Question

A nonconducting ring of radius r has charge Q. A magnetic field perpendicular to the plane of the ring changes at the rate dB/dt .The torque experienced by the ring is

(a) zero

(b) Qr2 (dB/dt)

(c) 1/2 Qr2 (dB/dt)

(d) πr2Q (dB/dt)

Answer 1

Correct Answer is: (c) 1/2 Qr2 (dB/dt)

The same emf is induced in a conducting and a nonconducting ring, and it is equal to πr2 dB/dt.

At any point on ring, E = 1/2 r dB/dt .

For any charge dQ on the ring, force = dF = EdQ, tangential to the ring

Torque about the centre due to dF = dτ = rdF = rEdQ.

Total torque = ΣrEdQ = r (1/2 r dB/dt) Q.