A thin non-conducting ring of mass m carrying a charge Q can freely rotate about its axis. Initially, the ring is rest and no magnetic field is present. Then a uniform field of magnetic induction was switched on, which was perpendicular to the plane of the ring and increased with time as a given function B(t). The angular velocity $\omega$(t) of the ring as a function of the field B(t) will be given by

detailed solution

Correct option is B

The changing magnetic flux through the ring (due to changing field B) produced an electrical field along the ring (tangential to the ring at every point), the magnitude of the electrical field being E→=−R2dBdt, the minus sign indicating that the electrical field opposes the change of the magnetic field. This electric field produces a tangential force F on the charge q carried by the ring, |F|=|E|⋅q=Rq2dBdt Angular acceleration =dωdt=|F|RI=|F|RmR2, where I is the moment of inertia of the ring (since the ring is non-conducting, the force on the charge tending to produce a current act on the ring itself)Hence, dωdt=1mRRq2dBdt=q2mdBdtIntegrating, |ω|=q2m|B→(t)|