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Question

(B) The changing magnetic field induces an electric field $E$ along the circumference of the ring. According to Faraday's law,$\oint E \cdot dl = -\fr

Vedclass · Accepted Answer

$\frac{QrR}{2mg}$

Vedclass · Answer

(B) The changing magnetic field induces an electric field $E$ along the circumference of the ring. According to Faraday's law,$\oint E \cdot dl = -\frac{d\phi}{dt}$.For a circular path of radius $r$,$E(2\pi r) = \pi r^2 \frac{dB}{dt} = \pi r^2 R$.Thus,the induced electric field is $E = \frac{rR}{2}$.The force on the charge $Q$ distributed on the ring is $F = QE = Q \frac{rR}{2}$.This force acts as a torque $	au$ about the center of the ring,tending to rotate it. However,the question asks for the condition of equilibrium against sliding or rotation. Assuming the force $F$ acts tangentially,the torque is $	au = F \cdot r = \frac{Qr^2R}{2}$.For the ring to remain in equilibrium against rotation,the frictional torque must balance this. The maximum frictional torque is $\mu N r = \mu mgr$.Equating,$\mu mgr \geq \frac{Qr^2R}{2}$,which simplifies to $\mu \geq \frac{QrR}{2mg}$.

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