A non-conducting ring of radius R and mass m | vedclass

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(C) According to Faraday's law of induction,the induced electric field $E$ is given by $\oint E \cdot d\ell = -\frac{d\phi}{dt}$.For a circular path o

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$\frac{8qR}{mg}$

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(C) According to Faraday's law of induction,the induced electric field $E$ is given by $\oint E \cdot d\ell = -\frac{d\phi}{dt}$.For a circular path of radius $R$,$E(2\pi R) = \pi R^2 \frac{dB}{dt}$.Given $B = 4t^2$,so $\frac{dB}{dt} = 8t$.Thus,$E(2\pi R) = \pi R^2 (8t) \implies E = 4Rt$.The torque $	au$ acting on the ring due to the induced electric field is $	au = (qE)R = q(4Rt)R = 4qR^2t$.The ring starts to rotate when the torque due to the induced electric field overcomes the maximum static frictional torque. The frictional force $f = \mu mg$ acts at the point of contact,but for the ring to rotate about its center,the torque provided by friction is $	au_f = fR = \mu mgR$.At $t = 2 \, s$,the torque due to the electric field is $	au = 4qR^2(2) = 8qR^2$.Equating the torques: $8qR^2 = \mu mgR$.Solving for $\mu$: $\mu = \frac{8qR}{mg}$.

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