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A non-conducting ring of mass m, radius R, with total charge q uniformly distributed on its circumference is placed on a rough horizontal surface. A uniform vertical magnetic field B = 8t² (tesla) is switched on at t = 0. The ring starts rotating at t = 3 s. Find the coefficient of friction between the ring and the table.

1. qR/24mg
2. qR/16mg
3. 16qR/mg
4. 24qR/mg

Correct answer: 24qR/mg

Solution

By Faraday's law, induced E at radius R: E*2*pi*R = pi*R²*(dB/dt), so E = R*dB/dt/2 = R*16t/2 = 8Rt. Force on charge q: F = qE = 8qRt. Torque = FR = 8qR²*t. At t=3: torque = 24qR². Max static friction torque = mu*mg*R. At the moment rotation starts: 24qR² = mu*mg*R => mu = 24qR/mg.

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