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A non-conducting ring of mass m, radius r and total charge Q rests on a rough horizontal surface in a region of perpendicular (transverse) magnetic field whose magnitude increases with time at the rate R (dB/dt = R). If mu is the coefficient of friction between the ring and the surface, find the minimum mu required for the ring to stay in rotational equilibrium (not start rotating).

1. Q*r*R/(2*m*g)
2. Q*r*R/(m*g)
3. Q*r*R/(3*m*g)
4. 2*Q*r*R/(m*g)

Correct answer: Q*r*R/(2*m*g)

Solution

The increasing field induces a circulating electric field E = (r/2)*(dB/dt) = (r/2)*R along the ring. The force on the ring's charge is Q*E, giving a driving torque tau_drive = Q*E*r = Q*(r/2)*R*r = Q*R*r²/2. The maximum static friction torque available is tau_fric = mu*N*r = mu*m*g*r (friction acts at radius r). For equilibrium tau_fric >= tau_drive: mu*m*g*r >= Q*R*r²/2, so mu >= Q*r*R/(2*m*g).

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