Question

A non-conducting ring having charge q uniformly distributed over its circumference is placed on a rough horizontal surface. A vertical time varying magnetic field $B=4t^2$ is switched on at time $t=0$. Mass of the ring is 'm' and radius is R. The ring starts rotating after 2 s. The coefficient of friction between the ring and the table is:

• A. $\frac{4qmR}{g}$
• B. $\frac{2qmR}{g}$
• C. $\frac{8qR}{mg}$
• D. $\frac{qR}{2mg}$

Answer 1

Answer: (C)

$\frac{8qR}{mg}$

$e=\frac{-d\varphi }{dt}=-\frac{d(B.A)}{dt}=\pi R^2\frac{d\left(4t^2\right)}{dt}=-8\pi R^{2}t$

min. force required to move ring is $=\mu mg$

where $\mu$ is coeff. of friction

at $t=2Sec$ $emf,V=16\pi R^2$

Force on ring = $qE=\frac{qV}{2\pi R}=\frac{16q\pi R^2}{2\pi R}=8qR=\mu mg$

So, $\mu =\frac{8qR}{mg}$