A non-conducting ring of mass `m` and radius `R` has a charge `Q` uniformly distributed over its circumference. The ring is placed on a rough horizontal surface such that plane of the ring is parallel to the surface. A vertical magnetic field `B = B_0t^2` tesla is switched on. After 2 a from switching on the magnetic field the ring is just about to rotate about vertical axis through its centre.
(a) Find friction coefficient `mu` between the ring and the surface.
(b) If magnetic field is switched off after `4 s`, then find the angle rotated by the ring before coming to stop after switching off the magnetic field.

Explanation

(a) `E = ( R)/(2)(dB)/(dt)` rarr `E = b_(0)Rt`
Force on the ring `F = QE = B_(0)QRt`. This force is tangential to the ring. The ring starts rotating when torque of this force is greater than the torque due to maximum friction `(f_(max) = mumg)`.
`tau_(F) ge tau_(fric, max), FR gt = mumgR` rarr `F gt mumg`

`B_(0)QRt = mumg`
Hence, `mu = (B_(0)QRt)/(mg)`
Given, `t = 2 s` rarr `mu = (2B_(0)QR)/(mg)`
(b) After `2s`
Net torque `tau = tau_(F) - tau_(f max) = B_(0)QR^(2)t - mugR`
`= B_(0)QR^(2)t - (2B_(0)QR)/(mg)mgR`
rarr `tau = B_(0)QR^(2)(t - 2)` rarr `Ialpha = B_(0)QR^(2)(t - 2)`
`mR^(2)((domega)/(dt)) = B_(0)QR^(2)(t - 2)`
rarr `domega = (B_(0)QR^(2))/(mR^(2))(t - 2)dt`
rarr `int_(0)^(omega)domega = (B_(0)Q)/(m)int_(2)^(4)(t - 2)dt`
rarr `omega = (2B_(0)Q)/(m)`
( c ) If magnetic field is switched off after `4s`, only force present is frictional force which will retard the motion.
Retarding torque `tau = tau_("friction")`
Angular retardation `alpha = (tau_("friction"))/(I)` rarr `alpha = (mumgR)/(mR^(2)) = (mug)/(R)`
Using `omega^(2) = omega_(0)^(2) + 2alpha theta` rarr `0 = (2(B_(0)Q)/(m))^(2) - 2((mug)/(R))theta`
rarr `theta = 2((B_(0)Q)/(m))^(2) (R)/(mug)`