Question

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.

Answer 1

Dipole moment of the ring
`p=[int_(-pi//2)^(pi//2)(lambda rd theta2r cos theta)]`, where `lambda=(q)/(pi r)`
`=4r^(2)l=(4q r)/(pi)`
From the configuration shown in fig. `vec(tau)=vec(p)xxvec(E )=pE sin theta`
or, `mr^(2)(d^(2)theta)/(dt^(2))=(4 qr)/(mr) E theta`
or `(d^(2)theta)/(dt^(2))=(4 qr)/(mr) E theta` Angular frequency of oscillation is
`omega=sqrt((4qE)/(mr))`
or `T=(2 pi)/(omega)=2pi sqrt((mr)/(4 qE)`