consider a nonconducting ring of radius r and mass m which has a total charge q distributed uniformly on it the ring is rotated about its axis with an angular speed `omega`. (a) Find the equivalent electric current in the ring (b) find the magnetic moment `mu` of the . (c) show that`mu=(q)/(2m)` where l is the angular momentum of the ring about its axis of rotation.

Explanation

(a) `I = (q)/(T) = (qw)/(2 pi) [:. W = (2pi)/(T) rArr T = (2 pi)/(w)]`
`(b) m = nia = ia`
`= (qw)/(2 pi) pi r^(2) = (qwr^(2))/(2)`
Magnetic moment
`mu = ia = (qL)/(2m).