consider a nonconducting ring of radius r and mass m which has a totoal chargej 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)L` where l is the angular momentum of the ring about its axis of rotation.

Explanation

Let ring is rotated in anticlockwise direction about its axis.
(a) Current in the ring
`i = (q)/(T) = (q)/(2pi//omega) = (qomega)/(2pi)`, anticlockwise

(b) Magnetic moment of the ring
`mu = iA = (q omega)/(2pi).pi r^(2) = (q omega r^(2))/(2), o.`
( c) Angular momentum of ring
`L = I omega, I: M.I.` of ring about its axis `= m r^(2) omega, o.`
`(mu)/(L) = (q omega r^(2))/(2mr^(2) omega) = (q)/(2m)`
`mu = (q)/(2m)L`
Since `mu` and `L` are in the same direction,
`vec(mu) = (q)/(2m)vec(L)`