Charge Q is uniformly distributed over a nonconducting ring of radius R and mass m. It is made to rotate about an axis passing through its centre and perpendicular to its plone with an angular velocity of rotation `omega`. Find its magnetic dipole moment and its ratio with angular momentum.

Explanation

Angular velocity ` = omega`, Total charge = Q ,
Radius = R, Mass of the ring = m
Magnetic moment , ` mu = iA`
Where I = current and A = area of coil ` = piR^(2)`
Also , ` I = Q/t = (Qomega)/(2pi) * piR^(2) = 1/2 Q omegar^(2)`
Angular momentum of coil , `L = I omega`
I = moment of inertia of ring ` = mR^(2)`
` L = mR^(2) omega`
` :. ("Magnetic moment "(mu))/("Angular momentum "(L)) = 1/2 (QomegaR^(2))/(mR^(2)omega) = Q/(2m) `