Consider a non conducting plate of radius `a` and mass `m` which has a charge `q` distributed uniformly over it, The plate is rotated about its own axis with a angular speed `omega`. Show that the magnetic moment `M` and the angular momentum `L` of the plate are related as `(M)/(L)=(q)/(2m)`.

Explanation

If `sigma` is the surface charge density , then `q=sigmapia^(2)` Current `i=sigmaomegar dr`. The magnetic moment of the element ring `dM=(i d A)=sigmaomegar dr(pir^(2))=pi sigma omegar^(3)dr`
`M=pisigmaomegaunderset(0)overset(a)intr^(3)dr=(pisigmaomegar^(4))/(4)`
`=(pi a^(3)sigma)(omegaa^(2))/(4)=(qomegaa^(2))/(4)`
The angular momentum of the disc about its axis
`L=(ma^(2))/(2)omega.` The ratio `(M)/(L)=(4)/((omegama^(2))/(2))=(q)/(2m)`