A non-conducting thin disc of radius `R` and mass `m` having charge uniformly over one side with surface density `sigma` rotates about its axis with an angular velocity `omega`. Find :
(a) the magnetic induction at the centre of the disc,
(b) the magnetic moment of the disc.
( c) the ratio of magnetic moment and angular momentum of disc.

Explanation

(a) Consider a ring of radius `x` and thickness `dx`

Area `dA = 2pi x dx`
Charge in this area
`dq = sigma dA = sigma. 2pi x dx`
Current `i = (q)/(T) = (sigma. 2 pi x dx)/(2pi//omega)`
`= sigma omega x dx`
Magnetic field due to ring
`dB = (mu_(0)i)/(2x) = (mu_(0) sigma omega x d x)/(2x)`
`B_(0) = (mu_(0) sigma omega)/(2) int_(0)^(R ) dx =(mu_(0) sigma omega R)/(2)`
(b) Magnetic moment of ring
`dM = i. pi x^(2)`
`M = sigma omega pi int_(0)^(R )x^(3) dx`
`= (sigma omega pi R^(4))/(4)`
( c) Angular momentum of disc
`L = Iomega = (1)/(2) mR^(2) omega, I:M.I`. of disc about its axis
`(M)/(L) = (sigma omega pi R^(4)//4)/(mR^(2) omega//2) = (1)/(2) (sigma pi R^(2))/(m) = (q)/(2m)`
where `q(= sigma. pi R^(2))`: total charge on disc.