A non-conducting thin disc of radius R charged uniformly over one side with surface density `sigma`, rotates about its axis with an angular velocity `omega`. Find (a) the magnetic field induction at the centre of the disc (b) the magnetic moment of the disc.

Explanation

(a) Let us take a ring element of radius `r` and thickness `dr`, then charge on the ring element,
`d q = sigma 2 pi r dr`
and current, due to this elemtn, `di = ((sigma 2pi r dr) omega)/(2pi) = sigma omega r dr`
So, magentic induction at the centre, due to this element : `dB = (mu_(0))/(2) (di)/(r)`
and hence , from symmetry : `B = int dB = int_(0)^(R) (mu_(0) sigma omega r dr)/(r) = (mu_(0))/(2) sigma omega R`
(b) Magnetic moment of the element, considered,
`dp_(m) = (dl) pi r^(2) = sigma omega dr pi r^(2) = sigma pi omega r^(3) dr`
Hence, the sought magentic moment,
`p_(m) = int dp_(m) = int_(0)^(R) sigma pi omega r^(3) dr = sigma omega pi (R^(4))/(4)`