# Surface current density and magnetic dipole moment

An insulating thin disk of radius R carries a uniformly distributed electric surface charge density. The total charge is Q, and the mass of the disk is M. The disk rotates with angular velocity ($\hat \omega_z$) about its center.

Please show me how to find the surface current density K(s), and the Magnetic Dipole Moment m. (Please do not just post a solution from Griffiths or some other book. Please explain, as I’ve done this problem myself but I think I’m getting the wrong solution.)

The infinitesimal current is

$d I =Q/(S*t)*d S$

where $d S= r*d L$ is a the area between circles with radius $r$ and $(r+d L)$

Thus $t =T$ which means one period of the rotation

The infinitesimal density of current as a function of $r$ is

$d J(r) = d I/d L = Q/(S*t) *(r*d l)/d L =Q/(S*t) * r$

the total density of current is

$J =\int _0 ^R {[Q/(S*T)]} *r d r = (QR^2)/(2*S*T) =QR^2/(2\pi*R^2*T) =$

$=Q/(2*\pi*T) =(Q*\omega)/(2*\pi)$

The dipole moment magnitude

$m = I*S =(J*S)*S = Q*\omega*S^2/(2*\pi) =(Q*\omega*\pi*R^4)/2$

(of course the dipole is perpendicular to the plane of the disk, in the direction of $\hat z$.)