A rod has a total charge Q uniformly distributed along its length L. If the rod rotates with angular velocity `omega` about its end, compute its magnetic moment.

Explanation

We can visualize the rod to consist of differential element dQ, which constitute a series of concentric current loop. The charge per unit length of the rod is `lambda`,
`lambda=Q/L`
So the charge on a differential element of length dl,
`dq=lambda dl`
The current `dl` due to rotation of this charge is given by
`dI=(dq)/((2pi/omenga)) =omega/(2pi)dq =omega/(2pi)lambdadl`

The magnetic moment of this differential current loop,
`dM=dI(pil^2)=(omega/(2pi) lambdadl)pil^2=(omega lambda)/2l^2dl`
To find total magnetic moment, we integrate
`M=(omega lambda)/2int+0^Ll^2dl=(omega lambdaL^3)/6`
Subtituting for `lambda`, we obtain `M=(QomegaL^2)/(6)`.