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.

Answer

6

Explanation

We can visualize the rod to consist of differential elements Dq, which constitute a series of concentric current loops. The charge per unit length of the rod `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
`dl=(dq)/((2pi//omega))=(omega)/(2pi)dq=(omega)/(2pi)lambda dl`
To find total magnetic moment, we integrate
`mu=(omegalambda)/(2)int_(0)^(L)l^(2)dl=(omegalambdaL^(3))/(6)`
Substituting for `lambda` , we obtain `mu=(QomegaL^(2))/(6)`