A long cylinder of radius `a` carrying a unifrom surface charge rotates about its axis with an anglur velocity `omega`. Find the magnetic field energy per unit length of the cylinder if the linear charge density equals `lambda` and `mu = 1`.

Explanation

When the cylinder with a linear charge densiy `lambda` rotates with a circular frequency `omega`, a surface current density `(charg e//l eng th xx time)` of `I = (lambda omega)/(2pi)` is set up.
The direction of the surface current is normal to the plane of paper at `Q` and the contribution of this current to the magnetic field at `P` is
`d vec(B) = (mu_(0))/(4pi) (i (vec(e) xx vec(r)))/(r^(3)) dS` where `vec(e)` is the direction of the current to the magnitude, `|vec(r) xx vec(r)| = r`, since `vec(e)` is normal to `vec(r)` and the direction of `d vec(B)` is shown.
It's component, `d vec(B)_(||)` cancels out by cylindrical symmetry. The component that surves is,
`|vec(B)_(_|_)| = (mu_(0))/(4pi) int (id S)/(r^(2)) cos theta = (mu_(0)i)/(4pi) int d Omega = mu_(0) i`,
where we have used `(dS cos theta)/(r^(2)) = d Omega` and `int d Omega = 4pi`, the total solid angle around any point.
The magnetic field vanishes outside the cylinder by similar argument.
The total energy per unit lenght of the cylinder is,
`W_(1) = (1)/(2 mu_(0)) mu_(0)^(2) ((lambda omega)/(2pi))^(2) xx pi a^(2) = (mu_(0))/(8pi) a^(2) omega^(2)`