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

Answer

A

Explanation

In this case, convection current due to mechanic transfer of charge is along the surface of cylinder perpendicular to the length of the cylinder.
We consider a ring element of thickness `dx` of the cylinder The charge on the considered element is `dq=lambdadx`
`therefore` The convection current is due to motion of considered element is `dl=(dq)/T=(dq)/((2pi)/omega)`
`(omegadq)/(2pi)=(omegalambdadx)/(2pi)`
The magnetic field at point `p` due to considered ring element is `B=(mu_(0)dia^(2))/(2(a^(2)+x^(2))^(3//2))`
or `B=(mu_(0)a^(2))/(2(a^(2)+x^(2))^(3//2))(omegalambda)/(2pi)dx=(mu_(0)a^(2)omegalambda)/(4pi(a^(2)+x^(2))^(3//2))dx`
`therefore dphi=(mu_(0)a^(2)omegalambdapia^(2))/(4pi).(dx)/((a^(2)+x^(2))^(3//2))`