A long, hollow insulating cylinder of radius R and negligible thickness has a uniform surface charge density `sigma`. The cylinder rotates about its central axis at a constant angular speed `omega`. Choose the correct option(s):

Options

A. The charge crossing an imaginary line of length L on the surface of the cylinder, parallel to its axis, per unit time, is `omega sigma RL`.
B. The magnetic moment of a length L of the cylinder is `pi omega sigma R^(3) L`
C. The magnetic field intensity at a point on the axis of the cylinder, inside the cylinder, is `mu_(0)omega sigma R^(2)`
D. If the direction of rotation of the cylinder is reversed, the magnetic field at a point on the axis of the cylinder will also reverse direction

Answer

A, B, D

Explanation

Consider a length L of the cylinder. All the charge on this length will pass through the given line of length L parallel to the axis of the cylinder in the time it takes the cylinder to complete one full rotation.
Total charge on this length L of the cylinder,
`Q=sigma(2piR)L`
So, change crossing per unit time `=Q/(((2pi)/(omega)))=omega sigma RL`
This is like an equivalen current `I_(eq)` carried by the this length L of the cylinder.
so its magnetic moment `m=I_(eq)(piR^(2))=pi omeg sigma R^(3)L`
Now, this rotating cylinder is exactly equivalent to a long solenoid carrying a current.
Magnetic field at a point on the axis of a solenoid with N turns per unit length carrying a current I,
`B=mu_(0)NI`
This field can be looked at as `B=mu_(0)(NI)` , where NI is the “total current” carried by a unit length of the solenoid. In other words, it is the current per unit length. The current per unit length carried by the rotating cylinder is `omega sigma R`. So, field due at a point on the axis of the cylinder `B=mu_(0)(omega sigma R)`