- #1

Argelium

- 25

- 7

## Homework Statement

A cylinder with radius ##R## and height ##h## which has a distributed charge on its surface with density ##\sigma## spins over its axis with angular velocity ##\omega##.

If the cylinder has a mass density ##\rho##, find the relationship between magnetic momentum and angular momentum.

## Homework Equations

$$L=I\omega$$

$$m_{mag}=IA$$

## The Attempt at a Solution

So my professor gave a hint that there should be no radius or height terms. However I attempted this:

First of all we must calculate two magnetic momentums, the one due to lateral surface area and another due to the cylinder heads.

For the first one consider a ##dI## on the surface area.

$$dI=\frac{dq \omega}{2\pi} = \frac{\omega\sigma}{2\pi}Rd\theta dz$$

Then the magnetic momentum of the sides is

$$m=\omega\sigma R^2h \int d\theta dz$$

$$m=\omega RhQ$$

Then for each heads we can find the magnetic momentum to be

$$\frac{1}{4}\omega R^2 Q$$

So summing them all (both vectors have the same direction: parallel to angular velocity) we obtain

$$m_{tot}= \omega R(hQ+\frac{RQ}{2}$$

Doing the calculations yields that

$$L=\frac{R}{2h+R}\frac{M}{Q} m$$

Where ##M## is the mass.

Am I missing something or is my work correct?