- #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?