# Magnetic Moment of a spinning shell

1. Feb 25, 2008

### Ataman

Find the magnetic moment of a spinning shell of charge Q, radius R, and angular velocity $$\vec{\omega}$$.

My solution:

I split the sphere into infinitely small loops of current (from top to bottom) and add them up. I set my origin on the center of the sphere and integrate with respect to y from 0 to R times two to get the entire sphere.

The start of the problem:

$$\vec{\mu} \,= A\vec{I}$$
$$d\vec{\mu} \,= A(d\vec{I})$$

Solving for $$d\vec{I}$$:

$$d\vec{I} = \frac{dQ}{t}$$
$$d\vec{I} = \frac{\sigma dA}{t}$$
$$d\vec{I} = \frac{\sigma dA}{(\frac{2\pi}{\omega})}$$
$$d\vec{I} = \frac{\sigma \omega dA}{2\pi}$$

Plugging $$d\vec{I}$$ back:

$$d\vec{\mu} \,= \frac{\sigma \omega A dA}{2\pi}$$

Changing A and dA, so I could integrate in terms of dy:

$$A \,= \pi r^2$$
$$A \,= \pi (R^2-y^2)$$
$$dA \,= -2 \pi y dy$$

Then setting up the integral:
$$d\vec{\mu} \,= - \frac{2 \sigma \omega \pi^2}{2\pi} y (R^2-y^2) dy$$
$$d\vec{\mu} \,= - \sigma \omega \pi y (R^2-y^2) dy$$
$$\vec{\mu} \,= - 2 \sigma \omega \pi \int_{0}^{R} y (R^2-y^2) dy$$

Without sigma:
$$\vec{\mu} \,= - \frac {Q \omega}{2R^2} \int_{0}^{R} y (R^2-y^2) dy$$

This is incorrect, and I have a suspicion it has to do with the "dA" part.

If you don't mind his handwriting, http://nebula.deanza.fhda.edu/physics/Newton/4B/MagneticMomentSphere.jpg [Broken] is the correct way to do the problem. I am rather uncomfortable integrating over angles, so I opted for a different way.

I would like to know why my way does not work.

-Ataman

Last edited by a moderator: May 3, 2017