- #1

Ataman

- 18

- 0

**The problem reads:**

Find the magnetic moment of a spinning shell of charge

**Q**, radius

**R**, and angular velocity [tex]\vec{\omega}[/tex].

**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:

[tex]\vec{\mu} \,= A\vec{I}[/tex]

[tex]d\vec{\mu} \,= A(d\vec{I})[/tex]

Solving for [tex]d\vec{I}[/tex]:

[tex]d\vec{I} = \frac{dQ}{t}[/tex]

[tex]d\vec{I} = \frac{\sigma dA}{t}[/tex]

[tex]d\vec{I} = \frac{\sigma dA}{(\frac{2\pi}{\omega})}[/tex]

[tex]d\vec{I} = \frac{\sigma \omega dA}{2\pi}[/tex]

Plugging [tex]d\vec{I}[/tex] back:

[tex]d\vec{\mu} \,= \frac{\sigma \omega A dA}{2\pi}[/tex]

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

[tex]A \,= \pi r^2[/tex]

[tex]A \,= \pi (R^2-y^2)[/tex]

[tex]dA \,= -2 \pi y dy[/tex]

Then setting up the integral:

[tex]d\vec{\mu} \,= - \frac{2 \sigma \omega \pi^2}{2\pi} y (R^2-y^2) dy[/tex]

[tex]d\vec{\mu} \,= - \sigma \omega \pi y (R^2-y^2) dy[/tex]

[tex]\vec{\mu} \,= - 2 \sigma \omega \pi \int_{0}^{R} y (R^2-y^2) dy[/tex]

Without sigma:

[tex]\vec{\mu} \,= - \frac {Q \omega}{2R^2} \int_{0}^{R} y (R^2-y^2) dy[/tex]

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 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: