# Griffiths 5.42

1. Apr 8, 2008

### ehrenfest

1. The problem statement, all variables and given/known data

Calculate the magnetic force of attraction between the northern and southern hemispheres of a spinning charged spherical shell. (ex. 5.11)

2. Relevant equations

3. The attempt at a solution

I am stuck figuring out how to proceed. Should I calculate the B-field due to one hemisphere and then integrate over the Lorentz force on the other? Calculating he- B-field due to one hemisphere does not seem fun...

Last edited: Apr 9, 2008
2. Apr 9, 2008

anyone?

3. Apr 9, 2008

### Ben Niehoff

Hmm...I'm too lazy to look at Griffiths right now, but here's one approach: consider each hemisphere to be a series of circular current loops. Do you have an expression for the force between two parallel, co-axial current loops? If so, then integrate that over both hemispheres.

4. Apr 9, 2008

### Ben Niehoff

Actually, that's probably a rather complicated method.

You don't want to look at the B field of each hemisphere separately; you actually want the B field of the entire sphere, because each individual hemisphere exerts a self-force as well.

You should know, from some other formulas in Griffiths, that the B field on the interior of the spinning, charged sphere is uniform, right? It seems to me that you should be able to simply use this uniform field and apply the Lorentz force law.

5. Apr 9, 2008

### ehrenfest

But I thought that is exactly why we should look at each hemisphere individually--we don't want to include the "self-force" in our answer!

6. Apr 10, 2008

### Ben Niehoff

OK, it was bothering me that I couldn't immediately see how to do this problem, so I dug out Griffiths and looked at it. Here is how to find the solution:

First, read pages 211-212 about surface currents. Note that B is discontinuous at a surface current, and in order to find the force, you must average B_above and B_below. This is what I suspected, but I couldn't figure out why. Luckily, Griffiths points out why: go back and read section 2.5.3 (page 102), and he explains the argument for electric forces on surface charges.

Now, go back to the problem. You have a spinning, charged sphere. Therefore, it experiences both electric and magnetic forces. He asks for the "magnetic force of attraction", so maybe he wants you to ignore the electric force of repulsion; I'm not sure. Anyway, you can use a similar technique to calculate them both:

To find the force on the sphere, you follow these three steps:

1. Find the fields everywhere.
2. Find the force on a differential area element due to the total field.
3. Take only the vertical component of this force, and integrate over each hemisphere separately to get the (opposite and equal) forces on each.

Step 1 is easy: You know that B field is uniform on the interior of the sphere. There is probably a formula somewhere for it; I haven't dug around enough to find it. You don't actually need to calculate the external field, because you can use the boundary conditions for B to find the field immediately outside the sphere, which is all you really need.

For Step 2, apply the boundary conditions to get the field immediately outside the sphere, and average that with the field immediately inside. Now you know that

$$d{\vec F} = \vec K \times \vec B_{avg} da$$

For Step 3, simply take the Z component of the above, and integrate over the northern hemisphere. By symmetry, you know the integral over the southern hemisphere is equal and opposite.

Overall, the net force over the entire sphere is zero (after all, how could the sphere push itself in any particular direction?), but the top and bottom of the sphere do compress toward each other, and hopefully you should arrive at the formula given by Griffiths as the answer.

Last edited: Apr 10, 2008