# B Field Inside of Sphere using Sep. Variables

1. Oct 29, 2013

### bowlbase

Done editing I hope.

1. The problem statement, all variables and given/known data
If Jf = 0 everywhere, then (as we showed in class), one can express H as the gradient of a scalar potential, W. W satisfies Poisson’s equation with ∇⋅M as the source. Use this fact to find the field inside a uniformly magnetized sphere. (Griffiths has some additional verbiage
example 6.1 (p. 264-5), which is this problem solved by another method.

2. Relevant equations

$H^\perp_{above} - H^\perp_{below}=-(M^\perp_{above}- M^\perp_{below})$

3. The attempt at a solution

My question is with the constraints and in particular the one I have in the equations section. I had this as $∇W^\perp_{in} - ∇W^\perp_{out}=M^\perp$ Since at r=R they are equivalent. I know that $M^\perp$ must writable as some version of M but I cannot determine what. I know the solutions manual has $M\hat{z}\hat{r}= Mcos(θ)$ but I don't understand how they have $\hat{z}$ in spherical coordinates...

Thanks for any clarification.

Last edited: Oct 29, 2013
2. Oct 29, 2013

### marcusl

Can you be more precise about your confusion regarding $\hat{z}$?

3. Oct 29, 2013

### bowlbase

More than the z vector I'm just confused how they got from M orthogonal to Mcos(θ). What is the reasoning or the process?

4. Oct 29, 2013

### marcusl

I don't own a copy of Griffiths and you haven't shown the whole problem, so I'm going to make a guess that the magnetization is along the z axis? The component of M normal to the boundary (at r=R) is proportional to cos(theta). You can see this intuitively: at theta=0, M is normal to the boundary so Mnormal=M. At 90 deg, M is tangential to the circle so Mnormal=0.