# Sphere with permanent radial magnetization

1. Apr 9, 2014

### maxverywell

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

We have a sphere of radius $a$ with permanent magnetization $\mathbf{M}=M\hat{e}_{\mathbf{r}}$.
Find the magnetic scalar potential.

2. Relevant equations

$$\Phi_M(\mathbf{x})=-\frac{1}{4\pi}\int_V \frac{\mathbf{\nabla}'\cdot\mathbf{M}(\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|}d^3 x' +\frac{1}{4\pi}\int_S \frac{\mathbf{n}'\cdot\mathbf{M}(\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|}da'$$

3. The attempt at a solution

$$\mathbf{\nabla}'\cdot\mathbf{M}(\mathbf{x}')=\frac{2M}{r'}$$

$$\mathbf{n}'\cdot\mathbf{M}(\mathbf{x}')=M$$

$$\Phi_M(\mathbf{x})=-\frac{M}{2\pi}\int_{0}^{a} r'dr'\int\int\frac{1}{|\mathbf{x}-\mathbf{x}'|}d\Omega' +\frac{Ma^2}{4\pi}\int \int\frac{1}{|\mathbf{x}-\mathbf{x}'|}d\Omega'$$

I expanded the $1/|\mathbf{x}-\mathbf{x}'|$ in terms of spherical harmonics (and because of the spherical symmetry we have $m=0,\ell=0$) and solved the integrals. What I got is:
$$\Phi_M(\mathbf{x})=-2M\int_{0}^{a}\frac{r'}{r_{>}}dr'+\frac{Ma^2}{r_{>}}$$

where $r_{>}=max(r,a)$

Inside the sphere we have $r_{>}=a$, therefore:
$$\Phi_M(r)=\frac{-2M}{a}\int_{0}^{a}r'dr'+Ma=0$$

This is constant. However the $\Phi_M(r)$ inside the sphere has to satisfy the Poisson equation:
$$\nabla^2 \Phi_M(r)=\nabla\cdot \mathbf{M}=\frac{2M}{r}$$

This is not true for the potential that I found..

Last edited: Apr 9, 2014
2. Apr 10, 2014

### maxverywell

We could also compute the vector potential $\mathbf{A}$

$$\mathbf{A}(\mathbf{x})=\frac{\mu_0}{4\pi}\int_V \frac{\mathbf{\nabla}'\times\mathbf{M}(\mathbf{x}')}{|\mathbf{x}-\mathbf{x}'|}d^3 x' +\frac{\mu_0}{4\pi}\int_S \frac{\mathbf{M}(\mathbf{x}')\times\mathbf{n}'}{|\mathbf{x}-\mathbf{x}'|}da'$$

$$\mathbf{\nabla}'\times\mathbf{M}(\mathbf{x}')=M\mathbf{\nabla}'\times \hat{e}_{\mathbf{r}'}=0$$

$$\mathbf{M}(\mathbf{x}')\times\mathbf{n}'=M\hat{e}_{\mathbf{r}'}\times \hat{e}_{\mathbf{r}'}=0$$

So $\mathbf{A}=0$.

Both methods give zeros. Where is the problem?

3. Apr 10, 2014

### TSny

Interesting. I think such a sphere is impossible.

Note that the divergence of M is undefined at the center of the sphere (r = 0). So, you have a singularity there that would need to be handled when integrating over the volume of the sphere to find $\Phi_M$.

It is easy to see that $\vec{B} = 0$ everywhere. If $\vec{B} \neq 0$ at some point located a distance r from the center of the magnetized sphere, then by spherical symmetry $\vec{B}$ is radial at that point and there would exist a radial $\vec{B}$ at every point on the surface of a sphere of radius r. Thus, there would be a nonzero magnetic flux through a closed surface which would violate the law $\vec{\nabla} \cdot \vec{B} = 0$ everywhere.