Does a magnetized sphere contain energy

Question

I believe this question is similar to "does a permanent magnet contain energy", which I understand to be no, but I just want to be sure. Say we have uniformly magnetized sphere with magnetization $M_0\hat{z}$ and radius $R$. I understand the resulting fields to be: $$\vec{H}_{in} = -\frac{M_0}3\hat{z} $$ $$\vec{B}_{in} = \frac{2\mu_0M_0}3\hat{z} $$ For r < R, and $$\vec{H}_{out} = \frac{M_0}3\frac{R^3}{r^3}[2\cos(\theta)\hat{r}+\sin(\theta)\hat{\theta}] $$ $$\vec{B}_{out} = \mu_0 \vec{H}_{out} $$ for r > R. So to obtain total energy, I need to solve the volume integral: $$U_M=\frac{1}2 \int{\vec{H} \cdot \vec{B}dV} $$ --EDIT: Here are my intermediate steps if anyone sees any obvious errors. $$U_{in} = \frac{1}2(\frac{4}3 \pi R^3)(-\frac{2 \mu_0 M_0^2}9)= -\frac{4}{27} \pi R^3 \mu_0 M_0^2$$ $$U_{out} = \frac{1}2 \int_0^{2\pi} { \int_0^{\pi} { \int_{R}^{\infty}{ \mu_0 (\frac{M_0}3)^2 (\frac{R}{r})^6 [3\cos^2(\theta)+1] r^2 \sin{\theta} dr d\theta d\phi} } }$$ $$ U_{out} = \frac{\pi \mu_0 M_0^2 R^6}{9} \int_0^{\pi} { \int_{R}^{\infty}{ \frac{1}{r^4} [3\cos^2(\theta)+1] \sin{\theta} dr d\theta} }$$ --END EDIT

This solves as: $$U_{in} = -\frac{4}{27} \pi R^3 \mu_0 M_0^2$$ $$U_{out} = \frac{4}{27} \pi R^3 \mu_0 M_0^2$$ Meaning the total energy of the permanently magnetized system is zero. Is this just a specific case demonstrating that a permanent magnet has no energy? My confusion comes from trying to solve a problem with a magnetic material in a uniform magnetic field. Instead of solving the problem the traditional way, I wanted to find the energy stored in the uniform field, the energy stored in the magnetized sphere, and the "energy coupling" between these two bodies. Similar to a mutual inductance, if you will. But my argument quickly falls apart.

Another way of phrasing my question: I'm confused why one can't define a mutual inductance term with a permanent magnet, a source of magnetic flux.

Answer 1

Ah, I found the problem.

The analog, a polarized sphere is described here

I used the equation: $$ U_M = \frac{1}2 \int{ \vec{H} \cdot \vec{B} dV } $$ Which only described the energy produced by free currents. But here the energy is stored entirely in bound currents. So I must use the equation: $$ U_M = \frac{1}{2\mu_0} \int{ | \vec{B}|^2 dV } $$ Which accounts for all currents, and does result in a positive net energy.

EDIT:

Stratton is actually careful to define the energy of a magnetic material in unambiguous terms (Electromagnetic Theory, sections 2.15-2.18).

The energy of a magnetic body in a Magnetostatic field is given by: $$ U_M = \frac{1}2 \int{ \vec{M} \cdot \vec{B} dV } $$

By this, the energy of just the magnetic material of the sphere is given by: $$U_M = \frac{3}2 V_p \mu_0 K H_0^2 $$ Where Vp is the volume of the particle, H0 is the applied field, and K is the Clausius-Mossotti Factor defined as: $$K=\frac{\mu_r-1}{\mu_r+2}$$