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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.

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  • $\begingroup$ I think that in general magnetized systems have a positive total energy (with respect to the unmagnetized state) if the magnetic energy both inside and outside of the magnet are taken into account. So I'm puzzled why the result here is zero. But even if there are no errors in the calculation, then the zero energy result must be just for the special case of a sphere because I'm pretty sure that the cases of a long prolate spheroid or a flat oblate spheroid will give positive total magnetic energies. $\endgroup$ – Samuel Weir Feb 27 '17 at 2:10
  • $\begingroup$ That is good to know, thank you! It is quite possible I have made a mistake in my calculations. I'll add more steps if anyone cares to double check my work. $\endgroup$ – Kthaxt Feb 27 '17 at 2:41
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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: $$K=\frac{\mu_r-1}{\mu_r+2}$$

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  • $\begingroup$ So what is your total energy, for your sphere ? And can you give some links about the interpretation of your expressions above ? $\endgroup$ – Cham Jul 15 '17 at 13:00
  • $\begingroup$ The best reference I found was Electromagnetic Theory by Stratton. See Chapter II, sections 2.15-2.18. The total energy of the system is given by my second expression, but the energy if the magnetic body (i.e. the sphere) is given by a separate expression, which I will include in an edit. $\endgroup$ – Kthaxt Oct 30 '17 at 23:47

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