# What is the correct expression for the magnetic energy density inside matter?

I'll use a magnetized sphere as an example, of radius $R$, with a magnetization density $\vec{M}$. The magnetic moment of the sphere is $\vec{\mu} = \vec{M} \, V$. The magnetic field inside and outside is well known : \begin{align}\tag{1} \vec{B}_{\text{inside}} &= \frac{\mu_0 \, \vec{\mu}}{2 \pi R^3}, \\[12pt] \vec{B}_{\text{outside}} &= \frac{\mu_0}{4 \pi} \Big( \frac{3 (\vec{\mu} \cdot \vec{r}) \vec{r}}{r^5} - \frac{\vec{\mu}}{r^3} \Big). \tag{2} \end{align} Then what is the correct way of calculating the total energy of that sphere ? \begin{align}\tag{3} U_{\text{magn}} &= \int \frac{B^2}{2 \mu_0} \, d^3 x \\[12pt] &= \frac{1}{2 \mu_0} \int_{\mathcal{V}_{\text{inside}}} B_{\text{inside}}^2 \, d^3 x + \frac{1}{2 \mu_0} \int_{\mathcal{V}_{\text{outside}}} B_{\text{outside}}^2 \, d^3 x, \tag{4} \end{align} or this ? \begin{align}\tag{5} U_2 &= \frac{1}{2}\int \vec{B} \cdot \vec{H} \, d^3 x, \end{align} Or another expression involving the material properties of the sphere ?

What is the proper interpretation of this equation : $$\tag{6} \int_{\text{all space}} \vec{B} \cdot \vec{H} \, d^3 x = 0,$$ as shown in Jackson (second edition), page 207 (exercise 5.13).

• this is not "physics" it is just vector analysis for if $\textbf{curl u}=0$ and $\text{div}\textbf{v}=0$ with $r^2 |\textbf{u}|$ and $r^2 |\textbf{v}|$ bounded at infinity then it is always true that $\int \textbf{u}\cdot\textbf{v} dV=0$ – hyportnex Jul 15 '17 at 17:16
• @hyportnex, well ok, but this doesn't give the physical interpretation of (5) (maybe I should remove (6) from the question, since it's a bit irrelevent). I think that the total magnetic energy in a material is given by (4), and that (5) gives only a part of it. – Cham Jul 15 '17 at 17:35
• my interpretation is that (6) is a warning: for magnetostatic fields H and B somewhere must be sufficiently oppositely directed otherwise the integral cannot be zero. And sure enough, for a uniformly polarized sphere the H and B fields inside are exactly opposite (anti-parallel) while outside are parallel with each other. – hyportnex Jul 15 '17 at 18:59

The energy of the magnetic field is the work required to establish a general steady-state distribution of currents and fields. This work is, in infinitesimal form,

$$\label{0}\tag{0} \delta W = \frac 1 2 \int (\delta \mathbf A \cdot \mathbf J) \ d^3 x$$

where $\mathbf J$ is the current density.

If we are interested in work done on the free (macroscopic) currents, we have (a):

\begin{align} \delta W & = \frac 1 2 \int (\delta \mathbf A \cdot \mathbf J_f) \ d^3 x\\ &=\int \delta \mathbf A \cdot (\nabla \times \mathbf H) \ d^3x\\ &=\int [\mathbf H \cdot (\nabla \times \delta \mathbf A) +\nabla \cdot(\mathbf H \times \delta \mathbf A)] \ d^3x \end{align}

Where $\mathbf A$ is the vector potential and $\mathbf H$ is the magnetic field (b). Assuming a localized field distribution, the second term in the integral vanishes, and using $\mathbf B = \nabla \times \mathbf A$ we get

$$\delta W = \int \mathbf H \cdot \delta \mathbf B \ d^3x$$

If we assume that the material is linear, i.e. that $\mathbf B = \mu \mathbf H$, we have

$$\mathbf H \cdot \delta \mathbf B = \frac 1 2 \delta( \mathbf H \cdot \mathbf B)$$

Therefore we finally get the following expression for the energy of the magnetic field in the presence of linear materials:

$$U = \frac 1 2 \int \mathbf H \cdot \mathbf B \ d^3 x = \frac 1 {2 \mu} \int B^2 \ d^3 x$$

where the magnetic energy density is written in the form

$$\tag{1}\label{1} u = \frac 1 2 \mathbf H \cdot \mathbf B = \frac {B^2} {2 \mu}$$

To derive \ref{1}, we made use of the macroscopic form of Maxwell's equations. In particular, we are assuming using for the 4th equation the form (neglecting the displacement current):

$$\nabla \times \mathbf H = \mathbf J_f$$

where $\mathbf J_f$ is the free (macroscopic) current.

This means that \ref{1} represents the work done on free currents when establishing the steady-state current distribution of currents and fields.

It would also be possible to use the microscopic form of Maxwell's equations, in particular

$$\nabla \times \mathbf B = \mu_0 \mathbf J$$

where $\mathbf J$ is the total current, i.e. the sum of free and bound currents:

$$\mathbf J = \mathbf J_f + \mathbf J_b$$

In this case, there is no $\mathbf H$ vector and the magnetic energy density is (c)

$$\tag{2}\label{2} u' = \frac{B^2}{ 2 \mu_0}$$

This represents the work done on every current when establishing the magnetic field, including the bound currents, i.e.

$$\begin{split} u & = u_f\\ u'& = u_f + u_b\\ \end{split}$$

This means that the energy required to establish the bound currents can be calculated for a linear material as

$$\tag{3}\label{3} u_b = u'-u = \frac{B^2}2 \left( \frac 1 {\mu_0} - \frac 1 {\mu} \right) = \frac{B^2} 2 \left( \frac{\mu-\mu_0}{\mu \mu_0} \right)$$

Since (for a linear material) we have

$$\mathbf M = \left( \frac{\mu-\mu_0}{\mu \mu_0} \right) \mathbf B$$

where $\mathbf M$ is the magnetization, we can write \ref{3} as

$$\tag{4}\label{4} u_b = \frac 1 2 \mathbf M \cdot \mathbf B$$

And indeed this expression is valid for every material, not just for linear ones (d).

As for the expression

$$\int \mathbf H \cdot \mathbf B \ d^3 x = 0$$

on Jackson's book the full text says:

A magnetostatic field is due entrirely to a localized distribution of permanent magnetization. Show that $$\int \mathbf H \cdot \mathbf B\ d^3 x = 0$$ provided the integral is taken over all space.

So this is a statement that (in the presence of linear materials) the work done on free currents when establishing a magnetostatic field is 0. I suspect this to be more general, i.e. valid for any kind of relation between $\mathbf H$ and $\mathbf B$, but at the moment I cannot prove it.

(a) J.D. Jackson, Classical Electrodynamics (1962) 6.2

(b) I adopt the nomenclature in which $\mathbf H$ is the magnetic field and $\mathbf B$ the magnetic-flux density (or magnetic induction), which is the one used by Jackson.

(c) D.J. Griffiths, Introduction to Electrodynamics, 3rd ed. (1999), 7.2.4 and 8.1.2. It is especially interesting to read the footnote at page 348.

(d) B.D. Popovic Evaluation of magnetic energy density in magnetised matter, PROC. IEE, Vol. 113, No. 7, July 1966.

• In the case of the magnetized sphere defined in the question, we do know the bound currents. They are all on the surface, and this is why I considered energy (3). By the way, you should add a tag to your main equations. – Cham Jul 16 '17 at 18:58
• @Cham I refined the answer. I think I finally got how it works. – valerio Jul 17 '17 at 8:59
• In the case of my sphere, it's very easy to calculate de "bound" energy from your (4), and it does give exactly the same as the total energy from your (2) (and the calculations are so much easier, it's not even funny !). So It's consistent. – Cham Jul 17 '17 at 12:00
• Is there a way to remove the assumption that the material is linear? It appears there should be since you mention some of the resulting equations are true in general. How does that work? Note that the question involves a permanently magnetized material, so this is not a linear material. – JJMalone Oct 9 '17 at 17:10
• @JJMalone Equation (2) is true in general and there is no assumption of linearity of the material. The problem is to calculate $u_b$ and $u_f$ separately. You can use either (4) (which is general, see reference (d)) and find $u_b$ or take the general expression for $\delta W$ and find $u_f$, but you need the constitutive relation between $M$ and $B$ in the first case, or $H$ and $B$ in the second case. In general things will be complicated because the integral can depend on the history of the sample (hysteresis). See also farside.ph.utexas.edu/teaching/em/lectures/node78.html – valerio Oct 11 '17 at 9:12