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32  Magnetization and the Field H

    Michael Fowler, UVa

Magnetization

Just as dielectrics respond to an imposed electric field by becoming polarized, that is, creating a local electric dipole density $\overrightarrow{P}\left(\overrightarrow{r}\right),$ so an imposed magnetic field induces what is equivalent to a magnetic dipole density, very small in most materials but very large in the ferromagnetic materials considered here. This dipole density, $\overrightarrow{M}\left(\overrightarrow{r}\right),$  is called the magnetization.  It is a more complicated phenomenon than polarization, we’ll look at it in more detail in the next lecture.  For now, though, we’ll just assume that a small volume $\Delta V\left({\overrightarrow{r}}^{\prime }\right)$ of material acquires a magnetic dipole moment $\Delta V\left({\overrightarrow{r}}^{\prime }\right)\overrightarrow{M}\left({\overrightarrow{r}}^{\prime }\right),$ identical in character to the dipole moment of a small local current distribution, and therefore (as discussed in the previous lecture) having a field with vector potential

$$\Delta \overrightarrow{A}\left(\overrightarrow{r}\right)=\frac{{\mu }_0}{4\pi }\frac{\overrightarrow{M}\left({\overrightarrow{r}}^{\prime }\right)\times \left(\overrightarrow{r}-{\overrightarrow{r}}^{\prime }\right)}{{\left|\overrightarrow{r}-{\overrightarrow{r}}^{\prime }\right|}^3}\Delta V\left({\overrightarrow{r}}^{\prime }\right).$$

For a more general magnetostatic system having steady electric currents plus some magnetization density, the total vector potential has contributions from both, so integrating over the whole space:

$$\overrightarrow{A}\left(\overrightarrow{r}\right)=\frac{{\mu }_0}{4\pi }{\displaystyle \int d^3{r}^{\prime }}\left[\frac{\overrightarrow{j}\left({\overrightarrow{r}}^{\prime }\right)}{\left|\overrightarrow{r}-{\overrightarrow{r}}^{\prime }\right|}+\frac{\overrightarrow{M}\left({\overrightarrow{r}}^{\prime }\right)\times \left(\overrightarrow{r}-{\overrightarrow{r}}^{\prime }\right)}{{\left|\overrightarrow{r}-{\overrightarrow{r}}^{\prime }\right|}^3}\right].$$

We can transform that second contribution (from the magnetization density) to an equivalent current distribution as follows:

$${\displaystyle \int \frac{\overrightarrow{M}\left({\overrightarrow{r}}^{\prime }\right)\times \left(\overrightarrow{r}-{\overrightarrow{r}}^{\prime }\right)}{{\left|\overrightarrow{r}-{\overrightarrow{r}}^{\prime }\right|}^3}{d}^3{r}^{\prime }={\displaystyle \int \overrightarrow{M}\left({\overrightarrow{r}}^{\prime }\right)\times }{\overrightarrow{\nabla }}^{\prime }\left(\frac{1}{\left|\overrightarrow{r}-{\overrightarrow{r}}^{\prime }\right|}\right)}{d}^3{r}^{\prime },$$ 

and integrating by parts (assuming the magnetization is zero at infinity) puts the gradient operator on to the magnetization (the sign change is compensated by switching the order of $\overrightarrow{M},\overrightarrow{\nabla }$ )  giving

$$\overrightarrow{A}\left(\overrightarrow{r}\right)=\frac{{\mu }_0}{4\pi }{\displaystyle \int \frac{\overrightarrow{j}\left({\overrightarrow{r}}^{\prime }\right)+{\overrightarrow{\nabla }}^{\prime }\times \overrightarrow{M}\left({\overrightarrow{r}}^{\prime }\right)}{\left|\overrightarrow{r}-{\overrightarrow{r}}^{\prime }\right|}{d}^3{r}^{\prime }}$$ .

This can be interpreted as representing the magnetization by an effective current density

$${\overrightarrow{j}}_M=\overrightarrow{\nabla }\times \overrightarrow{M},$$

so we can write (adding the subscript $f,$ meaning free, to the ordinary macroscopic currents):

$$\overrightarrow{A}\left(\overrightarrow{r}\right)=\frac{{\mu }_0}{4\pi }{\displaystyle \int \frac{{\overrightarrow{j}}_f\left({\overrightarrow{r}}^{\prime }\right)+{\overrightarrow{j}}_M\left({\overrightarrow{r}}^{\prime }\right)}{\left|\overrightarrow{r}-{\overrightarrow{r}}^{\prime }\right|}{d}^3{r}^{\prime }}.$$ 

Remember we began from the Helmholtz expression

$$\overrightarrow{B}\left(\overrightarrow{r}\right)=\overrightarrow{\nabla }\times \frac{1}{4\pi }{\displaystyle \int d^3{r}^{\prime }}\frac{{\overrightarrow{\nabla }}^{\prime }\times \overrightarrow{B}\left({\overrightarrow{r}}^{\prime }\right)}{\left|\overrightarrow{r}-{\overrightarrow{r}}^{\prime }\right|},$$ 

so

$$\overrightarrow{\nabla }\times \overrightarrow{B}={\mu }_0\left({\overrightarrow{j}}_f+{\overrightarrow{j}}_M\right)={\mu }_0\left({\overrightarrow{j}}_f+\overrightarrow{\nabla }\times \overrightarrow{M}\right).$$ 

Example:  think of a uniformly magnetized cylinder bar magnet, with no free currents.  Where is nonzero $\overrightarrow{\nabla }\times \overrightarrow{M}$?  On the cylindrical surface.  Convince yourself it's equivalent to a solenoid. 

The Field H

First a reminder: Recall that in electrostatics the fundamental field was the electric field $\overrightarrow{E},$ but in analyzing the electric field in continuous media it proved useful to introduce another field $\overrightarrow{D}$ defined by

$\overrightarrow{D}={\epsilon }_0\overrightarrow{E}+\overrightarrow{P},$

$\overrightarrow{P}$ being the local polarization of the medium.  A spatially varying polarization generates a nonzero charge density (as was discussed in detail earlier) and consequently contributes to the electric field.

The field $\overrightarrow{D}$ is the part of the electric field whose divergence comes from just the free charges, $\overrightarrow{\nabla }\cdot \overrightarrow{D}={\rho }_f,$ with no contribution from those rearranged charges bound in the polarized molecules.  But that is not the end of the story$—$from Helmholtz’ theorem, we know that there is another source term in the equation for $\overrightarrow{D},$ the $\overrightarrow{\nabla }\times \overrightarrow{D}$ term.  No such term contributed to $\overrightarrow{E},$ since $\overrightarrow{\nabla }\times \overrightarrow{E}=0$ in electrostatics.

Now for the parallel in magnetostatics:  The fundamental magnetic field is $\overrightarrow{B},$ with curl generated by macroscopic electric currents and by magnetized matter.  It is therefore natural to define another field (analogous to $\overrightarrow{D}$ ) that is the just the part of the magnetic field having curl generated by the actual macroscopic currents.

Recall now the definition of the "magnetic currents": ${\overrightarrow{j}}_M=\overrightarrow{\nabla }\times \overrightarrow{M}.$ 

Now from Helmholtz’ theorem and $\overrightarrow{\nabla }\cdot \overrightarrow{B}=0,$

$$\overrightarrow{B}\left(\overrightarrow{r}\right)=\frac{1}{4\pi }\overrightarrow{\nabla }\times {\displaystyle \int \frac{\overrightarrow{\nabla }\times \overrightarrow{B}\left({\overrightarrow{r}}^{\prime }\right)}{\left|\overrightarrow{r}-{\overrightarrow{r}}^{\prime }\right|}{d}^3{r}^{\prime }}=\frac{{\mu }_0}{4\pi }\overrightarrow{\nabla }\times {\displaystyle \int \frac{{\overrightarrow{j}}_f\left({\overrightarrow{r}}^{\prime }\right)+{\overrightarrow{j}}_M\left({\overrightarrow{r}}^{\prime }\right)}{\left|\overrightarrow{r}-{\overrightarrow{r}}^{\prime }\right|}{d}^3{r}^{\prime }}.$$

If we define the new field $\overrightarrow{H}$ by

$$\overrightarrow{H}=\frac{1}{{\mu }_0}\overrightarrow{B}-\overrightarrow{M},$$ 

 then

$$\overrightarrow{\nabla }\times \overrightarrow{H}=\overrightarrow{\nabla }\times \left(\overrightarrow{B}/{\mu }_0\right)-\overrightarrow{\nabla }\times \overrightarrow{M}={\overrightarrow{j}}_f$$ 

from which one might be tempted to think that $\overrightarrow{H}$ is just given by the same expression as $\overrightarrow{B}\left(\overrightarrow{r}\right)$, but with the ${\overrightarrow{j}}_M\left({\overrightarrow{r}}^{\prime }\right)$ term dropped.

However, this is wrong.  The expression for $\overrightarrow{B}$ above comes directly from the Helmholtz theorem, but it's only half the usual Helmholtz formula, which has another term, the div term, that vanished for $\overrightarrow{B}$  since $\overrightarrow{\nabla }\cdot \overrightarrow{B}=0$ everywhere. 

In contrast, the Helmholtz expression for $\overrightarrow{H}$ has to include a div term, because from $\overrightarrow{H}=\overrightarrow{B}/{\mu }_0-\overrightarrow{M},$ $\overrightarrow{\nabla }\cdot \overrightarrow{H}=-\overrightarrow{\nabla }\cdot \overrightarrow{M},$ which is definitely nonzero, for example on surfaces of magnetized material. 

Therefore,

$\begin{array}{c}\overrightarrow{H}\left(\overrightarrow{r}\right)=\frac{1}{4\pi }\overrightarrow{\nabla }\times {\displaystyle \int \frac{{\overrightarrow{\nabla }}^{\prime }\times \overrightarrow{H}\left({\overrightarrow{r}}^{\prime }\right)}{\left|\overrightarrow{r}-{\overrightarrow{r}}^{\prime }\right|}{d}^3{r}^{\prime }-}\frac{1}{4\pi }\overrightarrow{\nabla }{\displaystyle \int d^3{r}^{\prime }\frac{{\overrightarrow{\nabla }}^{\prime }\cdot \overrightarrow{H}\left({\overrightarrow{r}}^{\prime }\right)}{\left|\overrightarrow{r}-{\overrightarrow{r}}^{\prime }\right|}},\\ =\frac{1}{4\pi }\overrightarrow{\nabla }\times {\displaystyle \int \frac{{\overrightarrow{j}}_f\left({\overrightarrow{r}}^{\prime }\right)}{\left|\overrightarrow{r}-{\overrightarrow{r}}^{\prime }\right|}{d}^3{r}^{\prime }+}\frac{1}{4\pi }\overrightarrow{\nabla }{\displaystyle \int d^3{r}^{\prime }\frac{{\overrightarrow{\nabla }}^{\prime }\cdot \overrightarrow{M}\left({\overrightarrow{r}}^{\prime }\right)}{\left|\overrightarrow{r}-{\overrightarrow{r}}^{\prime }\right|}}.\end{array}$

As an example, consider a uniformly magnetized solid.  The second integral becomes a surface integral of $\overrightarrow{M}\cdot {\widehat{\overrightarrow{n}}}^{\prime }$ (check by using the divergence theorem for a small pillbox on the boundary).  This is just like the discussion of the uniformly polarized sphere in electrostatics, which could be visualized as two uniform spheres of charge, one positive and one negative, slightly displaced relative to each other in the direction of the polarization.  Exactly the same analysis works here, giving a surface layer of “magnetic pole” of intensity $\overrightarrow{M}\cdot {\widehat{\overrightarrow{n}}}^{\prime }$.  If there are no free currents, the field $\overrightarrow{H}$ comes entirely from this “magnetic pole distribution” just as the $\overrightarrow{E}$ field from a polarized dielectric comes from a surface bound charge distribution. So here $\overrightarrow{H}$ is the gradient of a magnetic scalar potential.

Jackson’s Unfortunate Field Notation

At the bottom of page 192, Jackson tells us that $\overrightarrow{H}$ is the “magnetic field”, he mentioned earlier (p 174) that $\overrightarrow{B}$ is sometimes called the magnetic induction, and that’s what he calls it. He does add (p 193): “We emphasize that the fundamental fields are $\overrightarrow{E}$ and $\overrightarrow{B}.$ ”

What is he thinking? 

It seems likely he’s thinking like a magnet engineer$—$a typical electromagnet has a solenoid with iron inside. If you take out the iron, and supply a current, there will be a magnetic field. (Jackson’s $\overrightarrow{H}.$ )  But, practically speaking, it’s probably not strong enough for what you want. So you put back in the iron, and the induced magnetism gives you a far bigger field$—$Jackson’s induction$—$typically hundreds or thousands of times stronger, in fact the original field is now just a tiny contribution$—$although of course crucial. (Introducing the iron changes$\overrightarrow{H}$ too, but we're just looking at motivation here.)

The utility of $\overrightarrow{H}$ from a practical point of view is that it is easy to control: it depends completely on voltages and currents easily controlled from outside. (In contrast, the electrostatic field $\overrightarrow{D}$ is much less easily managed: it depends on free charges, notoriously leak-prone.)

The problem with Jackson's choice of notation is that you’re stuck with the word “induction” to describe, say, the magnetic field around a long straight wire carrying a current, where there’s nothing being induced. 

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