# Magnetic moment - magnetic field relation without free currents

I'm trying to understand magnetostatics in the presense of ferromagnetic material. But I'm ending up in a contradiction:

Lets take a piece of iron: Assuming that we don't care about the hysteresis and non-linearity, the relationship between $B$ an $H$ inside the iron piece can be described as

$$B = \mu H.$$

In the magnetostatic case, $\nabla \times H = J_f$, where $J_f$ are the free currents.

If there are no free currents, it follows that $H=0$ and $B=0$ inside the iron piece.

But what if we place a strong permanent magnet close to our iron piece? I would expect that the external magnetic field causes a magnetization $M$, and that $B$ field inside the iron piece should not be zero.

• I believe it's $\nabla \times H = J_f$ – user3814483 Jul 28 '14 at 18:29
• you're right, I've fixed it. – Stefan Jul 28 '14 at 18:32
• Does that also address your logical inconsistency? I think it should, but I'd be happy to explain... – user3814483 Jul 28 '14 at 18:33
• No it doesn't. It was just a typo. – Stefan Jul 28 '14 at 18:36
• I starting to feel that the answer will involve $\nabla \cdot H$ at the material boundary, but I still can't wrap my head around it. – Stefan Jul 28 '14 at 18:49

The definition of $\mathbf{H}$ is $$\mathbf{H} = \frac{\mathbf{B}}{\mu_0} - \mathbf{M}$$ where $\mathbf{B}$ is the magnetic field in which the object is immersed in and $\mathbf{M}$ is the magnetisation of the object, i.e. the "field" ($\propto$ to a field) caused by the internal magnetic properties of the object.

If there are no free currents, then $\nabla \times \mathbf{H} = 0$ which means that $\mathbf{H}$ can be expressed a gradient of a scalar. Now, physically, $\mathbf{H}$ is the magnetic field composed by the contribution of free currents of the object (=0) and any external field. If you take the external field to be 0, then you can take $\mathbf{H} = 0$ since there would be no reason as to why it should be non-zero.

So even if $\mathbf{H} = 0$, $\mathbf{M}$ can be non-zero, and it will be non-zero if iron had been previously magnetised.

If you now place the iron piece in the vicinity of a strong magnet, then the domains inside the iron piece will align with the external field and result in a magnetisation $\mathbf{M}$. To get $\mathbf{H}$, you need to superpose $\mathbf{M}$ to the actual $\mathbf{B}$ generated by the strong magnet.

The resutling net magnetic field in the iron piece is going to be stronger than the field caused by the exteral magnet, since iron is ferromagnetic and has a very strong response to external fields (i.e. $\mathbf{M}$ is very big, because all the domains in the iron piece align with it).

extra

The relationship between $\mathbf{M}$ and $\mathbf{H}$ is not trivial for ferromagnets, because it is governed by magnetic hysteresis, but for dia/*para*magnets is just $\mathbf{M} = \chi_m \mathbf{H}$.

• Thank you, I think I found my error that lead to the contradiction in my question: I implicitly assumed $\nabla \times H = 0 \rightarrow H = 0$ but unlike $B$, $H$ can have a divergence. So what is $\nabla \cdot H$ ? – Stefan Jul 28 '14 at 19:11
• You can see from the definition of $\mathbf{H}$ that $\nabla \cdot \mathbf{H} = \frac{\nabla \cdot \mathbf{B}}{\mu_0} - \nabla \cdot \mathbf{M}$. So, since $\nabla \cdot \mathbf{B}$ is always $0$ (from Maxwell's equations), you need to know the expression for $\mathbf{M}$, which depends on the external field and the material's response to it. See the last paragraph of my answer. – SuperCiocia Jul 28 '14 at 20:06
• You're right! $M$ has a strong discontinuity at the material border of the permanent magnet, effectively sourcing & sinking $H$, which will in turn magnetize the iron piece. I think I actually got it now. Thanks! – Stefan Jul 28 '14 at 20:27
• Since $\mathbf M$ has jump on the surface of the iron body, $\nabla \cdot \mathbf M$ does not vanish there and hence generally $\mathbf H$ does not vanish either. – Ján Lalinský Jul 28 '14 at 20:29

I think the best explanation of electromagnets/Ferro magnetism is in the Feynman lectures Vol II. (chapter 36.) He makes up his own units so that might be a bit confusing. But work through the electromagnet problem. (and even use a simple linear relationship B = uH) That helped me a lot.

• I was reading that, actually :) Excellently written. I did have the missconception that $H$ (which non-Feynman's sometimes call 'magnetic field') had no monopoles. I don't find his units confusing, quite contrary, I find the units everyone else is using confusing. – Stefan Jul 28 '14 at 20:50
• Excellent, no better physics reading (IMHO). Hysteresis, is a bit of a complication to think about afterwards. We can permanently polarize dielectrics too. Doing the totally ideal case of just some huge permeability is nice for the physics mind. One common misconception I find is that the magnetic material makes the B field in the gap bigger, (more mu means more field.. in proportion to mu.) No, but it can really concentrate all those amp-turns from over there,into a big field here. – George Herold Jul 28 '14 at 23:39

Your error is that you cannot understand ferromagnetism while neglecting hysteresis and nonlinearity.

Consider the case of some iron cooled from above the Curie temperature in the absence of any magnetic field. As the iron crystallizes, it forms strongly magnetized domains — a classic case of spontaneously broken symmetry. Because there is no external field, the domains are oriented in random directions. So macroscopically you do have $B=\mu H = 0$, as you say.

Now you apply a field $H$. Your iron responds by moving the domain boundaries: domains aligned with the field grow, while domains opposed to the field shrink. At the point where they stabilize, you can define some proportionality constant $\mu$ so that $B=\mu H$. That's fine, too.

Now remove the applied field. There's nothing to make the domains move back to where they were! Your iron will remain magnetized until it is exposed to another field in another direction, or heated above the Curie temperature again. This is where your analogy breaks down.

Griffiths writes:

… it is misleading to speak of the susceptibility or permeability of a ferromagnet. The terms are used for such materials, but they refer to the proportionality factor between a differential increase in $\vec H$ and the resulting differential change in $\vec M$ (or $\vec B$); moreover, they are not constants, but functions of $\vec H$.

• I don't understand how hysteresis and nonlinearity are relevant for the contradiction I'm getting. Maybe it is easier to assume an unusually strong paramagnetic material? – Stefan Jul 28 '14 at 19:16
• Hysteresis and nonlinearity are the defining features of ferromagnetism. Your simple model $B = \mu H$ with constant $\mu$ doesn't describe reality. That's why you're confused. – rob Jul 28 '14 at 19:19
• I agree. I should have formulated the question more carefully. The confusion I have is not directly related to ferromagnetism, I think. It is about magnetization in general, about the interaction of magnetizable / magnetized matter in the absence of free currents. I used ferromagnetism as an example only. More specifically, in my example I never remove a field, so your answer does not apply. – Stefan Jul 28 '14 at 19:24
• Hi guys, @Stefan, in some ways I totally agree with you. Hysteresis and saturation are what you really need to think about in magnetics, if you are designing something. (I've only done a bit of design with magnetic materials.. and I find them a bit magical at times.) But as a first step for someone coming from a physics background, you can first just try and understand a huge mu, let mu go to infinity... and then add in all the complications. (At least that kinda works for me.) – George Herold Jul 28 '14 at 23:48

I'm not sure if this was clear in the previous answers, but you cannot use ${\bf B}=\mu{\bf H}$ in ferromagnetic materials.