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.
 A: 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 $\mathbf{M} = \chi_m \mathbf{H}$.
A: 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.         
A: 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 $\vec H$.

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