When can we use Ampère's Law methods to find $\mathbf{H}$ field?

Question

I was reading David J. Griffiths Introduction to Electrodynamics and came across this question today.

Part (b) of the question can be solved by realising that $\mathbf{H}$ (the auxiliary field), by symmetry, only points in the $\hat{z}$ direction.

Since $$\oint \mathbf{H} \cdot d\mathbf{s}=I_{enc,f}$$ and there is no free current in this case, $\mathbf{H}$ must be $0$, which then allow us to find the magnetic field, $\mathbf{B}$ through $$\mathbf{H}=\frac{\mathbf{B}}{\mu_0}-\mathbf{M}$$

Initially, the solution made sense to me. However, in the following section of the book which explains why we cannot always use Ampere's Law methods to find $\mathbf{H}$, it says

consider the example of the bar magnet—a short cylinder of iron that carries a permanent uniform magnetization $\mathbf{M}$ parallel to its axis. In this case there is no free current anywhere, and a naïve application of $\oint \mathbf{H} \cdot d\mathbf{s}=I_{enc,f}$ might lead you to suppose that $\mathbf{H}=0$, and hence that $\mathbf{B}=\mu_0 \mathbf{M}$ inside the magnet and $\mathbf{B}=0$ outside, which is nonsense. It is quite true that the curl of $\mathbf{H}$ vanishes everywhere, but the divergence does not. (Can you see where $\nabla \cdot \mathbf{M} \neq 0$?) Advice: When you are asked to find $\mathbf{B}$ or $\mathbf{H}$ in a problem involving magnetic materials, first look for symmetry. If the problem exhibits cylindrical, plane, solenoidal, or toroidal symmetry, then you can get $\mathbf{H}$ directly from $\oint \mathbf{H} \cdot d\mathbf{s}=I_{enc,f}$ by the usual Ampère’s law methods. (Evidently, in such cases $\nabla \cdot \mathbf{M}$ is automatically zero, since the free current alone determines the answer.)

I got quite confused after reading this and have a few questions:

• What is the difference in the $\mathbf{H}$ and $\mathbf{B}$ field between a short cylinder and a infinitely long one?

• Even though both scenarios exhibit cylindrical symmetry, why can we only use Ampere's Law method to calculate $\mathbf{H}$ for the case of an infinitely long cylinder?

• Moreover, for Problem 6.12, $\nabla \cdot \mathbf{M}$ is not $0$ at the boundary of the cylinder. However, the book says that in cases where we can obtain $\mathbf{H}$ directly from Ampere's methods, $\nabla \cdot \mathbf{M}$ is automatically zero, since the free current alone determines the answer. What is the reason for this contradition? If $\nabla \cdot \mathbf{M}$ is indeed $0$ everywhere (including at the boundary), please explain why.

In conclusion, I am very confused by the book's explanation and have trouble identifying when we can use Ampere's Law method to find $\mathbf{H}$ field and when it is incorrect to do so.

Help would be much appreciated.

Answer 1

Suppose that the H-field was composed of two parts. One of which had a curl of zero and the other which had a divergence of zero. Call them $\vec{H}_{c=0}$ and $\vec{H}_{d=0}$ respectively, such that $$\vec{H} = \vec{H}_{c=0} + \vec{H}_{d=0}\ .$$

In which case Amperes law $$J_f = \nabla \times (\vec{H}_{c=0} + \vec{H}_{d=0}) = \nabla \times \vec{H}_{d=0} $$ tells us nothing about the curl-free part of the H-field.

The Helmholtz decomposition theorem tells us that any vector field can be represented by two such fields, so Amperes law will only tell us about the total H-field when it consists only of a divergence-free component. i.e. when $\vec{H} = \vec{H}_{d=0}$.

In what circumstances is the H-field not entirely divergence free? When the magnetisation has a divergence. $$ \vec{H} = \frac{B}{\mu_0} - \vec{M}$$ $$ \nabla \cdot \vec{H}_{c=0} = - \nabla \cdot \vec{M}$$ In an infinitely long cylinder it is safe to assume the divergence of the magnetisation, and hence the H-field, is zero, and so the H-field only has a curl and is given by Ampere's law.

A short cylinder has a divergence in magnetisation at the ends (and possibly also at parts of the curved boundary, if there is any component of the magnetisation unaligned with the cylinder), so the H-field has an additional curl-free term that isn't given by Ampere's law.