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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.

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  • $\begingroup$ Pretty sure that this Q about Griffiths has been asked before. $\endgroup$
    – ProfRob
    Nov 3, 2021 at 6:58
  • $\begingroup$ This probably helps physics.stackexchange.com/questions/396586/… $\endgroup$
    – ProfRob
    Nov 3, 2021 at 7:07
  • $\begingroup$ @ProfRob Thanks! But the question talks about "a very long cylindrical bar magnet" where ampere's law method should be applicable? $\endgroup$ Nov 3, 2021 at 7:21

1 Answer 1

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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.

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  • $\begingroup$ Hi, great explanation! I understood everything until the part about the difference between long and short cylinders. Why won't there be a divergence in magnetization at the curved surface of the cylinder (but only at the ends)? $\endgroup$ Nov 3, 2021 at 9:46
  • $\begingroup$ @bobthelegend The magnetisation in the question is exclusively directed along the cylinder so has no divergence. Indeed if there is any component unaligned with the cylinder axis then that too will cause a divergence in H-field at the boundary. $\endgroup$
    – ProfRob
    Nov 3, 2021 at 10:06
  • $\begingroup$ If magnetisation (M) is directed along the z axis (along the cylinder), at the curved surface of the cylinder, won't there be a sudden change in magnetisation along the z axis from M $\hat{z}$ to $0$? @ProfRob $\endgroup$ Nov 3, 2021 at 10:40
  • $\begingroup$ What do you mean by 'parts of the curved boundary'? Divergence of a cylinder is non zero at the top and at the bottom, and that's it, isn't it? $\endgroup$
    – lalala
    Nov 3, 2021 at 10:44
  • $\begingroup$ @lalala a cylinder doesn't have a divergence. You mean that the magnetisation doesn't have a divergence? Perhaps, for a perfect magnetic material. $\endgroup$
    – ProfRob
    Nov 3, 2021 at 10:59

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