1
$\begingroup$

If a cylindrical magnet has a uniform magnetization $\vec{M}$ along its axis its magnetic field $\vec{B}$ looks approximately like this:

Permanent magnet

There seems to be a contradiction to me. Since there are no free currents anywhere, $\nabla \times \vec H = \vec{0}$ everywhere. But also, $\vec{B} = \mu_0(\vec{H} + \vec{M})$ everywhere. Then, because $\vec{M} = \vec{0}$ outside of the magnet, $\vec{H}$ has to be nonzero outside of the magnet, because clearly $\vec{B}$ is not zero there. But according to the Griffith's textbook, $\vec{H}$ is the magnetic field only due to free currents, or alternatively, the field that remains if one were to remove all magnetization. So if $\vec{H}$ is nonzero outside of the magnet, and it is not caused by free currents (because there are none), and it also is not caused by the magnetization, then what is $\vec H$ and why does it exist in this case?

$\endgroup$
5
  • 1
    $\begingroup$ I do not believe that Griffiths would say anywhere that $\vec H$ is due to free currents only, that is nonsense. What is true is that the curl of $\vec H$ is the free current. There is no causation implied. Since $div \vec B=0$ always, would that imply that $\vec B$ must be zero, too? In electrostatics $curl \vec E = 0$, does that imply.... $\endgroup$
    – hyportnex
    Jun 24, 2023 at 1:51
  • $\begingroup$ In that case, what is the physical significance of $\vec{H}$? Is it just an intermediate quantity that is useful when we know $\nabla \cdot \vec{M} = 0$ to find the actual physical magnetic field (the one that determines the force on particles)? $\endgroup$ Jun 24, 2023 at 2:55
  • 1
    $\begingroup$ this and this may help, especially if you know about the Helmholtz decomposition theorem $\endgroup$
    – hyportnex
    Jun 24, 2023 at 3:27
  • 1
    $\begingroup$ The conventional view is that the sources of H are the induced magnetic poles and the sources of B are the moving electric charges but this is really untenable inside matter because both are macroscopic averages of microscopic fields and there is no macroscopic current, neither is more fundamental than the other. Probably the best and safest to say is that in a vacuum there really is no difference between them, and in a magnetically polarizable matter B is measurable as the vacuum field in a slim disk crevice perpendicular to M, and H is the field in narrow needle crevice parallel with M. $\endgroup$
    – hyportnex
    Jun 24, 2023 at 3:39
  • $\begingroup$ In the magnet, B and H are oriented almost in opposite directions. See the figure titled "Comparison of B, H and M inside and outside a cylindrical bar magnet." in Wikipedia. $\endgroup$
    – HEMMI
    Jun 24, 2023 at 8:06

2 Answers 2

2
$\begingroup$

The equations satisfied by $\vec{H}$ are : $\vec{\nabla} \times \vec{H}=\vec{0}$ and $\vec{\nabla} \cdot \vec{H} = - \vec{\nabla} \cdot \vec{M} $

One might think that $\vec{\nabla} \cdot \vec{M} = 0$ because $\vec{M} $ is uniform. But it's wrong. We must not forget that $\vec{M} $ is discontinuous on the surface. Using the distribution's formalism, we can show that $\vec{\nabla} \cdot \vec{M} $ corresponds to two surface layers $\sigma = -M$ on the upper face and $-\sigma = + M$ on the underside (The "magnetic charges" at the poles of the magnet). Finally, the vector $\vec{H} $ has the same shape as the electrostatic field of a plane capacitor. But it is directed downwards: it is the "demagnetizing" field which tends to demagnetize an open magnet.

If the magnet is very long, we find that $\vec{H} \rightarrow \vec{0}$ except near the edges. The magnetic field $\vec{B} $ then tends towards $\mu_0 \vec{M} $ inside the magnet and towards $\vec{0} $ outside.

In electrical engineering, it is sometimes stated that the field $\vec{H}$ is controlled by free currents but this is only true if the demagnetizing field is zero. In particular, this is true for a torus or a very long cylinder. It is this type of geometry which is used to obtain the hysteresis cycle of a magnetic material and indeed, in this case, the vector $\vec{H}$ is directly controlled by the intensity which circulates in the winding of the primary. (If the secondary is traversed by a negligible current)

We could also obtain $\vec{B} $ directly by using the equivalent magnetization currents: $\vec{j_m} = \vec{\nabla} \times \vec{M}$ .

We will have : $\vec{\nabla} \times \vec{B}=\mu_0 \vec{j_m}$ and $\vec{\nabla} \cdot \vec{B} =0$

Again, one might think that $\vec{j_m} = \vec{\nabla} \times \vec{M} = \vec{0}$ But it's wrong. Again, we must not forget that $\vec{M} $ is discontinuous on the surface. Using the distribution's formalism, we can show that $\vec{j_m} = \vec{\nabla} \times \vec{M}$ corresponds to surface current $\vec{j_m} = \vec{M} \times \vec{n} = M \vec{e_\theta}$ . Clearly, the vector $\vec{B} $ has the same shape as the magnetic field of a solenoid.

One can note that with this equivalence, one proves "rigorously" that the magnetic field outside an "infinite" solenoid tends towards $0$. But it is a bit tortuous method!

Hope it can help and sorry for my poor english.

$\endgroup$
2
$\begingroup$

The curl of ${\bf H}$ is caused by free currents.

The H-field itself is defined as $({\bf B} - {\bf M})/\mu_0$ and since ${\bf M}=0$ outside the magnet then ${\bf H}={\bf B}/\mu_0$ and is curl-free.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.