1
$\begingroup$

Suppose I have a linear homogeneous isotropic magnetic material, such that $$\textbf{M} = \chi \textbf{H}\tag{1},$$ where $\chi$ is the magnetic susceptibility, $\textbf{M}$ the magnetization, and $\textbf{H}$ is the magnetic field.

In some articles/webpages, for exemple [1], [2] and [3], equation $(1)$ says how the material is magnetized when it is placed in an external magnetic field $\textbf{H}$ .

However, in Jackson's book and in this article (page 64), the magnetic field $\textbf{H}$ is the one inside the material, not an external field.

I always thought the correct $\textbf{H}$-field in equation $(1)$ is inside the material, because it leads to $\textbf{B}=\mu_{0}(\textbf{H}+\textbf{M}) \implies \textbf{B}=\mu_{0}(\textbf{H}+ \chi\textbf{H}) \implies \textbf{B}=\mu\textbf{H},$ where $\mu$ is the permeability of the material.

But I am really confused about what field I should use (internal or external) in equation $(1)$ because of the first three references.

Why some authors use $\textbf{H}$ as the applied field and others the internal field? For the former, is it a kind of approximation?

I'll appreciate any help.

$\endgroup$
3
  • $\begingroup$ The $\bf{H}$ in $\bf{M}=\chi \bf{H}$ is what you call the internal field not the applied field. The two fields, internal and applied are the same for a coil tightly wound over a ferromagnetic toroid core. $\endgroup$
    – hyportnex
    Commented May 19, 2020 at 13:03
  • $\begingroup$ @hyportnex, Ok, but why lot of authors use $\textbf{H}$ as the applied field? See the references that I cited and also this answer. I can cite several other articles if we want. I don't understand. Is it a kind of approximation? :( $\endgroup$
    – Alex Silva
    Commented May 19, 2020 at 13:11
  • $\begingroup$ Th external or better called applied field and the internal field are linearly related in a linear isotropic material, a "soft magnet". What your reference this answer says is wrong in general but correct for a toroid. The difference between the two fields, applied and internal, is usually called the demagnetization field and is caused by surface poles of which there are none in the case of a toroid. $\endgroup$
    – hyportnex
    Commented May 19, 2020 at 13:27

1 Answer 1

4
$\begingroup$

"Lots of authors" are very careless. The external applied ${\bf H}$ is only the same as the internal ${\bf H}$ for samples that are long and thin and oriented parallel to the external field. Direct measurements of $\chi$ use samples of this shape for this reason. Any other shape requires computations of demagnetizing factors: https://en.wikipedia.org/wiki/Demagnetizing_field.

One can get away with ignoring the demagnetizing effect when $\chi$ is small, as it often is. Hence the confusing discussions.

$\endgroup$
2
  • $\begingroup$ Thanks for the answer. A side question: if the material is ferromagnetic (in this case $\textbf{M}=\chi^{(d)}\textbf{H}$, where $\chi^{(d)}$ is the differential susceptibility), do we need to compute the demagnetizing field to obtain $\textbf{M}$? $\endgroup$
    – Alex Silva
    Commented May 19, 2020 at 14:06
  • $\begingroup$ So is Jackson right (as always) or not in this case? $\endgroup$
    – lalala
    Commented May 19, 2020 at 14:10

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.