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.

  • $\begingroup$ 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


"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.

  • $\begingroup$ 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

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