I am trying to re-write the constitutive equation of linear permeable media in a way that is more convenient for its numerical implementation. From most books I have seen that the constitutive relation is expressed as


being $\mu$ the media magnetic permeability.

However, I would like to express $\mathbf{B}$ as a funcion of $\mathbf{M}$ using the relation $\mathbf{B}=\mu_{0}\left(\mathbf{H}+\mathbf{M}\right)$. If it is right the fact that for linear media the field induces a magnetization, I could write


which gives


now being $\bar{\chi}$ a susceptibility measure.

With this form, $\mathbf{M}$ plays the role of an induced magnetization. So, if permanent materials are considered, a new $\mathbf{M}_{0}$ should be added as


Is this an accepted way to express the constitutive relation? Can you see any inconsistency?

Thanks in advance.


1 Answer 1


What you have written down does not look right. The relation ${\bf B}=\mu_{0}({\bf H}+{\bf M})$ always holds exactly; indeed, it gives the definition of the auxiliary field ${\bf H}$. So your notation already has a problem.

However, it seems that what you really want to do is to split the magnetization into two pieces: a linear piece and another constant piece supplied by a ferromagnetic background. Unfortunately, this is not generally a good approximation for ferromagnets. In a ferromagnet, the constitutive relation ${\bf M}={\bf M}({\bf B})$ is not typically an affine function (constant piece plus linear piece). The relation is actually extremely complicated and nonlinear, with saturation effects and dependence on the history of the sample (hysteresis). So your method is probably not going to be very useful.

  • $\begingroup$ Thanks for your answer @Buzz! What I am trying to do is to obtain a law that can handle both permeable and magnetized media simultaneously. If we consider a ferromagnetic material then M0 would be equal to zero and the law reduces to the classical one. If we consider a permanent magnet then M is equal to zero. I agree with you that the last equation may be wrong... however I cannot see the problem with the relation between B and M. $\endgroup$
    – nodarkside
    Commented Dec 12, 2018 at 3:32

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