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I have started studying electrodynamics for a couple of weeks and I came across a basic concept that I can not understand well, it is about the relationship between the magnetic field $H$ and the magnetization $M$ of an isotropic body ($M = \chi H$).

1- If $M$ is uniform, can I infer that $H$ is always constant inside the body independently of its shape?

2- Can I also say that the Laplacian of the scalar potential $U$, such that $H=-\nabla U$, vanishes everywhere, since $$\nabla \cdot B = \mu \nabla \cdot H = \mu \chi^{-1}\nabla \cdot M = 0?$$

Thanks for any help!

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As long as $\mathbf{M}$ is uniform within the body and $\mathbf{M}=\chi \mathbf{H}$ and $\chi$ is constant within the body you may say that $\mathbf{H}=\frac{1}{\chi} \mathbf{M}$ is also uniform within the body. It is not true that outside the body, that is where $\mathbf{M}=0$, $\mathbf{H}$ and $\mathbf{B}=\mu_0 \mathbf{H}$ are uniform. The reason is the poles on the surface of the magnetized body.

While it is true that $\text{div} \mathbf{B} = 0$ always, and if you write $\mathbf{H}=-\textrm{grad} U$, it will not follow that $\textrm{div grad} U=0$ becasue $\chi$ is not constant, it has a jump across the magnetized material.

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  • $\begingroup$ I saw on the wikipedia page about demagnetization field that $\nabla^{2} U = - \nabla \cdot H =\nabla \cdot M $ inside the body and $ \nabla^{2} U = 0$ outside. However, if $\chi$ is constant for the body, then $\nabla \cdot M = 0$ because $B = \mu\chi^{-1}M$ and $\nabla \cdot B =0$, which implies actually that $\nabla^{2}U =0$ everywhere. Is it correct or am I missing something? $\endgroup$ – Alex Silva May 10 at 15:32
  • $\begingroup$ let $f(x)=1 \textrm{ when } |x|<1 \textrm{ and } f(x)=0 \textrm{ when } $|x|>1$. What is the 2nd derivative of this function? Is it really zero everywhere or there are some exceptional points? $\endgroup$ – hyportnex May 10 at 16:28
  • $\begingroup$ I got it. Thanks. $\endgroup$ – Alex Silva May 10 at 16:31

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