# Unifom magnetization of an isotropic body

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!

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.
• 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? – Alex Silva May 10 at 15:32
• 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? – hyportnex May 10 at 16:28