Showing that B has no discontinuities at the surface

  • #1
Mike400
59
6
Consider a magnetic dipole distribution in space having magnetization ##\mathbf{M}##. The potential at any point is given by:

##\displaystyle\psi=\dfrac{\mu_0}{4 \pi} \int_{V'} \dfrac{ \rho}{|\mathbf{r}-\mathbf{r'}|} dV' + \dfrac{\mu_0}{4 \pi} \oint_{S'} \dfrac{\sigma}{|\mathbf{r}-\mathbf{r'}|} dS'=\psi^{V}+\psi^{S}##

The ##\mathbf{H}## field is:

##\displaystyle\mathbf{H}=\dfrac{\mu_0}{4 \pi} \int_{V'} \rho \dfrac{\mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} dV' + \dfrac{\mu_0}{4 \pi} \oint_{S'} \sigma \dfrac{ \mathbf{r}-\mathbf{r'}}{|\mathbf{r}-\mathbf{r'}|^3} dS'=\mathbf{H}^{V}+\mathbf{H}^{S}##

The ##\mathbf{B}## field is:

##\mathbf{B}=\mathbf{H} + \mu_0 \mathbf{M}=\mathbf{H}^{V} + \mathbf{H}^{S} + \mu_0 \mathbf{M}##

##\mathbf{H}^{V}## has no discontinuity.

##\mathbf{H}^{S}## has discontinuity of ##\mu_0 \mathbf{M} \cdot \hat{n}## at the surface ##S'##

##\mu_0 \mathbf{M}## has discontinuity of ##\mu_0 \mathbf{M}## at the surface ##S'##

From these knowledge, how shall one deduce that ##\mathbf{B}## is continuous at the surface?

My try: (I am getting a contradiction)

We need to show that ##\mu_0 \mathbf{M} \cdot \hat{n}+\mu_0 \mathbf{M}=0##, i.e. ##\mathbf{M} \cdot \hat{n}= -\mathbf{M}##
Since the surface could be oriented at any angle w.r.t. ##\mathbf{M}## at the surface, this is a contradiction. Where am I going wrong?
 
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  • #2
I guess what you mean is that due to
$$\vec{\nabla} \cdot \vec{B}=0$$
##\vec{B}##'s normal component at the surface must be continuous.

I'm also a bit lost what ##\rho## and ##\sigma## have to do with ##\vec{M}##. Without a surface magnetization (which is a bit unusual; I'm not sure, where one would find such a thing in nature) the correct solution of magnetostatics of a (hard) ferro magnet is
$$\vec{H}=-\vec{\nabla} \phi$$
with
$$\phi(\vec{x})=-\vec{\nabla}_x \cdot \int_{\mathbb{R}} \mathrm{d}^3 x' \frac{\vec{M}(\vec{x}')}{4 \pi |\vec{x}-\vec{x}'|}.$$
 
  • Informative
Likes Dale
  • #3
vanhees71 said:
##\vec{B}##'s normal component at the surface must be continuous.
I know that and can be simply deduced from the equation ##\mathbf{B}=\mathbf{H} + \mu_0 \mathbf{M}=\mathbf{H}^{V} + \mathbf{H}^{S} + \mu_0 \mathbf{M}##. But should the tangential component of ##\vec{B}## must be continuous too? The equation shows a discontinuity in the tangential component.

vanhees71 said:
I'm also a bit lost what ##ρ## and ##σ## have to do with ##\vec{M}##
##ρ=-\nabla \cdot \mathbf{M}## and ##σ=\mathbf{M} \cdot \hat{n}##
 
Last edited:
  • #4
The tangential component of ##\vec{H}## can have a singularity due to a surface current, and usually also ##\vec{B}## has one there too.

The standard example is a homogeneously magnetized body. The magnetization is equivalent to a current density
$$\vec{j}_{\text{m}}=\vec{\nabla} \times \vec{M},$$
and in this approximation that's a surface current density.

E.g., take a homogeneously magnetized sphere of radius ##a## around the origin of the coordinate system. Then we can write (with ##r=|\vec{x}|##)
$$\vec{M}(\vec{x})=M \vec{e}_3 \Theta(a-r).$$
The curl is
$$\vec{\nabla} \times \vec{M}=-M \vec{e}_3 \times \vec{\nabla} \Theta(a-r).$$
Now
$$\vec{\nabla} \Theta(a-r)=\frac{\mathrm{d}}{\mathrm{d} r} \Theta(a-r) \vec{\nabla r} = -\frac{\vec{x}}{r} \delta(a-r)=-\frac{\vec{x}}{a} \delta(a-r)$$
and thus
$$\vec{j}_{\text{m}}=\frac{M}{a} \vec{e}_3 \times \vec{x} \delta(r-a),$$
i.e., you have ##\delta##-function like singularity across the surface of the sphere which means that there's a surface-current density.

It's a good example to calculate the magnetic displacement ##\vec{H}## and the magnetic field ##\vec{B}## for this example, which is analytically solvable. Then the concepts of charge and current densities as well as their singular cases, i.e., surface charge and current densities become clear.
 

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