- #36

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$$\vec{j}_m=0, \vec{K}_m=\vec{M} \times \hat{n},$$

where ##\hat{n}## is the surface-unit normal vector along the boundary of the magnetized body.

In my interpretation it's included in ##\vec{\nabla} \times \vec{M}##, because I take the surface current into account in terms of a Dirac-##\delta##-like contribution from generalized derivatives.

Both methods are equivalent, because the surface contribution can be also defined in terms of the socalled (in German) "Sprungrotation" (I don't know an expression in English for it, literally translated it means "jump curl"). The definition is hard to describe without a figure. So here it is

The blue area is the surface across which a vector field ##\vec{V}## may have singularities (in our case of a homogeneously magnetized body, the magnetization ##\vec{M}## has a jump from some finite value inside and 0 outside the body along its boundary).

Now you define the "Sprungrotation" ##\text{Curl} \vec{V}## along the surface by drawing the red curve with an arbitrary tangent-unit vector ##\vec{t}## along the surface and taking the limit of the red curve of contracting it to the point on the surface you want the operator ##\text{Curl}## define at, which goes as follows

$$\Delta h (\vec{t} \times \vec{n}) \cdot \text{Curl} \vec{V} = \Delta h \vec{t} \cdot (\vec{n} \times \text{Curl} \vec{V})= \Delta h \vec{t} \cdot (\vec{V}_2-\vec{V}_1).$$

Since this construction holds for all tangent unit vectors ##\vec{t}## along the boundary surface at the given point, you get

$$\vec{n} \times \text{Curl} \vec{V}=(\vec{V}_2-\vec{V}_1).$$

The two parts of the path along ##\vec{n}## don't contribute because they cancel in the limit, because there's only a jump across but not along the surface. Thus you have ##\mathrm{Curl} \vec{V} \cdot{n}=0## by definition and thus one gets

$$\vec{n} \times (\vec{n} \times \text{Curl} \vec{V})=-\text{curl} \vec{V} = \vec{n} \times (\vec{V}_2 -\vec{V}_1).$$

The surface current for our case of the homogeneously magnetized sphere (taking region 1 as the exterior, region 2 as the interior of this sphere) thus is

$$\vec{k}_m=\mathrm{Curl} \vec{M} = \vec{n} \times (\vec{M}_1-\vec{M}_2) = \frac{\vec{r}}{a} \times (-M \vec{e}_3)=M \cos \vartheta \vec{e}_{\varphi}.$$

So we can write the solution for the vector potential indeed as

$$\vec{A}(\vec{r})=\mu_0 \int_{\partial V} \mathrm{d}^2 f' \frac{\vec{M}(\vec{r}') \times \vec{n}'}{|4 \pi |\vec{r}-\vec{r}'|}.$$