## Show Solution

The solution of the problem can be found straight forward. First of all, we have to find what this special **surface current** implies for the fields in- and outside of the sphere. Then, we will use the **magnetostatic potential** to express the fields in both regions. Connecting both information we will be able to find the magnetic induction.

### Boundary Conditions

Starting from **Gauss's law for magnetism**, \(\nabla\cdot\mathbf{B} \left(\mathbf{r}\right)=0\) and **Ampères law**, \(\nabla\times\mathbf{H}\left(\mathbf{r}\right)=\mathbf{j}\left(\mathbf{r}\right)\), we know that the normal part of the magnetic induction is continuous and the tangential part of the magnetic field makes a jump because of the present current. Because of the **spherical symmetry**, we might express both conditions using the radial unit vector \(\mathbf{e}_{r}\): \[\begin{eqnarray*}\mathbf{e}_{r}\cdot\left(\mathbf{B}_{\mathrm{out}}\left(r=r_{s,}\theta,\varphi\right)-\mathbf{B}_{\mathrm{in}} \left(r=r_{s,}\theta,\varphi\right)\right)&=&0\ ,\\ \mathbf{e}_{r}\times\left(\mathbf{H}_{\mathrm{out}}\left(r=r_{s,}\theta,\varphi\right)-\mathbf{H}_{\mathrm{in}} \left(r=r_{s,}\theta,\varphi\right)\right) & = & \mathbf{j} \left(r=r_{s,}\theta,\varphi\right)\ .\end{eqnarray*}\]In **component form**, we find \[\begin{eqnarray*}B_{r,\mathrm{in}}\left(r=r_{s,}\theta,\varphi\right)&=&B_{r,\mathrm{out}}\left(r=r_{s,}\theta,\varphi\right)\ \mathrm{and}\\H_{\theta,\mathrm{out}}\left(r=r_{s,}\theta,\varphi\right)-H_{\theta,\mathrm{in}}\left(r=r_{s,}\theta, \varphi \right) & = &j_{\varphi}\left(r=r_{s,}\theta,\varphi\right)=j_{0}\sin\left(\theta\right)\end{eqnarray*} \] since \(\mathbf{e}_{r}\times\mathbf{e}_{\theta}=\mathbf{e}_{\varphi}\) meaning that only the \(\theta\) component of the magnetic field is affected.

### Using the Magnetostatic Potential

Without present currents, the **magnetostatic potential** \(\psi\left(\mathbf{r}\right)\) has to be a solution to the **Laplace equation** \(\Delta\psi\left(\mathbf{r}\right)=0\). In **axial symmetry** it is convenient to use the known solution in terms of the **Legendre Polynomals** \(P_{l}\left(\cos\theta\right)\). Then, both \(P_{l}\left(\cos\theta\right)\cdot\left\{ r^{l}\ \text{and}\ r^{-l-1}\right\}\) are solutions for all \(l=0\dots\infty\).

Now we want to have a solution that respects natural boundary conditions outside of the sphere - the magnetostatic potential has to vanish for \(r\rightarrow\infty\). Furthermore we require \(\psi\left(\mathbf{r}\right)\) to be finite inside of the sphere. Since there is **no current** present except for \(r\neq r_{s}\), we know already that the solution in terms of the given expansion be expressed as \[\begin{eqnarray*}\psi_{\mathrm{in}}\left(\mathbf{r}\right)&=&\sum_{l=0}A_{l}P_{l}\left(\cos\theta\right)r^{l}\ r<r_{s}\ \text{and}\\\psi_{\mathrm{out}}\left(\mathbf{r}\right)&=&\sum_{l=0}B_{l}P_{l}\left(\cos\theta\right)r^{-l-1}\ r>r_{s}\ .\end{eqnarray*}\]For convenience we might use the superscripts “in” and “out” from now on for \(r\lessgtr r_{s}\). In terms of the magnetic induction, the latter equation reads as \[\begin{eqnarray*}B_{r,\mathrm{in}}\left(\mathbf{r}\right)&=&-\mu_{0}\mu_{r}\partial_{r}\phi_{\mathrm{in}}\left(\mathbf{r}\right)=-\mu_{0}\mu_{r}\sum_{l=1}lA_{l}P_{l}\left(\cos\theta\right)r^{l-1}\\B_{r,\mathrm{out}}\left(\mathbf{r}\right)&=&\mu_{0}\sum_{l=1}\left(l+1\right)B_{l}P_{l}\left(\cos\theta\right)r^{-l-2}\end{eqnarray*}\]Then, using the continuity of \(B_{r}\), we find at \(r=r_{s}\) \[\begin{eqnarray*} -\mu_{0}\mu_{r}lA_{l}r_{s}^{l-1}&=&\mu_{0}\left(l+1\right)B_{l}r_{s}^{-l-2}\ \mathrm{or} \\ A_{l} & = & -\frac{l+1}{\mu_{r}lr_{s}^{2l+1}}B_{l}\ .\end{eqnarray*}\] Note that the latter equation is still independent of the current itself. We can incorporate it using the magnetic field. First, we write down the corresponding expression for the affected field component: \[\begin{eqnarray*}H_{\theta,\mathrm{in}}\left(\mathbf{r}\right)&=&-\frac{1}{r}\partial_{\theta}\phi\psi_{\mathrm{in}}\left(\mathbf{r}\right)=\sum_{l=1}A_{l}P_{l}^{\prime}\left( \cos\theta\right) \sin\left(\theta\right)r^{l-1}\\H_{\theta,\mathrm{out}}\left(\mathbf{r}\right)&=&\sum_{l=1}B_{l}P_{l}^{\prime}\left(\cos\theta\right)\sin\left(\theta\right)r^{-l-2}\ .\end{eqnarray*}\]The summations start here from \(l=1\) again since the derivation of \(P_{0}\left(x\right)=1\) vanishes.

Now we have to use the **boundary conditions** for \(H_{\theta}\). We know already that this field component jumps at the surface of the spere. So, although we only have an expression for the magnetostatic potential in- and outside, we can use the expansion in a **limiting sense** \(r\rightarrow r_{s}\) from both sides. Furthermore, \(P_{1}\left(\cos\theta\right)=\cos\theta\) such that \(\partial_{\theta}P_{1}\left(\cos\theta\right)=-\sin\theta\). So, the surface current distribution \(j_{\varphi}\left(r_{s},\theta,\varphi\right)=j_{0}\sin\theta\) corresponds to the \(l=1\) term and the boundary conditions read as \[ A_{l}r_{s}^{l-1}-B_{l}r_{s}^{-l-2} = -j_{0}\delta_{l,1}\ .\]When we compare our result to the one we obtained for the boundary conditions of the magnetic induction, we can see that \[A_{l} = \frac{1}{r_{s}^{2l+1}}B_{l}-\frac{j_{0}}{r^{l-1}}\delta_{l,1}\overset{!}{=}-\frac{l+1}{\mu_{r}lr_{s}^{2l+1}}B_{l}\ .\]Since \(l=1,2,\dots\), the only nonvanishing coefficients can be in case of \(l=1\). Here, \[\begin{eqnarray*}j_{0}&=&\left(\frac{1}{r_{s}^{3}}+\frac{2}{\mu_{r}r_{s}^{3}}\right)B_{1}\ , \\ B_{1} & = & j_{0}r_{s}^{3}\frac{\mu_{r}}{\mu_{r}+2}\ \mathrm{and}\\A_{1}&=&-j_{0}\frac{2}{\mu_{r}+2}\ .\end{eqnarray*}\]This translates to \[\begin{eqnarray*}\psi_{\mathrm{in}}\left(\mathbf{r}\right)&=&-j_{0}\frac{2}{\mu_{r}+2}r\cos\theta\\\psi_{\mathrm{out}}\left(\mathbf{r}\right)&=&j_{0}\frac{\mu_{r}}{\mu_{r}+2}\frac{r_{s}^{3}}{r^{2}}\cos\theta\ .\end{eqnarray*}\]Finally, with \(r\cos\theta=z\), we find\[\begin{eqnarray*} \mathbf{B}\left(\mathbf{r}\right)&=&-\mu\left(\mathbf{r}\right)\nabla\phi\left(\mathbf{r}\right)\\&=&\mu_{0}\mu_{r}j_{0}\begin{cases}\frac{2}{\mu_{r}+2}\mathbf{e}_{z} & r<r_{s}\\\frac{1}{\mu_{r}+2}\frac{r_{s}^{3}}{r^{3}} \left(3\frac{z}{r}\mathbf{e}_{r}-\mathbf{e}_{z}\right) & r>r_{s}\end{cases}\ .\end{eqnarray*}\]Hence, the field inside the sphere is uniform but has the character of a magnetic dipole on the outside. The relation to the experiment mentioned in the **background** can now be seen in the following sense: The whole particle was smaller than the wavelength, approximately 200 nm. Then, the incident excitation field, a plane wave, is basically a constant field along the particle. From A Dielectric Sphere in a Homogeneous Electric Field we know already that such a field causes a **dipolar electric field** for a dielectric sphere and the same holds for our magnetic sphere with induced dipolar magnetic field. So, we have just used the corresponding surface current of such a situation!

### General Surface Current

So far we considered the case of a very easy surface current, \(\mathbf{j}\left(\mathbf{r}\right)=j_{0}\sin\theta\delta\left(r-r_{s}\right)\mathbf{e}_{\varphi}\). We could solve the problem in full analyticity because the current is restricted to the surface only. If we have a more complicated surface distribution, we could still solve the problem with the given approach. Then, we would have to expand the potential in terms of **spherical harmonics** \(Y_{lm}\left(\theta,\varphi\right)\) instead of the Legendre Polynomials with certain pre-factors \(A_{lm}\) and \(B_{lm}\) and expand the current in the same function set, \(\mathbf{j}\left(r=r_{s},\theta,\varphi\right)=\sum_{l,m}j_{lm}Y_{lm}\left(\theta,\varphi\right)\). Then, we can again determine the \(A_{lm}\) and \(B_{lm}\) as we did in the previous calculations. Well, of course, for a considerably complicated \(\mathbf{j}\), the system of equations might be a little blown up...