Fall 2006 13 Em

Consider a thin spherical shell of radius R centered on the spherical coordinate system r, $\theta$,$\phi$. On the surface of the sphere is a current density K which produces a uniform magnetic field inside the shell directed along the polar axis, that is B = $B_{0} \hat{z}$ inside the shell. Find K.

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The only currents are on the sphere, so inside and outside $\vec{\nabla} \times \vec{B}=0$.

Hence, we can express $\vec{B}$ as the gradient of a scalar potential $\phi$: $\vec{B}=-\nabla \phi$

Since $\vec{\nabla}\cdot\vec{B}=0 \Rightarrow \nabla^2 \phi = 0$

So the problem reduces to the solution of Laplace's equation for $\phi$ (one of the boundary conditions involve the surface currents, so it will give us what we are looking for).

We know the solution of $\nabla^2 \phi=0$ in spherical symmetry can be expanded in spherical harmonics.

Because $\vec{B}_{in} = B_{0} \hat{z}$, the solution must have azimuthal symmetry, so we are left with:

(1)
\begin{align} \phi (r, \theta) = \sum^\infty_{l=0} \left ( A_l r^l + \frac{C_l}{r^{l+1}} \right ) P_l (\cos{\theta}) \end{align}

We then impose the following 4 Boundary Conditions:

1) $\vec{B} = B_0 \hat{z}$ as $r \rightarrow 0$
2) $\vec{B} \rightarrow 0$ as $r \rightarrow \infty$
3) $B_{\bot , out} - B_{\bot , in} = 0$
4) $B_{\parallel , out} - B_{\parallel, in} = \frac{4 \pi}{c} K$

Boundary Condition 1:

(2)
\begin{align} \phi (r, \theta) = \sum^\infty_{l=0} \left ( A_l r^l + \frac{C_l}{r^{l+1}} \right ) P_l (\cos{\theta}) \end{align}
(3)
\begin{align} -\nabla \phi_{in}= B_0 \hat{z} \end{align}
(4)
\begin{align} -\frac{\partial}{\partial r} \left [ \sum^\infty_{l=0} \left ( A_l r^l P_l (\cos{\theta}) \right ) \right ] \hat{r} - \frac{1}{r} \frac{\partial}{\partial \theta}\left [ \sum^\infty_{l=0} \left ( A_l r^l P_l (\cos{\theta}) \right ) \right ] \hat{\theta} = B_0 \cos{\theta} \hat{r} - B_0 \sin{\theta} \hat{\theta} \end{align}

Clearly this is satisfied only by $A_l = 0$ for $l \neq 1$, and $A_l = - B_0$ for $l = 1$.

(5)
\begin{align} \Rightarrow \: \phi_{in} (r,\theta) = - B_0 r \cos{\theta} \quad (r < R) \end{align}

Boundary Condition 2:

(6)
\begin{align} \phi_{out} (r, \theta) = \sum^\infty_{l=0} \frac{C_l}{r^{l+1}} P_l (\cos{\theta}) \quad (r > R) \end{align}

Boundary Condition 3:

(7)
\begin{align} \vec{B}_{out} \cdot \hat{r} |_{R^{+}} - \vec{B}_{in} \cdot \hat{r} |_{R^{-}} = 0 \end{align}
(8)
\begin{align} \Rightarrow \; - \nabla \phi_{out} \cdot \hat{r} |_{R^{+}} +\nabla \phi_{in} \cdot \hat{r} |_{R^{-}} = 0 \end{align}
(9)
\begin{align} \nabla \phi = \frac{\partial}{\partial r} \phi \hat{r} + \frac{1}{r} \frac{\partial}{\partial \theta} \phi \hat{\theta} \end{align}

But, since we are projecting on $\hat{r}$, we only keep the first component: $\frac{\partial}{\partial r} \phi \hat{r}$

(10)
\begin{align} \nabla \phi_{out} \cdot \hat{r} = \sum^\infty_{l=0} (-l-1) \frac{C_l}{r^{l+2}} P_l (\cos{\theta}) \end{align}
(11)
\begin{align} \nabla \phi_{in} \cdot \hat{r} = -B_0 \cos{\theta} \end{align}
(12)
\begin{align} \Rightarrow \; \sum^\infty_{l=0} (l+1) \frac{C_l}{R^{l+2}} P_l (\cos{\theta})=B_0 \cos{\theta} \end{align}

Hence,

(13)
\begin{align} C_l=0 \: \: \: \forall \: l \ne 1 \end{align}

and

(14)
\begin{align} C_1 = \frac{B_0 R^3}{2} \end{align}
(15)
\begin{align} \Rightarrow \phi_{out} (r,\theta) = \frac{B_0 R^3}{2 r^2} \cos{\theta} \end{align}

(Notice that $\phi$ is NOT continuous on the surface of the sphere!!)

Boundary Condition 4:

Write it in vector form:

(16)
\begin{align} \hat{r} \times \vec{B}_{out} |_{R^+} - \hat{r} \times \vec{B}_{in}|_{R^-} = \frac{4 \pi}{c} \vec{K} \end{align}
(17)
\begin{align} \vec{B}_{out} = - \nabla \phi_{out} = \frac{B_0 R^3}{r^3} \cos{\theta}\hat{r} + \frac{B_0 R^3}{2 r^3}\sin{\theta}\hat{\theta} \end{align}
(18)
\begin{align} \hat{r} \times \vec{B}_{out} |_{R^+} = \frac{B_0}{2} \sin{\theta} \hat{\phi} \end{align}
(19)
\begin{align} \hat{r} \times \vec{B}_{in} |_{R^-} = - B_0 \sin{\theta} \hat{\phi} \end{align}
(20)
\begin{align} \Rightarrow \: \: \frac{B_0}{2} \sin{\theta} \hat{\phi} + B_0 \sin{\theta} \hat{\phi} = \frac{4 \pi}{c} \vec{K} \end{align}
(21)
\begin{align} \vec{K} = \frac{c}{4 \pi} \frac{3}{2} B_0 \sin{\theta} \hat{\phi} \end{align}