Ferromagnetic sphere placed in uniform field Imagine a ferromagnetic sphere, which I assume that it originally contains no magnetisation, is placed inside a uniform magnetic field as shown in the following image:

In the image, B is externally applied magnetic field and φ is defined such that H = −∇φ.
I would like to solve for the value of φ both inside and outside the sphere.  Using Maxwell's second equation and assuming that the divergence of magnetisation is zero, I can write down the Laplace equation Δφ = 0.
I'm thinking about solving the Laplace equation by separating it into functions in spherical coordinates.  Is it valid to guess that the solution is in the form:

I've guessed that the potential created by the sphere is only dependent on cosθ, but not higher order terms of the Legendre expansion.  I've put it into a computer simulation and it, indeed, worked.  However, how do I know that higher order terms of the Legendre expansion vanishes?
 A: One can demonstrate this by applying boundary conditions to the general solution of Laplace's equation with azimuthal symmetry: 
$$ V(r,\theta ) = \sum ^{\infty} _{l=0} \left( A_l r^l + \frac{B_l}{r^{l+1}} \right) P_l (\cos \theta )) $$
We now insist that the potential remains finite everywhere and tends to $ -\frac{B_{0}}{\mu{0}}z = - \frac{B_{0}}{\mu_{0}}rcos(\theta) $ so that $\mathbf{H} = -\nabla{V} = \frac{B_{0}}{\mu_{0}}\hat{z} $, giving:
$$ V_{in} = \sum ^{\infty} _{l=0}  A_l r^l P_l (\cos \theta ) $$
$$ V_{out} = \sum ^{\infty} _{l=0} \frac{B_l}{r^{l+1}} P_l (\cos \theta ) - \frac{B_{0}}{\mu_{0}}rcos(\theta) $$
We now apply the following boundary conditions at the surface r = a:


*

*Continuity of the potential:  $ V(a,\theta )_{in} = V(a,\theta )_{out} $

*$ \mathbf{H_{in}^{\perp}} + \mathbf{M_{in}^{\perp}} = \mathbf{H_{out}^{\perp}}$ which follows from $\nabla\cdot\mathbf{B}=0$
The first boundary condition gives us:
$$B_{l}a^{-(l+1)}  = A_{l} $$  for $  l\neq 1$, and 
$$B_{1}a^{-3}  = A_{1} - \frac{B_{0}}{\mu_{0}} $$ for $  l= 1$
Before applying the second boundary condition, we must make the assumption that $\mathbf{M}$ is a constant vector pointing in the z-direction:
$$ \mathbf{M} = M\mathbf{\hat{z}} = M(cos\theta 
\mathbf{\hat{r}}  -sin\theta \hat{\mathbf{\theta}}) $$
The perpendicular components which feature in the second boundary condition are the $\mathbf{\hat{r}}$ - components of the vectors, so:
$$ -\frac{\partial{V_{in}}}{\partial{r}} + Mcos{\theta} = -\frac{\partial{V_{out}}}{\partial{r}} $$
We can now plug in our values for $ V_{in} $ ,$V_{out}$, and $ M $:
$$ -l A_{l}a^{l-1} = -(l+1)B_{l}a^{-(l+2)}   $$ for $  l\neq 1$, and 
$$- A_{1} +M = - \frac{B_{0}}{\mu_{0}} -2B_{1}a^{-3}  $$ for $  l= 1$
Looking at the equations concerning  $l\neq 1$, it follows that $A_{l} = B_{l} = 0$ for $l\neq 1$. However, the equations concerning $l=1$ terms in the potential can be simultaneously solved to find nonzero $A_{1},  B_{2} $ in terms of $B_{0}$ and $M$. Finally, to make contact with the question, call $B_{1} = C $ and $A_{1} = D$
