lf ${\mathbf{J}}_f=\mathbf{0}$ everywhere, the curl of $\mathbf{H}$ vanishes (Eq. 6.19), and we can express $\mathbf{H}$ as the gradient of a scalar potential W:

$\mathbf{H}\mathbf{=}\mathbf{-}\mathbf{\nabla }\mathbf{W}$

According to Eq. 6.23, then,

${\mathbf{\nabla }}^2\mathbf{W}\mathbf{=}\mathbf{(}\mathbf{\nabla }\mathbf{\cdot }\mathbf{M}\mathbf{)}$

So $\mathbf{W}$obeys Poisson's equation, with $\mathbf{\nabla }\mathbf{\cdot }\mathbf{M}$ as the "source." This opens up all the machinery of Chapter 3. As an example, find the field inside a uniformly magnetized sphere (Ex. 6.1) by separation of variables.

Consider the curl of $\mathrm{H}$vanishes (Eq. 6.19), and we can express $\mathrm{H}$ as the gradient of a scalar potential W.

Consider $\nabla \cdot \text{M}$everywhere except at the surface $(\text{r}=\text{R})$, so $\mathrm{W}$satisfies Laplace's equation in the regions $\text{r}<\text{R}$and $\text{r}>\text{R}$.

Write the formula of magnetic field inside a uniformly magnetized sphere by separation of variables.

$\overrightarrow{\mathbf{B}}\mathbf{=}{\mathbf{\mu }}_0\mathbf{(}\overrightarrow{\mathbf{H}}\mathbf{+}\overrightarrow{\mathbf{M}}\mathbf{)}$ …… (1)

Here, ${\mathbf{\mu }}_0$ is permeability, $\mathbf{H}$ is gradient of scalar potential and $\overrightarrow{\mathbf{M}}$ is magnetization.

First we start with the boundary conditions for the problem.

${\nabla }^2\mathrm{W}=\nabla \cdot \overrightarrow{\mathrm{M}}$ ; $\overrightarrow{\mathrm{H}}=-\nabla \mathrm{W}$

Let represent the sphere's interior or exterior. For the boundary, we may take an infinitesimal line integral of length $\mathrm{\epsilon }$, and we obtain

$\int \overrightarrow{\mathrm{H}}\cdot \mathrm{d}\overrightarrow{\mathrm{l}}=-\int \nabla \mathrm{W}\cdot \mathrm{d}\overrightarrow{\mathrm{l}}$

Using the fact that line integral is infinitesimal and the gradient theorem

$e\widehat{n}\cdot \overrightarrow{H}=-(W_2-W_1)$

In the limit $\text{ε}\to \text{0}$ we have

$W_1=W_2$

Second, we may set up a tiny Gaussian pillbox with base $A$and height $\epsilon$ at the boundary and calculate the integral below.

$\begin{array}{l}\int \nabla \cdot (\nabla W)dr=\int \nabla \cdot \overrightarrow{M}dr\\ \oint \nabla W\cdot d\overrightarrow{a}=\oint \overrightarrow{M}\cdot d\overrightarrow{a}\\ A(\nabla W_2-\nabla W_1)\cdot \widehat{n}=-A\overrightarrow{M}\cdot \widehat{n}\end{array}$

Since $\overrightarrow{M}=M\widehat{z}$, we have

$\frac{\partial W_2}{\partial r}{{|}_{r=R}-\frac{\partial W_1}{\partial r}|}_{r=R}=-M\cos \theta$

Let $\overrightarrow{M}=M\widehat{z}$be along the z axis. Due to axial $\left(\phi \right)$symmetry, potential is given by

$W=\left\{\begin{array}{l}{W}_1={\sum }_{l=0}^{\infty }{A}_l{r}^l{P}_l\left(\cos \theta \right)\\ W_2={\sum }_{l=0}^{{}^{\infty }}{B}_l{r}^{-l-l}{P}_l\left(\cos \theta \right)\end{array}\right\}$

Consider solve for $A_l$and $B_l$using boundary conditions. From (1) we have

$\begin{array}{c}{W}_1=W_2\\ A_l{R}^{2l+1}=B_l\end{array}$

Due to equation (2) and proportionality of $A_l$ and $B_l$ we see that only $l=1$terms will survive (orthogonlity of Legendre polynomials).

$\begin{array}{l}2B_l{R}^{-3}+A_l=M\\ \Rightarrow A_1=\frac{M}{3}\\ \Rightarrow B_1=\frac{M{R}^3}{3}\end{array}$

So, we have

$W=\left\{\begin{array}{l}{W}_2=-\frac{M}{3}r\cos \theta r<R\\ W_1=-\frac{M{R}^3}{3r^2}\cos \theta r>R\end{array}\right\}$

For the filed inside the sphere we have

$\begin{array}{c}\overrightarrow{H}=-\nabla W\\ =-\nabla \left(\frac{M}{3}r\cos \theta \right)\\ =-\frac{M}{3}\widehat{z}\\ =-\frac{M}{3};z=r\cos \theta \end{array}$

Determine the magnetic field inside a uniformly magnetized sphere by separation of variables.

Substitute$\frac{2}{3}\overrightarrow{M}$ for $(\overrightarrow{H}+\overrightarrow{M})$ into equation (1).

$\overrightarrow{B}={\mu }_0\frac{2}{3}\overrightarrow{M};r<R$

Therefore, the value of magnetic field inside a uniformly magnetized sphere by separation of variables is$\overrightarrow{B}={\mu }_0\frac{2}{3}\overrightarrow{M}$ .