91Ó°ÊÓ

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 $∇â‹\dots \text{M}$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.

$\stackrel{\mathbf{→}}{\mathbf{B}}\mathbf{=}{\mathbf{μ}}_0\mathbf{(}\stackrel{\mathbf{→}}{\mathbf{H}}\mathbf{+}\stackrel{\mathbf{→}}{\mathbf{M}}\mathbf{)}$ …… (1)

Here, ${\mathbf{μ}}_0$ is permeability, $\mathbf{H}$ is gradient of scalar potential and $\stackrel{\mathbf{→}}{\mathbf{M}}$ is magnetization.

First we start with the boundary conditions for the problem.

${∇}^2\mathrm{W}=∇â‹\dots \stackrel{→}{\mathrm{M}}$ ; $\stackrel{→}{\mathrm{H}}=−∇\mathrm{W}$

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

$∫\stackrel{→}{\mathrm{H}}â‹\dots \mathrm{d}\stackrel{→}{\mathrm{l}}=−∫∇\mathrm{W}â‹\dots \mathrm{d}\stackrel{→}{\mathrm{l}}$

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

$e\widehat{n}â‹\dots \stackrel{→}{H}=−(W_2−W_1)$

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

$W_1=W_2$

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

$\begin{array}{l}∫∇â‹\dots (∇W)dr=∫∇â‹\dots \stackrel{→}{M}dr\\ âˆ\circledR ∇Wâ‹\dots d\stackrel{→}{a}=âˆ\circledR \stackrel{→}{M}â‹\dots d\stackrel{→}{a}\\ A(∇W_2−∇W_1)â‹\dots \widehat{n}=−A\stackrel{→}{M}â‹\dots \widehat{n}\end{array}$

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

$\frac{∂W_2}{∂r}{{|}_{r=R}−\frac{∂W_1}{∂r}|}_{r=R}=−M\cos \mathrm{θ}$

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

$W=\left\{\begin{array}{l}{W}_1={∑}_{l=0}^{\mathrm{∞}}{A}_l{r}^l{P}_l\left(\cos \mathrm{θ}\right)\\ W_2={∑}_{l=0}^{{}^{\mathrm{∞}}}{B}_l{r}^{−l−l}{P}_l\left(\cos \mathrm{θ}\right)\end{array}\right\}$

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).

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

So, we have

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

For the filed inside the sphere we have

$\begin{array}{c}\stackrel{→}{H}=−∇W\\ =−∇\left(\frac{M}{3}r\cos \mathrm{θ}\right)\\ =−\frac{M}{3}\widehat{z}\\ =−\frac{M}{3};z=r\cos \mathrm{θ}\end{array}$

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

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

$\stackrel{→}{B}={\mathrm{μ}}_0\frac{2}{3}\stackrel{→}{M};\text{ }r<R$

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