Question

A magnetic dipole $\mathit{m}$ is imbedded at the center of a sphere (radius $\mathit{R}$) of linear magnetic material (permeability $\mathit{\mu }$). Show that the magnetic field inside the sphere $\mathbf{0}\mathbf{<}\mathit{r}\mathbf{\le }\mathit{R}$ is

$\frac{\mathbf{\mu }}{\mathbf{4}\mathbf{\pi }}\left\{\frac{1}{r^3}\left[3(m.\widehat{r}\widehat{r}-m)\right]+\frac{2({\mu }_0-\mu )m}{(2{\mu }_0+\mu )R^3}\right\}$

What is the field outside the sphere?

Short answer

The filed inside the sphere is$\frac{\mu }{4\pi }\left[\frac{1}{r^3}(3(m\cdot \widehat{r})\widehat{r}-m)+\frac{2}{R^3}\frac{(\mu -{\mu }_0)}{(2{\mu }_0+\mu )}m\right]$ and outside the sphere is $\frac{{\mu }_0}{4\pi }\left(\frac{3\mu m}{2{\mu }_0+\mu }\right)\frac{1}{r^3}[3(m\cdot \widehat{r})\widehat{r}-m]$.

Step-by-step solution

Write the given data from the question.

The magnetic dipole is $m$.

The radius of sphere is $R$.

The magnetic permeability is $\mu$.

Determine the formulas to calculate the magnetic field inside the sphere.

The expression for the magnetic field at the centre of sphere due to dipole is given as follows.

${\mathbf{B}}_{\mathbf{centre}}\mathbf{=}\frac{\mathbf{\mu }}{\mathbf{4}\mathbf{\pi }}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{3}}}\mathbf{(}\mathbf{3}\mathbf{\left(}\mathbf{m}\widehat{\mathbf{r}}\mathbf{\right)}\widehat{\mathbf{r}}\mathbf{-}\mathbf{m}\mathbf{)}$

The expression for surface current density is given as follows.

${\mathbf{K}}_{\mathbf{b}}\mathbf{=}\mathbf{M}\mathbf{\left(}\mathbf{R}\mathbf{\right)}\mathbf{\times }\widehat{\mathbf{r}}$

Calculate the magnetic field inside the sphere.

The magnetic dipole moment due to embedded dipole is given by,

$m_b={\chi }_{m}m$

Here,${\chi }_m$is the magnetic susceptibility and magnetic moment of the dipole.

The net dipole at the centre of the sphere is given by,

$m_{centre}=m+m_b$

Substitute ${\chi }_{m}m$for $m_b$into above equation.

$\begin{array}{l}{m}_{centre}=m+{\chi }_{m}m\\ m_{centre}=m(1+{\chi }_m)\end{array}$

Substitute$\frac{\mu }{{\mu }_0}$for$1+{\chi }_m$into above equation.

$m_{centre}=\frac{\mu }{{\mu }_0}m$

The magnetic field at the centre of the sphere is given by,

$B_{centre}=\frac{\mu }{4\pi }\frac{1}{r^3}(3(m\cdot \widehat{r})\widehat{r}-m)$

The magnetic field at surface due to surface bound current is given by,

$B_{surface}=Am$

Here,$A$is the constant.

The magnetization is given by,

$\begin{array}{l}M={\chi }_{m}H\\ M={\chi }_m\frac{B}{\mu }\\ M={\chi }_m\frac{B_{centre}+B_{surface}}{\mu }\end{array}$

Substitute$\frac{\mu }{4\pi }\frac{1}{r^3}(3(m\cdot \widehat{r})\widehat{r}-m)$for $B_{centre}$and$Am$ for$B_{surface}$into above equation.

$M=\frac{{\chi }_m}{\mu }\left[\frac{\mu }{4\pi }\frac{1}{r^3}(3(m\cdot \widehat{r})\widehat{r}-m)+Am\right]$

$M=\frac{{\chi }_m}{4\pi }\frac{1}{r^3}(3(m\cdot \widehat{r})\widehat{r}-m)+\frac{{\chi }_m}{\mu }Am$ …… (1)

The magnetization due to bound current $J_b$is given by,

$\begin{array}{l}{J}_b=\nabla \times M\\ J_b=0\end{array}$

Because the first term curl less and second term is constant.

The expression for the surface charge density is given by,

$K_b=M\left(R\right)\times \widehat{r}$

From equation (1) and (2),

$K_b=\frac{{\chi }_m}{4\pi R^3}(-m\times \widehat{r})+\frac{A{\chi }_m}{\mu }(m\times \widehat{r})$

$K_b={\chi }_{m}m\left(-\frac{1}{4\pi R^3}+\frac{A}{\mu }\right)\sin \theta \widehat{\varphi }$ ……. (3)

The surface current due to spinning sphere is given by,

$K=\sigma v$

$K=\sigma \omega R\sin \theta \widehat{\varphi }$ …… (4)

From the equation (3) and (4).

$\sigma \omega R={\chi }_{m}m\left(-\frac{1}{4\pi R^3}+\frac{A}{\mu }\right)$

Filed produce by the surface current due for point inside the surface is given by,

$B_{surfacecentre}=\frac{2}{3}{\mu }_0\left(\sigma \omega R\right)$

Substitute ${\chi }_{m}m\left(-\frac{1}{4\pi R^3}+\frac{A}{\mu }\right)$for $\sigma \omega R$into above equation.

$B_{surfacecentre}=\frac{2}{3}{\mu }_0\left[{\chi }_{m}m\left(-\frac{1}{4\pi R^3}+\frac{A}{\mu }\right)\right]$ …… (5)

The equation (5) will be constant if,

$\begin{array}{c}A=\frac{2}{3}{\mu }_0\left[{\chi }_{m}m\left(-\frac{1}{4\pi R^3}+\frac{A}{\mu }\right)\right]\\ A\left[1-\frac{2}{3}{\mu }_0{\chi }_{m}m\right]=-\frac{2}{3}\frac{{\mu }_0{\chi }_m}{4\pi R^3}\end{array}$

Substitute $\frac{\mu }{{\mu }_0}-1$for ${\chi }_m$into above equation.

$\begin{array}{c}A\left[1-\frac{2}{3}\frac{{\mu }_0}{\mu }\left(\frac{\mu }{{\mu }_0}-1\right)\right]=-\frac{2}{3}\frac{{\mu }_0}{4\pi R^3}\left(\frac{\mu }{{\mu }_0}-1\right)\\ A\left[1-\frac{2}{3}+\frac{2}{3}\frac{{\mu }_0}{\mu }\right]=-\frac{2}{3}\frac{(\mu -{\mu }_0)}{4\pi R^3}\\ A\left[1+\frac{2{\mu }_0}{\mu }\right]=-\frac{2}{3}\frac{(\mu -{\mu }_0)}{4\pi R^3}\\ A=\frac{\mu }{4\pi }\frac{2}{R^3}\frac{(\mu -{\mu }_0)}{(2{\mu }_0+\mu )}\end{array}$

The magnetic field inside the surface is given by,

$B_{in}=B_{centre}+B_{suface}$

Substitute$\frac{\mu }{4\pi }\frac{1}{r^3}(3(m\cdot \widehat{r})\widehat{r}-m)$for$B_{centre}$and $Am$ for$B_{surface}$into above equation.

$B_{in}=\frac{\mu }{4\pi }\frac{1}{r^3}(3(m\cdot \widehat{r})\widehat{r}-m)+Am$

Substitute $\frac{\mu }{4\pi }\frac{2}{R^3}\frac{(\mu -{\mu }_0)}{(2{\mu }_0+\mu )}$for $A$into above equation.

$\begin{array}{l}{B}_{in}=\frac{\mu }{4\pi }\frac{1}{r^3}(3(m\cdot \widehat{r})\widehat{r}-m)+\frac{\mu }{4\pi }\frac{2}{R^3}\frac{(\mu -{\mu }_0)}{(2{\mu }_0+\mu )}m\\ B_{in}=\frac{\mu }{4\pi }\left[\frac{1}{r^3}(3(m\cdot \widehat{r})\widehat{r}-m)+\frac{2}{R^3}\frac{(\mu -{\mu }_0)}{(2{\mu }_0+\mu )}m\right]\end{array}$

Hence the magnetic field inside the sphere is

$\frac{\mu }{4\pi }\left[\frac{1}{r^3}(3(m\cdot \widehat{r})\widehat{r}-m)+\frac{2}{R^3}\frac{(\mu -{\mu }_0)}{(2{\mu }_0+\mu )}m\right]$

The magnetic field outside the sphere due to surface current is given by,

${B^{\text{'}}}_{surfacecentre}=\left(\frac{4}{3}\pi R^3\right)\left(\sigma \omega R\right)$

Substitute $\frac{3}{2{\mu }_0}{B}_{surface}$for $\sigma \omega R$into above equation.

${B^{\text{'}}}_{surfacecentre}=\left(\frac{4}{3}\pi R^3\right)\left(\frac{3}{2{\mu }_0}{B}_{surfacecentre}\right)$

Substitute $\frac{2\mu }{4\pi R^3}\frac{(\mu -{\mu }_0)}{(2{\mu }_0+\mu )}m$for $B_{surface}$into above equation.

$\begin{array}{l}{B^{\text{'}}}_{surfacecentre}=\left(\frac{4}{3}\pi R^3\right)\left(\frac{3}{2{\mu }_0}\frac{2\mu }{4\pi R^3}\frac{(\mu -{\mu }_0)}{(2{\mu }_0+\mu )}m\right)\\ {B^{\text{'}}}_{surfacecentre}=\frac{\mu }{{\mu }_0}\frac{(\mu -{\mu }_0)}{(2{\mu }_0+\mu )}m\end{array}$

The total dipole moment is given by,

$\begin{array}{l}{m}_{tot}=\frac{\mu }{{\mu }_0}m+\frac{\mu }{{\mu }_0}m\left(\frac{{\mu }_0-\mu }{2{\mu }_0+\mu }\right)\\ m_{tot}=\frac{3\mu m}{2{\mu }_0+\mu }\end{array}$

The field outside the sphere is given by,

$B_{out}\frac{{\mu }_0}{4\pi }{m}_{tot}\frac{1}{r^3}[3(m\cdot \widehat{r})\widehat{r}-m]$

Substitute $\frac{3\mu m}{2{\mu }_0+\mu }$for$m_{tot}$into above equation.

$B_{out}=\frac{{\mu }_0}{4\pi }\left(\frac{3\mu m}{2{\mu }_0+\mu }\right)\frac{1}{r^3}[3(m\cdot \widehat{r})\widehat{r}-m]$

Hence the filed inside the sphere is$\frac{\mu }{4\pi }\left[\frac{1}{r^3}(3(m\cdot \widehat{r})\widehat{r}-m)+\frac{2}{R^3}\frac{(\mu -{\mu }_0)}{(2{\mu }_0+\mu )}m\right]$ and outside the sphere is$\frac{{\mu }_0}{4\pi }\left(\frac{3\mu m}{2{\mu }_0+\mu }\right)\frac{1}{r^3}\left[3(m\cdot \widehat{r})\widehat{r}-m\right]$ .