Question

Notice the following parallel:

$$\begin{gathered} \{\begin{array}{l}\nabla ·D=0\nabla \times E=0,{\epsilon }_{0}E=D-P(Nofreecharge) \\ \nabla ·B=0\nabla \times H=0,{\mu }_{0}H=B-{\mu }_{0}M(Nofreecharge)\\ \end{array} \end{gathered}$$

Thus, the transcription $\mathit{D}\mathbf{\to }\mathit{B}\mathbf{,}\mathbf{}\mathit{E}\mathbf{\to }\mathit{H}\mathbf{,}\mathbf{}\mathit{P}\mathbf{\to }{\mathit{\mu }}_{\mathbf{0}}\mathit{M}\mathbf{,}\mathbf{}{\mathit{\epsilon }}_{\mathbf{0}}\mathbf{\to }{\mathit{\mu }}_{\mathbf{0}}$,, turns an electrostatic problem into an analogous magnetostatic one. Use this, together with your knowledge of the electrostatic results, to rederive.

(a) the magnetic field inside a uniformly magnetized sphere (Eq. 6.16);

(b) the magnetic field inside a sphere of linear magnetic material in an otherwise uniform magnetic field (Prob. 6.18);

(c) the average magnetic field over a sphere, due to steady currents within the sphere (Eq. 5.93).

Short answer

(a) The magnetic field inside the sphere is $\frac{2}{3}{\mu }_{0}M$.

(b) The magnetic field inside the linear magnetic material is $\frac{\left(1+{\mathcal{X}}_m\right)}{\left(1+{\displaystyle \frac{{\mathcal{X}}_m}{3}}\right)}{B}_0$.

(c) The average magnetic field over a sphere is $\frac{{\mu }_0}{4\mathrm{\pi }}\frac{2m}{R^3}$.

Step-by-step solution

Write the given data from the question:

When there is no charge in electrostatics.

$\begin{array}{rcl}\nabla ·D& =& 0\\ \nabla \times E& =& 0\\ {\epsilon }_{0}E& =& D-9\end{array}$

When there is no current in magnetostatics.

$\begin{array}{rcl}\nabla ·B& =& 0\\ \nabla \times H& =& 0\\ {\epsilon }_{0}H& =& B-{\mu }_{0}M\end{array}$

The transcription from electrostatics to magnetostatics.

$\begin{array}{rcl}D& \to & B\\ E& \to & H\\ P& \to & {\mu }_{0}M\\ {\epsilon }_0& \to & {\mu }_0\end{array}$

Calculate the magnetic field inside a uniformly magnetized sphere.

(a)

The electric field inside the uniformly polarized sphere is given by,

$E=-\frac{1}{3{\epsilon }_0}P$

Substitute ${\mu }_{0}M$for P, H for E and ${\mu }_0$for ${\epsilon }_0$into above equation.

$$\begin{gathered} H=-\frac{1}{3{\mu }_0}{\mu }_{0}M \\ H=-\frac{M}{3} \end{gathered}$$

The magnetic field inside the uniformly polarized sphere is given by,

$B={\mu }_0(H+M)$

Substitute $-\frac{M}{3}$ for H into above equation.

$$\begin{gathered} B={\mu }_0\left(-\frac{M}{3}+M\right) \\ B=\frac{2}{3}{\mu }_{0}M \end{gathered}$$

Hence the magnetic field inside the sphere is $\frac{2}{3}{\mu }_{0}M$.

Calculate the magnetic field inside sphere of linear magnetic material.

(b)

The electric field inside the linear dielectric sphere is given by,

$E=\frac{1}{\left(1+{\displaystyle \frac{{\mathcal{X}}_e}{3}}\right)E_0}$

Here, ${\mathcal{X}}_e$ is the electric susceptibility and E₀ is the uniform electric field.

The magnetic field inside the sphere of linear magnetic material is given by,

$H=\frac{1}{\left(1+{\displaystyle \frac{{\mathcal{X}}_m}{3}}\right)}{H}_0$ …… (1)

Here, ${\mathcal{X}}_m$is the magnetic susceptibility and H₀ is the external magnetizing field.

The relationship between the magnetic field strength B₀ and magnetic field intensity is given by,

$$\begin{gathered} B_0={\mu }_0{H}_0 \\ {H}_0=\frac{B_0}{{\mu }_0} \end{gathered}$$

The magnetic field inside the linear magnetic material is given by,

$$\begin{gathered} B={\mu }_0(1+{\mathcal{X}}_m)H \\ H=\frac{B}{{\mu }_0(1+{\mathcal{X}}_m)} \end{gathered}$$

Substitute$\frac{B}{{\mu }_0\left(1+{\mathcal{X}}_m\right)}$ for H , and$\frac{B_0}{{\mu }_0}$ for H₀ into equation (1).

$$\begin{gathered} \frac{B}{{\mu }_0\left(1+{\mathcal{X}}_m\right)}=\frac{B}{\left(1+{\displaystyle \frac{{\mathcal{X}}_m}{3}}\right)}\frac{B_0}{{\mu }_0} \\ B=\frac{\left(1+{\mathcal{X}}_m\right)}{\left(1+\frac{{\mathcal{X}}_m}{3}\right)}{B}_0 \end{gathered}$$

Hence the magnetic field inside the linear magnetic material is $\frac{\left(1+{\mathcal{X}}_m\right)}{\left(1+\frac{{\mathcal{X}}_m}{3}\right)}{B}_0$.

Calculate the average magnetic field over a sphere.

(c)

The electric field due to charge in the sphere is given by,

$E=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\rho }{R^3}$ ……. (2)

Here, p is sphere polarization.

The dipole moment in case of no charge is given by,

$p=\int P\tau$

Substitute $\int P\tau$and ${\mu }_0$for ${\epsilon }_0$into equation (2).

$E=\frac{1}{4{\mathrm{\pi \mu }}_0}\frac{1}{R^3}\int P\tau$

The average magnetization field inside the sphere is given by

role="math" localid="1657720730875" $$\begin{gathered} H_{avg}=\frac{-1}{4{\mathrm{\pi \mu }}_0}\frac{1}{R^3}\int {\mu }_{0}Md\tau \\ {H}_{avg}=\frac{-1}{4\pi R^3}\int Md\tau \dots \dots .\left(3\right) \end{gathered}$$

The magnetic dipole is given by,

role="math" localid="1657720650276" $m=\int Md\tau$

Substitute m for $\int Md\tau$into equation (3).

$$\begin{gathered} H_{avg}=\frac{-1}{4\pi R^3}m \\ {H}_{avg}=\frac{-m}{4\pi R^3} \end{gathered}$$

The average magnetic field is given by,

$$\begin{gathered} M_{avg}=\frac{m}{V} \\ {M}_{avg}=\frac{m}{{\displaystyle \frac{4}{3}}\pi R^3} \\ {M}_{avg}=\frac{3m}{4\pi R^3} \end{gathered}$$

The average magnetic field with in the sphere is given by,

$B_{avg}={\mu }_0\left(H_{avg}+M_{avg}\right)$

Substitute $\frac{3m}{4{\mathrm{\pi R}}^3}$ for M_avg and $\frac{-m}{4{\mathrm{\pi R}}^3}$ for H_avg into above equation.

$$\begin{gathered} B_{avg}={\mu }_0\left(\frac{-m}{4{\mathrm{\pi R}}^3}+\frac{3m}{4{\mathrm{\pi R}}^3}\right) \\ {B}_{avg}={\mu }_0\left(\frac{2m}{4{\mathrm{\pi R}}^3}\right) \\ {B}_{avg}=\frac{{\mu }_0}{4\mathrm{\pi }}\frac{2m}{R^3} \end{gathered}$$

Hence the average magnetic field over a sphere is $\frac{{\mu }_0}{4\mathrm{\pi }}\frac{2m}{R^3}$.