91Ó°ÊÓ

When there is no charge in electrostatics.

$\begin{array}{rcl}∇·D& =& 0\\ ∇×E& =& 0\\ {\mathrm{Î\mu }}_{0}E& =& D-9\end{array}$

When there is no current in magnetostatics.

$\begin{array}{rcl}∇·B& =& 0\\ ∇×H& =& 0\\ {\mathrm{Î\mu }}_{0}H& =& B-{\mathrm{μ}}_{0}M\end{array}$

The transcription from electrostatics to magnetostatics.

$\begin{array}{rcl}D& →& B\\ E& →& H\\ P& →& {\mathrm{μ}}_{0}M\\ {\mathrm{Î\mu }}_0& →& {\mathrm{μ}}_0\end{array}$

(a)

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

$E=-\frac{1}{3{\mathrm{Î\mu }}_0}P$

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

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

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

$B={\mathrm{μ}}_0(H+M)$

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

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

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

(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,

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

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

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

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

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

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

(c)

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

$E=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{\mathrm{ÒÏ}}{R^3}$ ……. (2)

Here, p is sphere polarization.

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

$p=∫P\mathrm{τ}$

Substitute $∫P\mathrm{Ï„}$and ${\mathrm{μ}}_0$for ${\mathrm{Î\mu }}_0$into equation (2).

$E=\frac{1}{4{\mathrm{πμ}}_0}\frac{1}{R^3}∫P\mathrm{τ}$

The average magnetization field inside the sphere is given by

role="math" localid="1657720730875" $$\begin{gathered} H_{avg}=\frac{-1}{4{\mathrm{πμ}}_0}\frac{1}{R^3}∫{\mathrm{μ}}_{0}Md\mathrm{τ} \\ {H}_{avg}=\frac{-1}{4\mathrm{π}{R}^3}∫Md\mathrm{τ}…….\left(3\right) \end{gathered}$$

The magnetic dipole is given by,

role="math" localid="1657720650276" $m=∫Md\mathrm{τ}$

Substitute m for $∫Md\mathrm{τ}$into equation (3).

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

The average magnetic field is given by,

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

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

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

Substitute $\frac{3m}{4{\mathrm{Ï€¸é}}^3}$ for M_avg and $\frac{-m}{4{\mathrm{Ï€¸é}}^3}$ for H_avg into above equation.

$$\begin{gathered} B_{avg}={\mathrm{μ}}_0\left(\frac{-m}{4{\mathrm{Ï€¸é}}^3}+\frac{3m}{4{\mathrm{Ï€¸é}}^3}\right) \\ {B}_{avg}={\mathrm{μ}}_0\left(\frac{2m}{4{\mathrm{Ï€¸é}}^3}\right) \\ {B}_{avg}=\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\frac{2m}{R^3} \end{gathered}$$

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