91Ó°ÊÓ

Write the expression for the total electromagnetic momentum.

${\stackrel{\mathbf{Áåœ}}{\mathbf{i}}}_{\mathbf{t}\mathbf{o}\mathbf{t}}\mathbf{=}{\stackrel{\mathbf{Áåœ}}{\mathbf{i}}}_{\mathbf{i}\mathbf{n}}\mathbf{+}{\stackrel{\mathbf{Áåœ}}{\mathbf{i}}}_{\mathbf{o}\mathbf{u}\mathbf{t}}$ ….. (1)

Here, ${\stackrel{Áåœ}{i}}_{in}$is the electromagnetic momentum from inside the sphere and${\stackrel{Áåœ}{i}}_{out}$ is the electromagnetic momentum from outside the sphere.

Write the expression for the electromagnetic momentum from inside the sphere.

${\stackrel{\mathbf{Áåœ}}{\mathbf{i}}}_{\mathbf{i}\mathbf{n}}\mathbf{=}{\mathit{Î\mu }}_{\mathbf{0}}\mathbf{∫}\left(E_{in}×B_{in}\right)\mathit{d}\mathit{Ï„}$ ….. (2)

Write the expression for the electromagnetic momentum from outside the sphere.

${\stackrel{\mathbf{Áåœ}}{\mathbf{i}}}_{\mathbf{o}\mathbf{u}\mathbf{t}}\mathbf{=}{\mathit{Î\mu }}_{\mathbf{0}}\mathbf{∫}\left(E_{out}×B_{out}\right)\mathit{d}\mathit{Ï„}$ ….. (3)

Write the expression for the electric field inside the sphere of uniform polarization.

$E_{in}=\frac{1}{3{\mathrm{Î\mu }}_0}\mathrm{ÒÏ}$

Here, the polarization of the sphere is $p=\left(\frac{4}{3}{\mathrm{Ï€¸é}}^3\right)P$.

Write the expression for the magnetic field inside the sphere of uniform magnetization.

$B_{in}=\frac{2}{3}{\mathrm{μ}}_{0}M$

Here, Mis the magnetization of the sphere.

Substitute the known values in equation (2).

$$\begin{gathered} {\stackrel{Áåœ}{i}}_{in}={\mathrm{Î\mu }}_0∫-\left(\frac{p}{3{\mathrm{Î\mu }}_0}×\frac{2}{3}{\mathrm{μ}}_{0}M\right)d\mathrm{Ï„} \\ {\stackrel{Áåœ}{i}}_{in}={\mathrm{Î\mu }}_0\frac{2}{3{\mathrm{Î\mu }}_0}∫\left(-p×M\right)d\mathrm{Ï„} \\ {\stackrel{Áåœ}{i}}_{in}=\frac{-2}{3}\left(\frac{4}{3}{\mathrm{Ï€¸é}}^3\right)\left(P×M\right) \\ {\stackrel{Áåœ}{i}}_{in}=\frac{8}{27}{\mathrm{Ï€¸é}}^3\left(\mathrm{P}×\mathrm{M}\right) \end{gathered}$$

Write the expression for the electric field outside the sphere of uniform polarization.

$E_{out}=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{1}{3}\left[3\left(p.\stackrel{Áåœ}{r}\right)\stackrel{Áåœ}{r}-p\right]$

Write the expression for the magnetic field outside the sphere of uniform magnetization.

$B_{out}=\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\frac{1}{r^3}\left[3\left(m.\stackrel{Áåœ}{r}\right)\stackrel{Áåœ}{r}-m\right]$

Here, the magnetization is $m=\left(\frac{4}{3}{\mathrm{Ï€¸é}}^3\right)M$.

Substitute the known values in equation (3).

$$\begin{gathered} {\stackrel{Áåœ}{i}}_{out}={\mathrm{Î\mu }}_0∫\frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{1}{3}\left[3\left(p.\stackrel{Áåœ}{r}\right)\stackrel{Áåœ}{r}-p\right]×\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\frac{1}{r^3}\left[3\left(m.\stackrel{Áåœ}{r}\right)\stackrel{Áåœ}{r}-m\right]d\mathrm{Ï„} \\ {\stackrel{Áåœ}{i}}_{out}=\frac{{\mathrm{μ}}_0}{16{\mathrm{Ï€}}^2}∫\frac{1}{r^6}\left[\begin{array}{c}9\left(p.\stackrel{Áåœ}{r}\right)×\stackrel{Áåœ}{r}×\left(m.\stackrel{Áåœ}{r}\right)\stackrel{Áåœ}{r}-3\left(p.\stackrel{Áåœ}{r}\right)\stackrel{Áåœ}{r}×m-\\ p×3\left(m.\stackrel{Áåœ}{r}\right)\stackrel{Áåœ}{r}+p×m\end{array}\right]d\mathrm{Ï„} \\ {\stackrel{Áåœ}{i}}_{out}=\frac{{\mathrm{μ}}_0}{16{\mathrm{Ï€}}^2}∫\frac{1}{r^6}\left[0-3\left(p.\stackrel{Áåœ}{r}\right)\left(\stackrel{Áåœ}{r}×m\right)+3\left(m.\stackrel{Áåœ}{r}\right)\left(\stackrel{Áåœ}{r}×p\right)+p×m\right]d\mathrm{Ï„} \\ {\stackrel{Áåœ}{i}}_{out}=\frac{{\mathrm{μ}}_0}{16{\mathrm{Ï€}}^2}{∫}_0^{2\mathrm{Ï€}}{∫}_0^{\mathrm{Ï€}}{∫}_0^{\mathrm{a}}\frac{1}{r^6}\left[-3\left(p.\stackrel{Áåœ}{r}\right)\left(\stackrel{Áåœ}{r}×m\right)+3\left(m.\stackrel{Áåœ}{r}\right)\left(\stackrel{Áåœ}{r}×p\right)+p×m\right]r^2\sin \mathrm{θ}d\mathrm{θ}drd\mathrm{Ï•} \end{gathered}$$

On further solving,

role="math" localid="1657612741238" $$\begin{gathered} {\stackrel{Áåœ}{\mathrm{i}}}_{\mathrm{out}}=\frac{{\mathrm{μ}}_0}{16{\mathrm{Ï€}}^2}{∫}_0^{\mathrm{a}}{∫}_0^{\mathrm{Ï€}}{∫}_0^{2\mathrm{Ï€}}\frac{1}{{\mathrm{r}}^4}\left\{\left[3\left[\stackrel{Áåœ}{\mathrm{r}}.\left(\stackrel{Áåœ}{\mathrm{r}}.\left(\mathrm{p}×\mathrm{m}\right)\right)-\left(\mathrm{p}×\mathrm{m}\right)\right]+\mathrm{p}×\mathrm{m}\right]\right\}\mathrm{²õ¾\pm ²Ôθ»åθ»å°ù»åÏ•} \\ {\stackrel{Áåœ}{\mathrm{i}}}_{\mathrm{out}}=\frac{{\mathrm{μ}}_0}{16{\mathrm{Ï€}}^2}{∫}_0^{\mathrm{a}}{∫}_0^{\mathrm{Ï€}}{∫}_0^{2\mathrm{Ï€}}\frac{1}{{\mathrm{r}}^4}\left[-2\left(\mathrm{p}×\mathrm{m}\right)+3\stackrel{Áåœ}{\mathrm{r}}\left[\stackrel{Áåœ}{\mathrm{r}}.\left(\mathrm{p}×\mathrm{m}\right)\right]\mathrm{p}×\mathrm{m}\right]\mathrm{²õ¾\pm ²Ôθ»åθ»å°ù»åÏ•}......\left(4\right) \end{gathered}$$

Here,$\stackrel{Áåœ}{\mathrm{r}}=\mathrm{²õ¾\pm ²Ô賦´Ç²õÏ•}\stackrel{Áåœ}{\mathrm{x}}+\mathrm{²õ¾\pm ²Ôθ²õ¾\pm ²ÔÏ•}\stackrel{Áåœ}{\mathrm{y}}+\mathrm{³¦´Ç²õÏ•}\stackrel{Áåœ}{\mathrm{z}}\mathrm{and}\stackrel{Áåœ}{\mathrm{r}}.\left(\mathrm{p}×\mathrm{m}\right)=\left|\mathrm{p}×\mathrm{m}\right|\mathrm{³¦´Ç²õθ}$.

Substitute the known values in equation (4).

$$\begin{gathered} {\stackrel{Áåœ}{i}}_{out}=\frac{{\mathrm{μ}}_0}{16{\mathrm{Ï€}}^2}{∫}_R^a\left(\frac{1}{r^4}dr\right)\left\{-2\left(p×m\right){∫}_0^{2\mathrm{Ï€}}{∫}_0^{\mathrm{Ï€}}\sin \mathrm{θ}d\mathrm{θ}dp+3\left|p×m\right|\stackrel{Áåœ}{z}{∫}_0^{2\mathrm{Ï€}}{∫}_0^{\mathrm{Ï€}}{\cos }^2\mathrm{θ}\sin \mathrm{θ}d\mathrm{θ}d\mathrm{Ï•}\right\} \\ {\stackrel{Áåœ}{i}}_{out}=\frac{{\mathrm{μ}}_0}{16{\mathrm{Ï€}}^2}{\left(\frac{1}{3r^4}\right)}_r^a\left[-2\left(p×m\right)\left(4\mathrm{Ï€}\right)+3\left(p×m\right)\frac{4\mathrm{Ï€}}{3}\right] \\ {\stackrel{Áåœ}{i}}_{out}=\frac{{\mathrm{μ}}_0}{16{\mathrm{Ï€}}^2\left(3{\mathrm{R}}^3\right)}\left[-8\mathrm{Ï€}\left(\mathrm{p}×\mathrm{m}\right)+4\mathrm{Ï€}\left(\mathrm{p}×\mathrm{m}\right)\right] \\ {\stackrel{Áåœ}{i}}_{out}=\frac{{\mathrm{μ}}_0}{16{\mathrm{Ï€}}^2\left(3{\mathrm{R}}^3\right)}\left[-4\mathrm{Ï€}\left(\frac{4}{3}{\mathrm{Ï€¸é}}^3\mathrm{P}\right)×\left(\frac{4}{3}{\mathrm{Ï€¸é}}^3\mathrm{M}\right)\right] \end{gathered}$$

On further solving,

$$\begin{gathered} {\stackrel{Áåœ}{\mathrm{i}}}_{\mathrm{out}}=\frac{{\mathrm{μ}}_0}{3{\mathrm{R}}^3}\left[\frac{1}{9}{\mathrm{R}}^6\left(4\mathrm{Ï€}\right)\left(\mathrm{M}×\mathrm{P}\right)\right] \\ {\stackrel{Áåœ}{\mathrm{i}}}_{\mathrm{out}}=\frac{4{\mathrm{μ}}_0\mathrm{Ï€}}{27}{\mathrm{R}}^3\left(\mathrm{M}×\mathrm{P}\right) \end{gathered}$$

Substitute the values in equation (1).

$$\begin{gathered} {\stackrel{Áåœ}{i}}_{tot}=\left[\frac{8}{27}{\mathrm{Ï€¸é}}^3\left(\mathrm{P}×\mathrm{M}\right)\right]+\left[\frac{4{\mathrm{μ}}_0\mathrm{Ï€}}{27}{R}^3\left(M×P\right)\right] \\ {\stackrel{Áåœ}{i}}_{tot}=\frac{4}{9}{\mathrm{μ}}_0{R}^3\left(M×P\right) \end{gathered}$$

Therefore, the total electromagnetic momentum of the configuration is

${\stackrel{Áåœ}{i}}_{tot}=\frac{4}{9}{\mathrm{μ}}_0{R}^3\left(M×P\right)$