A sphere of radius R carries a uniform polarization P and a uniform magnetization M (not necessarily in the same direction). Find the electromagnetic momentum of this configuration. [Answer:$\frac{\mathbf{4}}{\mathbf{9}}\mathit{\tau }\mathit{\tau }{\mathit{\mu }}_{\mathbf{0}}{\mathit{R}}^{\mathbf{3}}\left(M\times P\right)$]

Write the expression for the total electromagnetic momentum.

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

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

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

${\stackrel{\mathbf{⏜}}{\mathbf{i}}}_{\mathbf{i}\mathbf{n}}\mathbf{=}{\mathit{\epsilon }}_{\mathbf{0}}\mathbf{\int }\left(E_{in}\times B_{in}\right)\mathit{d}\mathit{\tau }$ ….. (2)

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

${\stackrel{\mathbf{⏜}}{\mathbf{i}}}_{\mathbf{o}\mathbf{u}\mathbf{t}}\mathbf{=}{\mathit{\epsilon }}_{\mathbf{0}}\mathbf{\int }\left(E_{out}\times B_{out}\right)\mathit{d}\mathit{\tau }$ ….. (3)

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

$E_{in}=\frac{1}{3{\epsilon }_0}\rho$

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

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

$B_{in}=\frac{2}{3}{\mu }_{0}M$

Here, Mis the magnetization of the sphere.

Substitute the known values in equation (2).

$$\begin{gathered} {\stackrel{⏜}{i}}_{in}={\epsilon }_0\int -\left(\frac{p}{3{\epsilon }_0}\times \frac{2}{3}{\mu }_{0}M\right)d\tau \\ {\stackrel{⏜}{i}}_{in}={\epsilon }_0\frac{2}{3{\epsilon }_0}\int \left(-p\times M\right)d\tau \\ {\stackrel{⏜}{i}}_{in}=\frac{-2}{3}\left(\frac{4}{3}{\mathrm{\pi R}}^3\right)\left(P\times M\right) \\ {\stackrel{⏜}{i}}_{in}=\frac{8}{27}{\mathrm{\pi R}}^3\left(\mathrm{P}\times \mathrm{M}\right) \end{gathered}$$

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

$E_{out}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{1}{3}\left[3\left(p.\stackrel{⏜}{r}\right)\stackrel{⏜}{r}-p\right]$

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

$B_{out}=\frac{{\mu }_0}{4\mathrm{\pi }}\frac{1}{r^3}\left[3\left(m.\stackrel{⏜}{r}\right)\stackrel{⏜}{r}-m\right]$

Here, the magnetization is $m=\left(\frac{4}{3}{\mathrm{\pi R}}^3\right)M$.

Substitute the known values in equation (3).

$$\begin{gathered} {\stackrel{⏜}{i}}_{out}={\epsilon }_0\int \frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{1}{3}\left[3\left(p.\stackrel{⏜}{r}\right)\stackrel{⏜}{r}-p\right]\times \frac{{\mu }_0}{4\mathrm{\pi }}\frac{1}{r^3}\left[3\left(m.\stackrel{⏜}{r}\right)\stackrel{⏜}{r}-m\right]d\tau \\ {\stackrel{⏜}{i}}_{out}=\frac{{\mu }_0}{16{\mathrm{\pi }}^2}\int \frac{1}{r^6}\left[\begin{array}{c}9\left(p.\stackrel{⏜}{r}\right)\times \stackrel{⏜}{r}\times \left(m.\stackrel{⏜}{r}\right)\stackrel{⏜}{r}-3\left(p.\stackrel{⏜}{r}\right)\stackrel{⏜}{r}\times m-\\ p\times 3\left(m.\stackrel{⏜}{r}\right)\stackrel{⏜}{r}+p\times m\end{array}\right]d\tau \\ {\stackrel{⏜}{i}}_{out}=\frac{{\mu }_0}{16{\mathrm{\pi }}^2}\int \frac{1}{r^6}\left[0-3\left(p.\stackrel{⏜}{r}\right)\left(\stackrel{⏜}{r}\times m\right)+3\left(m.\stackrel{⏜}{r}\right)\left(\stackrel{⏜}{r}\times p\right)+p\times m\right]d\tau \\ {\stackrel{⏜}{i}}_{out}=\frac{{\mu }_0}{16{\mathrm{\pi }}^2}{\int }_0^{2\mathrm{\pi }}{\int }_0^{\mathrm{\pi }}{\int }_0^{\mathrm{a}}\frac{1}{r^6}\left[-3\left(p.\stackrel{⏜}{r}\right)\left(\stackrel{⏜}{r}\times m\right)+3\left(m.\stackrel{⏜}{r}\right)\left(\stackrel{⏜}{r}\times p\right)+p\times m\right]r^2\sin \theta d\theta drd\varphi \end{gathered}$$

On further solving,

role="math" localid="1657612741238" $$\begin{gathered} {\stackrel{⏜}{\mathrm{i}}}_{\mathrm{out}}=\frac{{\mathrm{\mu }}_0}{16{\mathrm{\pi }}^2}{\int }_0^{\mathrm{a}}{\int }_0^{\mathrm{\pi }}{\int }_0^{2\mathrm{\pi }}\frac{1}{{\mathrm{r}}^4}\left\{\left[3\left[\stackrel{⏜}{\mathrm{r}}.\left(\stackrel{⏜}{\mathrm{r}}.\left(\mathrm{p}\times \mathrm{m}\right)\right)-\left(\mathrm{p}\times \mathrm{m}\right)\right]+\mathrm{p}\times \mathrm{m}\right]\right\}\mathrm{sin\theta d\theta drd\varphi } \\ {\stackrel{⏜}{\mathrm{i}}}_{\mathrm{out}}=\frac{{\mathrm{\mu }}_0}{16{\mathrm{\pi }}^2}{\int }_0^{\mathrm{a}}{\int }_0^{\mathrm{\pi }}{\int }_0^{2\mathrm{\pi }}\frac{1}{{\mathrm{r}}^4}\left[-2\left(\mathrm{p}\times \mathrm{m}\right)+3\stackrel{⏜}{\mathrm{r}}\left[\stackrel{⏜}{\mathrm{r}}.\left(\mathrm{p}\times \mathrm{m}\right)\right]\mathrm{p}\times \mathrm{m}\right]\mathrm{sin\theta d\theta drd\varphi }......\left(4\right) \end{gathered}$$

Here,$\stackrel{⏜}{\mathrm{r}}=\mathrm{sin\theta cos\varphi }\stackrel{⏜}{\mathrm{x}}+\mathrm{sin\theta sin\varphi }\stackrel{⏜}{\mathrm{y}}+\mathrm{cos\varphi }\stackrel{⏜}{\mathrm{z}}\mathrm{and}\stackrel{⏜}{\mathrm{r}}.\left(\mathrm{p}\times \mathrm{m}\right)=\left|\mathrm{p}\times \mathrm{m}\right|\mathrm{cos\theta }$.

Substitute the known values in equation (4).

$$\begin{gathered} {\stackrel{⏜}{i}}_{out}=\frac{{\mu }_0}{16{\mathrm{\pi }}^2}{\int }_R^a\left(\frac{1}{r^4}dr\right)\left\{-2\left(p\times m\right){\int }_0^{2\mathrm{\pi }}{\int }_0^{\mathrm{\pi }}\sin \theta d\theta dp+3\left|p\times m\right|\stackrel{⏜}{z}{\int }_0^{2\mathrm{\pi }}{\int }_0^{\mathrm{\pi }}{\cos }^2\theta \sin \theta d\theta d\varphi \right\} \\ {\stackrel{⏜}{i}}_{out}=\frac{{\mu }_0}{16{\mathrm{\pi }}^2}{\left(\frac{1}{3r^4}\right)}_r^a\left[-2\left(p\times m\right)\left(4\mathrm{\pi }\right)+3\left(p\times m\right)\frac{4\mathrm{\pi }}{3}\right] \\ {\stackrel{⏜}{i}}_{out}=\frac{{\mu }_0}{16{\mathrm{\pi }}^2\left(3{\mathrm{R}}^3\right)}\left[-8\mathrm{\pi }\left(\mathrm{p}\times \mathrm{m}\right)+4\mathrm{\pi }\left(\mathrm{p}\times \mathrm{m}\right)\right] \\ {\stackrel{⏜}{i}}_{out}=\frac{{\mu }_0}{16{\mathrm{\pi }}^2\left(3{\mathrm{R}}^3\right)}\left[-4\mathrm{\pi }\left(\frac{4}{3}{\mathrm{\pi R}}^3\mathrm{P}\right)\times \left(\frac{4}{3}{\mathrm{\pi R}}^3\mathrm{M}\right)\right] \end{gathered}$$

On further solving,

$$\begin{gathered} {\stackrel{⏜}{\mathrm{i}}}_{\mathrm{out}}=\frac{{\mathrm{\mu }}_0}{3{\mathrm{R}}^3}\left[\frac{1}{9}{\mathrm{R}}^6\left(4\mathrm{\pi }\right)\left(\mathrm{M}\times \mathrm{P}\right)\right] \\ {\stackrel{⏜}{\mathrm{i}}}_{\mathrm{out}}=\frac{4{\mathrm{\mu }}_0\mathrm{\pi }}{27}{\mathrm{R}}^3\left(\mathrm{M}\times \mathrm{P}\right) \end{gathered}$$

Substitute the values in equation (1).

$$\begin{gathered} {\stackrel{⏜}{i}}_{tot}=\left[\frac{8}{27}{\mathrm{\pi R}}^3\left(\mathrm{P}\times \mathrm{M}\right)\right]+\left[\frac{4{\mu }_0\mathrm{\pi }}{27}{R}^3\left(M\times P\right)\right] \\ {\stackrel{⏜}{i}}_{tot}=\frac{4}{9}{\mu }_0{R}^3\left(M\times P\right) \end{gathered}$$

Therefore, the total electromagnetic momentum of the configuration is

${\stackrel{⏜}{i}}_{tot}=\frac{4}{9}{\mu }_0{R}^3\left(M\times P\right)$