Suppose you had an electric charge ${\mathit{q}}_{\mathbf{e}}$and a magnetic monopole ${\mathit{q}}_{\mathbf{m}}$. The field of the electric charge is

$\mathit{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{q}}{{\mathbf{r}}^{\mathbf{2}}}\hat{\mathbf{r}}$

(of course), and the field of the magnetic monopole is

$\mathit{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}}{\mathbf{4}\mathbf{\pi }}\frac{{\mathbf{q}}_{\mathbf{m}}}{{\mathbf{r}}^{\mathbf{2}}}\hat{\mathbf{r}}$.

Find the total angular momentum stored in the fields, if the two charges are separated by a distance d. [Answer: $\mathbf{\left(}\frac{{\mathbf{\mu }}_{\mathbf{0}}}{\mathbf{4}\mathbf{\pi }}\mathbf{\right)}{\mathit{q}}_{\mathbf{e}}{\mathit{q}}_{\mathbf{m}}$]20

By the given condition, draw the depicted situation.

Write the expression for an electric field due to an electric charge.

$\mathit{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\pi }{\mathbf{\epsilon }}_{\mathbf{0}}}\frac{{\mathbf{q}}_{\mathbf{e}}}{{\mathbf{r}}^{\mathbf{3}}}\mathit{r}$ ……. (1)

Write the expression for the magnetic field due to the magnetic monopole.

$\mathit{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}}{\mathbf{4}\mathbf{\pi }}\frac{{\mathbf{q}}_{\mathbf{m}}}{{\mathbf{r}}^{\mathbf{2}}}\frac{{\mathbf{r}}^{\mathbf{1}}}{{\mathbf{r}}^{{\mathbf{1}}^{\mathbf{3}}}}$ ……. (2)

Derive the expression for the radial component:

$$\begin{gathered} {\mathit{r}}^{\mathbf{1}}\mathbf{=}\mathit{r}\mathbf{-}\mathit{d}\hat{\mathbf{z}} \\ {\mathit{r}}^{\mathbf{1}}\mathbf{=}{\left(r^2+d^2-2rdcos\theta \right)}^{\frac{\mathbf{1}}{\mathbf{2}}} \end{gathered}$$

Substitute the known values in equation (2).

$\mathit{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}}{\mathbf{4}\mathbf{\pi }}\frac{{\mathbf{q}}_{\mathbf{m}}\mathbf{(}\mathbf{r}\mathbf{-}\mathbf{d}\hat{\mathbf{z}}\mathbf{)}}{{\mathbf{(}{\mathbf{r}}^{\mathbf{2}}\mathbf{+}{\mathbf{d}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{rd}\mathbf{}\mathbf{cos\theta }\mathbf{)}}^{{\displaystyle \frac{\mathbf{3}}{\mathbf{2}}}}}$

Write the expression for momentum density.

$\mathit{\rho }\mathbf{=}{\mathit{\epsilon }}_{\mathbf{0}}\mathbf{(}\mathbf{E}\mathbf{\times }\mathbf{B}\mathbf{)}$

Substitute the known values in the above expression.

$\mathit{\rho }\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{q}}_{\mathbf{e}}{\mathbf{q}}_{\mathbf{m}}}{{\mathbf{\left(}\mathbf{4}\mathbf{\pi }\mathbf{\right)}}^{\mathbf{2}}}\mathbf{·}\frac{\mathbf{(}\mathbf{-}\mathbf{d}\mathbf{)}\mathbf{(}{\displaystyle \stackrel{\boxed{\xcancel{}}}{r}}\mathbf{\times }\hat{\mathbf{z}}\mathbf{)}}{{\mathbf{r}}^{\mathbf{3}}{\mathbf{(}{\mathbf{r}}^{\mathbf{2}}\mathbf{+}{\mathbf{d}}^{\mathbf{2}}\mathbf{2}\mathbf{rd}\mathbf{}\mathbf{cos\theta }\mathbf{)}}^{{\displaystyle \frac{\mathbf{3}}{\mathbf{2}}}}}$

Write the expression for the angular momentum density.

$\mathit{I}\mathbf{=}\stackrel{\boxed{\xcancel{}}}{\mathbf{r}}\mathbf{\times }\mathit{\rho }$

Substitute the known values in the above expression.

$$\begin{gathered} \mathit{I}\mathbf{=}\stackrel{\boxed{\xcancel{}}}{\mathbf{r}}\left[\frac{{\mu }_0{q}_e{q}_m}{{\left(4\pi \right)}^2}·\frac{\left(-d\right)\left({\displaystyle \stackrel{\boxed{\xcancel{}}}{r}}\times \hat{z}\right)}{r^3{\left(r^2+d^{2}2rdcos\theta \right)}^{{\displaystyle \frac{3}{2}}}}\right] \\ \mathit{I}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{q}}_{\mathbf{e}}{\mathbf{q}}_{\mathbf{m}}\mathbf{d}}{{\left(4\pi \right)}^{\mathbf{2}}}\frac{{\displaystyle \stackrel{\boxed{\xcancel{}}}{\mathbf{r}}}\mathbf{\times }\left(\stackrel{\boxed{\xcancel{}}}{r}\times \hat{z}\right)}{{\mathbf{r}}^{\mathbf{3}}{\left(r^2+d^2-2rdcos\theta \right)}^{{\displaystyle \frac{\mathbf{3}}{\mathbf{2}}}}} \end{gathered}$$

Since,

$$\begin{gathered} \stackrel{\boxed{\xcancel{}}}{\mathbf{r}}\mathbf{\times }\left(\stackrel{\boxed{\xcancel{}}}{r}\times \hat{z}\right)\mathbf{=}\stackrel{\boxed{\xcancel{}}}{\mathbf{r}}\left(\stackrel{\boxed{\xcancel{}}}{r}·\hat{z}\right)\mathbf{-}{\mathit{r}}^{\mathbf{2}}\hat{\mathbf{z}} \\ \stackrel{\boxed{\xcancel{}}}{\mathbf{r}}\mathbf{\times }\left(\stackrel{\boxed{\xcancel{}}}{r}\times \hat{z}\right)\mathbf{=}{\mathit{r}}^{\mathbf{2}}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\mathbf{}\hat{\mathbf{r}}\mathbf{}\mathbf{-}{\mathit{r}}^{\mathbf{2}}\hat{\mathbf{z}} \end{gathered}$$

Substitute the known values in equation (3).

$\mathit{L}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{q}}_{\mathbf{e}}{\mathbf{q}}_{\mathbf{m}}\mathbf{d}}{{\left(4\pi \right)}^{\mathbf{2}}}\hat{\mathbf{z}}\mathbf{\int }\frac{{\mathbf{r}}^{\mathbf{2}}\left(co{s}^2\theta -1\right){\mathbf{r}}^{\mathbf{2}}\mathbf{}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }\mathbf{d}\mathbf{r}\mathbf{d}\mathbf{\theta }\mathbf{\varphi }}{{\mathbf{r}}^{\mathbf{3}}{\left(r^2+d^2-2rdcos\theta \right)}^{{\displaystyle \frac{\mathbf{3}}{\mathbf{2}}}}}$

Let’s assume,

$$\begin{gathered} \mathit{u}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{\theta } \\ \mathit{d}\mathit{u}\mathbf{=}\mathbf{-}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathit{\theta } \end{gathered}$$

Solve for the total angular momentum:

$$\begin{gathered} \mathit{L}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{q}}_{\mathbf{e}}{\mathbf{q}}_{\mathbf{m}}\mathbf{d}}{{\mathbf{\left(}\mathbf{4}\mathbf{\pi }\mathbf{\right)}}^{\mathbf{2}}}\hat{\mathbf{z}}{\mathbf{\int }}_{\mathbf{-}\mathbf{1}}^{\mathbf{1}}\underset{\mathbf{0}}{\overset{\mathbf{\infty }}{\mathbf{\int }}}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{2}\mathbf{\pi }}\frac{{\mathbf{r}}^{\mathbf{2}}\left(1-u^2\right)\mathbf{d}\mathbf{u}\mathbf{d}\mathbf{r}\mathbf{d}\mathbf{\varphi }}{{\left(r^2+r^2-2rd\cos \varphi \right)}^{{\displaystyle \frac{\mathbf{3}}{\mathbf{2}}}}} \\ \mathit{L}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{q}}_{\mathbf{e}}{\mathbf{q}}_{\mathbf{m}}\mathbf{d}}{{\mathbf{\left(}\mathbf{4}\mathbf{\pi }\mathbf{\right)}}^{\mathbf{2}}}\hat{\mathbf{z}}{\mathbf{\int }}_{\mathbf{-}\mathbf{1}}^{\mathbf{1}}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\pi }}\frac{\mathbf{r}\mathbf{(}\mathbf{1}\mathbf{-}{\mathbf{u}}^{\mathbf{2}}\mathbf{)}\mathbf{dudr}}{{\mathbf{(}{\mathbf{r}}^{\mathbf{2}}\mathbf{+}{\mathbf{r}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{rd}\mathbf{}\mathbf{cos\theta }\mathbf{)}}^{{\displaystyle \frac{3}{2}}}} \\ \mathit{L}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}\frac{\mathbf{rdr}}{{\mathbf{(}{\mathbf{r}}^{\mathbf{2}}\mathbf{+}{\mathbf{r}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{rdu}\mathbf{)}}^{{\displaystyle \frac{3}{2}}}} \\ {\overline{)\mathbf{L}\mathbf{=}\frac{\mathbf{ru}\mathbf{-}\mathbf{d}}{\mathbf{d}\mathbf{(}\mathbf{1}\mathbf{-}{\mathbf{u}}^{\mathbf{2}}\mathbf{)}\sqrt{{\mathbf{r}}^{\mathbf{2}}\mathbf{+}{\mathbf{d}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathbf{rdu}}}}}_{\mathbf{0}}^{\mathbf{\infty }} \end{gathered}$$

On further solving,

$$\begin{gathered} \mathit{L}\mathbf{=}\frac{\mathbf{u}}{\mathbf{d}\left(1-u^2\right)}\mathbf{+}\frac{\mathbf{d}}{\mathbf{d}\left(1-u^2\right)\mathbf{d}} \\ \mathit{L}\mathbf{=}\frac{\mathbf{u}\mathbf{+}\mathbf{1}}{\mathbf{d}\left(1-u^2\right)} \\ \mathit{L}\mathbf{=}\frac{\mathbf{1}}{\mathbf{d}\left(1-u\right)} \end{gathered}$$

Solve further as:

$$\begin{gathered} \mathit{L}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{q}}_{\mathbf{e}}{\mathbf{q}}_{\mathbf{m}}}{\mathbf{8}\mathbf{\pi }}\hat{\mathbf{z}}{\mathbf{\int }}_{\mathbf{-}\mathbf{1}}^{\mathbf{1}}\frac{\mathbf{1}\mathbf{-}{\mathbf{u}}^{\mathbf{2}}}{\left(1-u\right)}\mathit{d}\mathit{u} \\ \mathit{L}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{q}}_{\mathbf{e}}{\mathbf{q}}_{\mathbf{m}}}{\mathbf{8}\mathbf{\pi }}\hat{\mathbf{z}}{\mathbf{\int }}_{\mathbf{-}\mathbf{1}}^{\mathbf{1}}\left(1+u\right)\mathit{d}\mathit{u} \\ \mathit{L}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{q}}_{\mathbf{e}}{\mathbf{q}}_{\mathbf{m}}}{\mathbf{8}\mathbf{\pi }}\hat{\mathbf{z}}{\left[u+\frac{u^2}{2}\right]}_{\mathbf{-}\mathbf{1}}^{\mathbf{1}} \\ \mathit{L}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{q}}_{\mathbf{e}}{\mathbf{q}}_{\mathbf{m}}}{\mathbf{4}\mathbf{\pi }}\hat{\mathbf{z}} \end{gathered}$$

Therefore, the total angular momentum stored in the fields is $\mathit{L}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{q}}_{\mathbf{e}}{\mathbf{q}}_{\mathbf{m}}}{\mathbf{4}\mathbf{\pi }}\hat{\mathbf{z}}$.