Use the duality transformation (Prob. 7.64) to construct the electric and magnetic fields of a magnetic monopole ${\mathit{q}}_{\mathbf{m}}$in arbitrary motion, and find the “Larmor formula” for the power radiated.

Write the expression for the duality transformation equations.

$$\begin{gathered} {\mathit{E}}^{\mathbf{\text{'}}}\mathbf{=}\mathit{E}\mathit{c}\mathit{o}\mathit{s}\mathit{\alpha }\mathbf{+}\mathit{c}\mathit{B}\mathit{s}\mathit{i}\mathit{n}\mathit{\alpha } \\ \mathit{c}{\mathit{B}}^{\mathbf{\text{'}}}\mathbf{=}\mathit{c}\mathit{B}\mathit{c}\mathit{o}\mathit{s}\mathit{\alpha }\mathbf{-}\mathit{E}\mathit{s}\mathit{i}\mathit{n}\mathit{\alpha } \\ \mathit{c}{\mathit{q}}_{\mathbf{e}}^{\mathbf{\text{'}}}\mathbf{=}\mathit{c}{\mathit{q}}_{\mathbf{e}}\mathit{c}\mathit{o}\mathit{s}\mathit{\alpha }\mathbf{+}{\mathit{q}}_{\mathbf{m}}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathit{\alpha } \\ {\mathit{q}}_{\mathbf{m}}^{\mathbf{\text{'}}}\mathbf{=}{\mathit{q}}_{\mathbf{m}}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathit{\alpha }\mathbf{-}\mathit{c}{\mathit{q}}_{\mathbf{e}}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathit{\alpha } \end{gathered}$$

Substitute $\alpha ={90}^{°}$in all the duality transformation equations.

localid="1657524521051" $$\begin{gathered} E^{\text{'}}=E\cos \left({90}^{°}\right)+cB\sin \left({90}^{°}\right)=cB \\ c{B}^{\text{'}}=cB\cos \left({90}^{°}\right)-E\sin \left({90}^{°}\right)=-E \\ c{q}_e^{\text{'}}=c{q}_e\cos \left({90}^{°}\right)+q_m\sin \left({90}^{°}\right)=q_m \\ {q}_m^{\text{'}}=q_m\cos \left({90}^{°}\right)-c{q}_e\sin \left({90}^{°}\right)=-c{q}_e \end{gathered}$$

Now, write all the duality transformation equations in a proper form.

$$\begin{gathered} E^{\text{'}}=cB \\ c{B}^{\text{'}}=-E \\ c{q}_e^{\text{'}}=q_m \\ {q}_m^{\text{'}}=-c{q}_e \end{gathered}$$

Calculate the electric field due to a magnetic monopole.

$$\begin{gathered} E^{\text{'}}=c\left(\frac{1}{c}\hat{r}\times E\right) \\ {E}^{\text{'}}=c\left[\hat{r}\times \frac{E}{c}\right].......\left(1\right) \end{gathered}$$ …… (1)

Here, E is the electric field which is given as:

$E=\frac{q}{4\mathrm{\pi }}\frac{r}{{(r-u)}^3}\left[(c^2-v^2)u+r\times (u\times a)\right]$

Substitution $-\frac{E}{c}$for $B^{\text{'}}$in equation (1).

$E^{\text{'}}=-c\left[\hat{r}\times B^{\text{'}}\right]$

Calculate the magnetic field due to a magnetic monopole.

localid="1657524651730" $$\begin{gathered} B^{\text{'}}=-\frac{E}{c} \\ {B}^{\text{'}}=-\frac{1}{c}\frac{q_e}{4\mathrm{\pi }}\frac{r}{{(r.u)}^3}\left[(c^2-v^2)u+r\times (u\times a)\right] \\ {B}^{\text{'}}=-\frac{1}{c}\frac{\left(-{\displaystyle \frac{q_m}{C}}\right)}{4{\mathrm{\pi \epsilon }}_0}\frac{r}{{(r.u)}^3}\left[(c^2-v^2)u+r\times (u\times a)\right] \\ {B}^{\text{'}}=\frac{{\mu }_0{q}_m}{4\mathrm{\pi }}\frac{r}{{(r.u)}^3}\left[(c^2-v^2)u+r\times (u\times a)\right] \end{gathered}$$

Write the expression for the total power radiated by the decelerating charge.

$P=\frac{{\mu }_0{a}^2}{6\mathrm{\pi c}}{q}_e^{\text{'}2}$

Substitute the value of $q_e^{\text{'}}$in the above expression.

$$\begin{gathered} P=\frac{{\mu }_0{a}^2}{6\mathrm{\pi c}}{\left(\frac{-q_m}{c}\right)}^2 \\ p=\frac{{\mu }_0{a}^2{q}_m^2}{6{\mathrm{\pi c}}^3} \end{gathered}$$

Therefore, the electric field and magnetic field of a magnetic monopole is $E^{\text{'}}=-c\left[\hat{r}\times B^{\text{'}}\right]and{B}^{\text{'}}=\frac{{\mu }_0{q}_m}{4\mathrm{\pi }}\frac{r}{{(r.u)}^3}\left[(c^2-v^2)u+r(u\times a)\right]$respectively, and the “Larmor formula” for the power radiated is $p=\frac{{\mu }_0{a}^2{q}_m^2}{6{\mathrm{\pi c}}^3}$.