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We have to show that

(a)$鈭�路E\text{'}=\frac{{蚁}_e}{{蔚}_0}$

(b) $鈭�路B\text{'}={渭}_0{蚁}_m$

(c) $鈭�脳E\text{'}=-{渭}_0{J}_m-\frac{鈭�B}{鈭�t}$

(d) $鈭�脳B\text{'}={渭}_0{J}_e+{渭}_0{蔚}_0\frac{鈭�E}{鈭�t}$

Write the formula ofMaxwell鈥檚 equations with magnetic charge is invariant under the duality transformation.

$\mathbf{鈭�}\mathbf{鈥�}\mathit{E}\mathbf{\text{'}}$ 鈥︹�� (1)

Here, $\mathit{E}\mathbf{\text{'}}$is electrical field.

Write the formula of Maxwell鈥檚 equations with magnetic charge is invariant under the duality transformation.

localid="1657626502486" $\mathbf{鈭�}\mathbf{.}\mathbf{B}\mathbf{\text{'}}$ 鈥︹�� (2)

Here,$\mathbf{B}\mathbf{\text{'}}$is magnetic field.

Write the formula of Maxwell鈥檚 equations with magnetic charge is invariant under the duality transformation.

localid="1657626718393" $\mathbf{鈭�}\mathbf{脳}\mathit{E}\mathbf{\text{'}}$ 鈥︹�� (3)

Write the formula of Maxwell鈥檚 equations with magnetic charge is invariant under the duality transformation.

$\mathbf{鈭�}\mathbf{脳}\mathbf{B}\mathbf{\text{'}}$ 鈥︹�� (4)

Here,$\mathbf{B}\mathbf{\text{'}}$ is magnetic field.

Write the formula of force law.

$\mathbf{F}\mathbf{=}{\mathbf{q}}_{\mathbf{e}}\mathbf{(}\mathbf{E}\mathbf{脳}\mathbf{V}\mathbf{脳}\mathbf{B}\mathbf{)}\mathbf{+}{\mathbf{q}}_{\mathbf{m}}\left(B-\frac{1}{C^2}V脳E\right)$ 鈥︹�� (5)

Here,${\mathbf{q}}_{\mathbf{e}}$ is electric charge,${\mathbf{q}}_{\mathbf{m}}$ is magnetic charge,$\mathbf{B}$ is magnetic field and$\mathbf{v}$ is voltage.

Consider given equation as:

$$\begin{gathered} E\text{'}=Ecos伪+cBsin伪, \\ cB\text{'}=cBcos伪-Esin伪, \\ cq{\text{'}}_e=c{q}_{e}cos伪+q_{m}sin伪, \\ q{\text{'}}_m=q_{m}cos伪-c{q}_{e}sin伪, \end{gathered}$$

Determine the value of Maxwell鈥檚 equation with magnetic charge is invariant under the duality transformation.

Substitute $\mathrm{Ecos}伪+c\mathrm{Bsin}伪$for $\mathrm{E}\text{'}$into equation (1).

localid="1657686378180" $$\begin{gathered} 鈭�.\mathrm{E}\text{'}=鈭�.\left[\mathrm{Ecos}伪+c\mathrm{Bsin}伪\right] \\ =\left(鈭�.\mathrm{E}\right)\cos 伪+c\left(鈭�.B\right)\sin 伪 \\ =\frac{1}{{蔚}_0}\left({蚁}_e\right)\cos 伪+c{渭}_0{蚁}_m\sin 伪 \\ =\frac{1}{{蔚}_0}\left[{蚁}_e\cos 伪+\frac{1}{c}{蚁}_m\sin 伪\right] \end{gathered}$$

Solve further as,

$$\begin{gathered} 鈭�.E=\frac{1}{{蔚}_0}\left[{蚁}_e\cos 伪+\frac{1}{c}{蚁}_m\sin 伪\right] \\ =\frac{1}{{蔚}_0}{蚁}_e^{\text{'}} \end{gathered}$$

Therefore, the value of Maxwell鈥檚 equations with magnetic charge are invariant under the duality transformation is$鈭�脳B\text{'}={渭}_0{J}_c^{\text{'}}+{渭}_0{蔚}_0\frac{鈭�E\text{'}}{鈭�t}.$

Determine the value of Maxwell鈥檚 equation with magnetic charge is invariant under the duality transformation.

Substitute$B\cos 伪-\frac{E}{c}\sin 伪$for$B\text{'}$into equation (2).

$$\begin{gathered} 鈭�.B\text{'}=B\cos 伪-\frac{E}{c}\sin 伪 \\ =\left(鈭�.B\right)\cos 伪-\frac{\left(鈭�脳E\right)}{c}\sin 伪 \\ ={渭}_0{蚁}_m\cos 伪-\frac{{蚁}_e}{{蔚}_{0}c}\sin 伪 \\ ={渭}_0\left(f_m\cos 伪-c{蚁}_e\sin 伪\right) \end{gathered}$$

Solve further as,

$鈭�.B\text{'}={渭}_0{蚁}_m^{\text{'}}$

Therefore, the value of Maxwell鈥檚 equations with magnetic charge are invariant under the duality transformation is$鈭�.B\text{'}={渭}_0{蚁}_m.$

Determine the value of Maxwell鈥檚 equation with magnetic charge is invariant under the duality transformation.

Substitute$E\cos 伪+cBs\mathrm{in}伪$for$E\text{'}$into equation (3).

$$\begin{gathered} 鈭�脳E\text{'}=E\cos 伪+cB\sin 伪 \\ =\left(鈭�脳E\right)\cos 伪+c\left(鈭�脳B\right)\sin 伪 \\ =\left(-{渭}_0{J}_m-\frac{鈭�B}{鈭�t}\right)\cos 伪+c\left({渭}_0{J}_e+{渭}_0{蔚}_0\frac{鈭�E}{鈭�t}\right)\sin 伪 \\ =-{渭}_0\left(J_m\cos 伪-c{J}_e\sin 伪\right)-\frac{鈭�}{鈭�t}\left(B\cos 伪-\frac{E}{c}\sin 伪\right) \end{gathered}$$

Solve further as,

$鈭�脳E\text{'}=-{渭}_0{J}_m^{\text{'}}-\frac{鈭�B\text{'}}{鈭�t}$

Therefore, the The value of Maxwell鈥檚 equations with magnetic charge are invariant under the duality transformation is$鈭�脳E\text{'}=-{渭}_0{J}_m^{\text{'}}-\frac{鈭�B\text{'}}{鈭�t}.$

Determine the value of Maxwell鈥檚 equation with magnetic charge is invariant under the duality transformation.

Substitute$B\cos 伪-\frac{1}{c}E\sin 伪$for$B\text{'}$into equation (4).

$$\begin{gathered} 鈭�脳B\text{'}=鈭�脳\left(B\cos 伪-\frac{1}{c}E\sin 伪\right) \\ =\left(鈭�脳B\right)\cos 伪-\frac{1}{c}\left(鈭�脳E\right)\sin 伪 \\ =\left[{渭}_0{J}_{e+}{渭}_0{蔚}_0\frac{鈭�E}{鈭�t}\right]\cos 伪-\frac{1}{c}\left[-{渭}_0{J}_m-\frac{鈭�B}{鈭�t}\right]\sin 伪 \\ ={渭}_0\left(J_e\cos 伪+\frac{1}{c}{J}_m\sin 伪\right)+{渭}_0{蔚}_0\frac{鈭�}{鈭�t}\left(E\cos 伪+cB\sin 伪\right) \end{gathered}$$

Solve further as:

$鈭�脳B\text{'}={渭}_0{J}_e^{\text{'}}+{渭}_0{蔚}_0\frac{鈭�E\text{'}}{鈭�t}$

Therefore, the value of Maxwell鈥檚 equations with magnetic charge are invariant under the duality transformation is$鈭�脳B\text{'}={渭}_0{J}_e^{\text{'}}+{渭}_0{蔚}_0\frac{鈭�E\text{'}}{鈭�t}$

Here, the force law for a monopole$q_m$traveling across the electric and magnetic fields E and B at velocity $v$ is

Determine the force law.

localid="1657692041980" $$\begin{gathered} F=q_e\left(E+\left(V脳B\right)\right)+q_m\left(B-\frac{1}{c^2}V脳E\right) \\ ThenF\text{'}=q_e^{\text{'}}(E\text{'}+(V脳B\text{'}\left)\right)+q_m^{\text{'}}\left(B\text{'}-\frac{1}{c^2}V脳E\text{'}\right) \\ Substitute\left(q_e\cos 伪+\frac{1}{c}{q}_m\sin 伪\right)for{q}_e^{\text{'}},(E\cos 伪+c\sin 伪)forE\text{'}, \\ \left(B\cos 伪-\frac{1}{c}E\right)forB\text{'},\left(q_m\cos 伪-c{q}_e\sin 伪\right)for{q}_m^{\text{'}},\left(B\cos 伪-\frac{1}{c}E\sin 伪\right)forB \\ and\left(cE\cos 伪+cB\sin 伪\right)forE\text{'}intoequation\left(5\right). \end{gathered}$$

F'=
$$\begin{gathered} \left(q_e\cos 伪+\frac{1}{c}{q}_m\sin 伪\right)\left[\left(E\cos 伪+cB\sin 伪\right)+v脳\left(B\cos 伪-\frac{1}{c}E\sin 伪\right)\right]+(q_m\cos 伪-c{q}_e\sin 伪)\left(Bc\right) \\ =\left\{=q_e\left[\begin{array}{c}(E{\cos }^2伪+cB\sin 伪\cos 伪-cB\sin 伪\cos 伪-cB\sin 伪\cos 伪+E{\sin }^2伪+\\ v脳\left(B{\cos }^2伪-\frac{1}{c}E\sin 伪\cos 伪+\frac{1}{c}E\sin 伪\cos 伪+B{\sin }^2伪\right)\end{array}\right] \\ +q_m\left[\begin{array}{c}\left(\frac{1}{c}E\sin 伪\cos 伪+B{\sin }^2伪+B{\cos }^2伪-\frac{1}{c}E\sin 伪\cos 伪\right)+v\\ 脳\left(\frac{1}{c}B\sin 伪\cos 伪-\frac{E}{c^2}{\sin }^2伪-\frac{E}{c^2}{\cos }^2伪-\frac{B}{c}\sin 伪\cos 伪\right)\end{array}\right]\right\} \\ =q_e\left(E脳(V脳B)\right)+q_m\left(B-\frac{1}{c^2}v脳E\right) \\ =F \end{gathered}$$

Therefore, the value of force law is$F=q_e\left(E+\left(V脳B\right)\right)+q_m\left(B-\frac{1}{c^2}V脳E\right).$