91Ó°ÊÓ

Write the expression for Maxwell’s equation with a magnetic charge.

$\mathbf{∇}\mathbf{·}\mathit{F}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{Î\mu }}_{\mathbf{0}}}{\mathit{μ}}_{\mathbf{0}}$ …… (1)

Consider the second equation as:

$\mathbf{∇}\mathbf{·}\mathit{B}\mathbf{=}\mathit{p}{\mathit{μ}}_{\mathbf{0}\mathbf{m}}$ …… (2)

Consider the third equation as:

$\mathbf{∇}\mathbf{·}\mathit{E}\mathbf{=}\mathbf{-}{\mathit{μ}}_{\mathbf{0}}{\mathit{J}}_{\mathbf{m}}\mathbf{-}\frac{\mathbf{∂}\mathbf{B}}{\mathbf{∂}\mathbf{t}}$ …… (3)

Consider the fourth equation as:

$\mathbf{∇}\mathbf{×}\mathit{B}\mathbf{=}\mathbf{-}{\mathit{μ}}_{\mathbf{0}}{\mathit{J}}_{\mathbf{e}}\mathbf{+}{\mathit{μ}}_{\mathbf{0}}{\mathit{Î\mu }}_{\mathbf{0}}\frac{\mathbf{∂}\mathbf{E}}{\mathbf{∂}\mathbf{t}}$ …… (4)

Here, E is the electric field, ${\mathrm{Î\mu }}_0$is the permittivity of free space, ${\mathrm{ÒÏ}}_e$is the electric charge density, $J_e$is the electric current density, J_m is the magnetic current density, B is the magnetic field, t is the time and ${\mathrm{μ}}_0$is the permeability of free space.

Write the expression for the Lorentz force.

$\mathit{F}\mathbf{=}{\mathit{q}}_{\mathbf{e}}\left(E+v×B\right)\mathbf{+}{\mathit{q}}_{\mathbf{m}}\left(B-\frac{1}{c^2}v×E\right)$ …… (5)

Here, q is the charge, c is the speed of light, and v is the speed.

Multiply by dl in equation (5) to calculate the work done by the electromagnetic force in the interval dt.

$\mathit{F}\mathbf{·}\mathit{d}\mathit{l}\mathbf{=}\left[q_e\left(E+v×B\right)+q_m\left(B-\frac{1}{c^2}v×E\right)\right]\mathbf{·}\mathit{d}\mathit{t}$ ……. (6)

Write the expression for displacement in terms of velocity and time.

dl = vdt

Substitute the known values in equation (6).

$$\begin{gathered} \mathit{F}\mathbf{·}\mathit{d}\mathit{l}\mathbf{=}\mathbf{[}{\mathbf{q}}_{\mathbf{e}}\left(E+v×B\right)\mathbf{+}{\mathbf{q}}_{\mathbf{m}}\left(B-\frac{1}{c^2}v×E\right)\mathbf{]}\mathbf{·}\mathit{v}\mathit{d}\mathit{t} \\ \mathit{F}\mathbf{·}\mathit{d}\mathit{l}\mathbf{=}\mathbf{[}{\mathbf{q}}_{\mathbf{e}}\mathbf{(}\mathbf{E}\mathbf{+}\mathbf{v}\mathbf{×}\mathbf{B}\mathbf{)}\mathbf{·}\mathbf{vdt}\mathbf{+}{\mathbf{q}}_{\mathbf{m}}\mathbf{(}\mathbf{B}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{c}}^{\mathbf{2}}}\mathbf{v}\mathbf{×}\mathbf{E}\mathbf{)}\mathbf{·}\mathbf{vdt}\mathbf{]} \\ \mathit{F}\mathbf{·}\mathit{d}\mathit{l}\mathbf{=}\mathbf{[}\mathbf{(}\mathbf{(}{\mathbf{q}}_{\mathbf{e}}\mathbf{E}\mathbf{·}\mathbf{vdt}\mathbf{)}\mathbf{+}{\mathbf{q}}_{\mathbf{e}}\mathbf{(}\mathbf{v}\mathbf{×}\mathbf{B}\mathbf{)}\mathbf{·}\mathbf{vdt}\mathbf{)}\mathbf{+}\mathbf{(}{\mathbf{q}}_{\mathbf{m}}\mathbf{B}\mathbf{·}\mathbf{vdt}\mathbf{-}\frac{{\mathbf{q}}_{\mathbf{m}}}{{\mathbf{c}}^{\mathbf{2}}}\mathbf{(}\mathbf{v}\mathbf{×}\mathbf{E}\mathbf{)}\mathbf{·}\mathbf{vdt}\mathbf{)}\mathbf{]} \\ \mathit{F}\mathbf{·}\mathit{d}\mathit{l}\mathbf{=}\mathbf{[}\left(q_{e}E·\mathrm{vdt}\right)\mathbf{+}\mathbf{(}{\mathbf{q}}_{\mathbf{m}}\mathbf{B}\mathbf{·}\mathbf{vdt}\mathbf{)}\mathbf{]} \end{gathered}$$ ……. (7)

Since,

Substitute the known values in equation (7).

$\frac{\mathbf{d}\mathbf{W}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{∫}\left[E·J_e+B·J_m\right]\mathit{d}\mathit{Î\P }$

Use equations (3) and (4) to remove ${\mathit{J}}_{\mathbf{e}}$and ${\mathit{J}}_{\mathbf{m}}$in the above equation.

$$\begin{gathered} \frac{\mathbf{d}\mathbf{W}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{∫}\left[E·\left(\frac{∇×B}{{\mathrm{μ}}_0}-{\mathrm{Î\mu }}_0\frac{∂E}{∂∂}\right)+B·\left(-\frac{∆×E}{{\mathrm{μ}}_0}-\frac{1}{{\mathrm{μ}}_0}\frac{∂B}{∂t}\right)\right] \\ \frac{\mathbf{d}\mathbf{W}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{∫}\left[\left(\frac{E·\left(∇×B\right)}{{\mathrm{μ}}_0}-{\mathrm{Î\mu }}_{0}E·\frac{∂E}{∂∂}\right)+\left(-\frac{B·\left(∆×E\right)}{{\mathrm{μ}}_0}-\frac{1}{{\mathrm{μ}}_0}\frac{∂B}{∂t}\right)\right] \end{gathered}$$ ……. (9)

Since,

$\mathbf{∇}\mathbf{·}\left(E×B\right)\mathbf{=}\left(∇×E\right)\mathbf{-}\mathit{E}\mathbf{·}\left(∇×B\right)$

On further solving equation (8),

$$\begin{gathered} \frac{\mathbf{d}\mathbf{W}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{∫}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{μ}}_{\mathbf{0}}}\mathbf{∇}\mathbf{·}\left(E×B\right)\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\frac{\mathbf{∂}}{\mathbf{∂}\mathbf{t}}\left({\mathrm{Î\mu }}_0{E}^2+\frac{1}{{\mathrm{μ}}_0}{B}^2\right) \\ \frac{\mathbf{d}\mathbf{W}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{-}\frac{\mathbf{d}}{\mathbf{d}\mathbf{t}}\mathbf{∫}\frac{\mathbf{1}}{\mathbf{2}}\left({\mathrm{Î\mu }}_0{E}^2+\frac{1}{{\mathrm{μ}}_0}{B}^2\right)\mathit{d}\mathit{Î\P }\mathbf{-}\frac{\mathbf{1}}{{\mathbf{μ}}_{\mathbf{0}}}\boxed{\xcancel{}}\mathbf{}\mathbf{∫}\left(E×B\right)\mathbf{·}\mathit{d}\mathit{a} \end{gathered}$$

Hence, the energy stored in the field will be,

$\mathit{u}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\left({\mathrm{Î\mu }}_0{E}^2+\frac{1}{{\mathrm{μ}}_0}{B}^2\right)$

So, the Poynting vector will be,

$\mathit{S}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{μ}}_{\mathbf{0}}}\left(E×B\right)$

Write the expression for electromagnetic momentum density.

$\mathit{g}\mathbf{=}{\mathit{μ}}_{\mathbf{0}}{\mathit{Î\mu }}_{\mathbf{0}}\mathit{S}$

Substitute the known values in the above equation.

$$\begin{gathered} \mathit{g}\mathbf{=}{\mathit{μ}}_{\mathbf{0}}{\mathit{Î\mu }}_{\mathbf{0}}\mathbf{\left(}\frac{\mathbf{1}}{{\mathbf{μ}}_{\mathbf{0}}}\left(E×B\right)\mathbf{\right)} \\ \mathit{g}\mathbf{=}{\mathit{Î\mu }}_{\mathbf{0}}\mathbf{(}\mathbf{E}\mathbf{×}\mathbf{B}\mathbf{)} \end{gathered}$$

Write the expression for the total electromagnetic force on the charges for volume.

$\mathit{F}\mathbf{=}\mathbf{∫}\left[\left(E+V×B\right){\mathrm{ÒÏ}}_e+\left(B-\frac{1}{c^2}v×E\right){\mathrm{ÒÏ}}_m\right]\mathit{d}\mathit{Î\P }$

Write the expression for force per unit volume.

$$\begin{gathered} \frac{\mathbf{F}}{\mathbf{d}\mathbf{Î\P }}\mathbf{=}\mathbf{∫}\mathbf{[}\left(E+V×B\right){\mathbf{ÒÏ}}_{\mathbf{e}}\mathbf{+}\left(B-\frac{1}{c^2}v×E\right){\mathbf{ÒÏ}}_{\mathbf{m}}\mathbf{]} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{f}\mathbf{=}\mathbf{[}\mathbf{(}{\mathbf{ÒÏ}}_{\mathbf{e}}\mathbf{E}\mathbf{+}{\mathbf{ÒÏ}}_{\mathbf{e}}\mathbf{v}\mathbf{×}\mathbf{B}\mathbf{)}\mathbf{+}\mathbf{(}{\mathbf{ÒÏ}}_{\mathbf{m}}\mathbf{B}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{c}}^{\mathbf{2}}}{\mathbf{ÒÏ}}_{\mathbf{m}}\mathbf{v}\mathbf{×}\mathbf{E}\mathbf{)}\mathbf{]} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{f}\mathbf{=}\mathbf{[}\mathbf{(}{\mathbf{ÒÏ}}_{\mathbf{e}}\mathbf{E}\mathbf{+}{\mathbf{J}}_{\mathbf{e}}\mathbf{×}\mathbf{B}\mathbf{)}\mathbf{+}\mathbf{(}{\mathbf{ÒÏ}}_{\mathbf{m}}\mathbf{B}\mathbf{-}\frac{\mathbf{1}}{{\mathbf{c}}^{\mathbf{2}}}{\mathbf{J}}_{\mathbf{m}}\mathbf{×}\mathbf{E}\mathbf{)}\mathbf{]} \end{gathered}$$

From equations (1), (2), (3) and (4),

$\mathit{f}\mathbf{=}{\mathit{Î\mu }}_{\mathbf{0}}\left[\left(∇·E\right)E-\frac{1}{2}∇\left(E^2\right)+\left(E·∇\right)E\right]\mathbf{+}\frac{\mathbf{1}}{{\mathbf{μ}}_{\mathbf{0}}}\left[\left(∇·B\right)B-\frac{1}{2}∇\left(B^2\right)+\left(B·∇\right)B\right]\mathbf{-}{\mathit{Î\mu }}_{\mathbf{0}}\frac{\mathbf{∂}}{\mathbf{∂}\mathbf{t}}\left(E×B\right)$

So, the stress tensor will be,

${\mathit{T}}_{\mathbf{ij}}\mathbf{=}{\mathit{Î\mu }}_{\mathbf{0}}\mathbf{(}{\mathbf{E}}_{\mathbf{i}}{\mathbf{E}}_{\mathbf{j}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{δ}}_{\mathbf{ij}}{\mathbf{E}}^{\mathbf{2}}\mathbf{)}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{μ}}_{\mathbf{0}}}\mathbf{(}{\mathbf{B}}_{\mathbf{i}}{\mathbf{B}}_{\mathbf{j}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{δ}}_{\mathbf{ij}}{\mathbf{B}}^{\mathbf{2}}\mathbf{)}$

Therefore, the energy stored in the field is $\mathit{u}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{(}{\mathbf{Î\mu }}_{\mathbf{0}}{\mathbf{E}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{μ}}_{\mathbf{0}}}{\mathbf{B}}^{\mathbf{2}}\mathbf{)}$, the Poynting vector is $\mathit{S}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{μ}}_{\mathbf{0}}}\mathbf{(}\mathbf{E}\mathbf{×}\mathbf{B}\mathbf{)}$, the electromagnetic momentum density is $\mathit{g}\mathbf{=}{\mathit{Î\mu }}_{\mathbf{0}}\mathbf{(}\mathbf{E}\mathbf{×}\mathbf{B}\mathbf{)}$, and the stress tensor is ${\mathit{T}}_{\mathbf{ij}}\mathbf{=}{\mathit{Î\mu }}_{\mathbf{0}}\mathbf{(}{\mathbf{E}}_{\mathbf{i}}{\mathbf{E}}_{\mathbf{j}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{δ}}_{\mathbf{ij}}{\mathbf{E}}^{\mathbf{2}}\mathbf{)}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{μ}}_{\mathbf{0}}}\mathbf{(}{\mathbf{B}}_{\mathbf{i}}{\mathbf{B}}_{\mathbf{j}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{δ}}_{\mathbf{ij}}{\mathbf{B}}^{\mathbf{2}}\mathbf{)}$. Thus, the formulas for the momentum density and the stress are unchanged.