91Ó°ÊÓ

Write the Maxwell’s equation.

$\begin{array}{l}∇â‹\dots \mathbf{E}\mathbf{=}\frac{\mathrm{ÒÏ}}{{\mathbf{ÃŽ}}_0}\text{ â¶Ä‰â¶Ä‰â¶Ä‰}\left(Gauss\text{ }law\right)\\ ∇×E=0\\ ∇â‹\dots B=0\end{array}$

Write the amperes law

$∇â‹\dots \mathbf{B}\mathbf{=}{\mathbf{μ}}_0\mathbf{J}$

As the magnetic monopole exists then there will be no change in Ampere’s law and gauss law.

Actually $∇â‹\dots B=0$implies there will be no magnetic monopoles.Then if magnetic monopoles exist, then

$∇â‹\dots B={\mathrm{Î\pm }}_0{\mathrm{ÒÏ}}_m$

Here${\mathrm{ÒÏ}}_m$,is the density of magnetic change and${\mathrm{Î\pm }}_0$is the same constant.

Rewrite the Maxwell’s equation as

$∇×E={\mathrm{β}}_0{J}_m$

Here, $J_m$is the magnetic current density and${\mathrm{β}}_0$is another constant.

Thus, magnetic charge is conserved. Hence, ${\mathrm{ÒÏ}}_m$and$J_m$satisfy continuity equation and is written as

$∇â‹\dots J_m+\frac{∂{\mathrm{ÒÏ}}_m}{∂t}=0$

Write the expression for force on magnetic monopole.

$F=q_m[B+(\mathrm{Ï\dots }×E)]$

Consider both the equation directionally not correct.

Here, $E$has same units as$\mathrm{Ï\dots }B$ .

Hence, divide $\mathrm{Ï\dots }×E$with dimensions of velocity squared and rewrite the equation as

$F=q_e[E+(\mathrm{Ï\dots }×B)]+q_m[B−\frac{1}{c^2}(\mathrm{Ï\dots }×E)]$

Now, write the expression for magnetic field along the lines ofCoulomb’s law.

.

$F=\frac{{\mathrm{Î\pm }}_0}{4\mathrm{Ï€}}\frac{q_{m1}{q}_{m2}}{r^2}\widehat{r}$