Mind you, we still have electric charge and electric currents. But, what would Maxwell's equations look like if we had to take magnetic charges and magnetic currents into consideration? Would there be any sign changes?
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2 Answers
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By analogy (between $\mathbf{E}$ and $\mathbf{B}$ as they are pretty much equivalent) then the divergeance of $\mathbf{B}$ field wouldn't be 0 anymore, instead: $$\nabla \cdot \mathbf{B}= \frac{\rho_{\rm magnetic}}{\mu_0} $$ With $\rho_{\rm magnetic}$ the magnetic charge density, and $\mu_0$ the permeability in vacuum, to interpret it, the divergence of the magnetic field at a point in space is equal to the magnetic charge density divided by the permeability of space.
As for the curl equations, with magnetic charges, curl of $\mathbf{E}$ should also give a non-zero density current of magnetic charges, i.e.: $$\nabla \times\mathbf{E}= -\mu_0 \mu \frac{\partial \mathbf{H}}{\partial t} + \sigma_m \mathbf{B} $$
Where $\sigma_m$ would then be the magnetic conductivity.
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A slight correction to the answer posted above:
Gauss' Law for Magnetic field will now take the form:
$$ \nabla \cdot \vec{B}=\mu_0 \rho_m\\ $$ i.e. $$\bigcirc \!\!\!\!\!\!\!\! \iint \vec{B}\cdot\text{d}\vec{A}=\mu_0 \iiint \rho_m \text{d}V $$
The expression for the curl of the electric field becomes: $$ \nabla\times\vec{E}=-\frac{\partial \vec{B}}{\partial t}-\vec{J}_m\\ i.e. \oint \vec{E}\cdot\text{d}\vec{l}=-\frac{\partial}{\partial t}\bigg(\iint\vec{B}\cdot\text{d}\vec{A}\bigg)-\mu_0 I_m $$
The remaining two equations stay the same.
($\rho_m$=magnetic charge density, $\vec{J}_m$=Magnetic current, $I_m$=magnetic current)
