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We know the magnetic field produced by electric current density...if we add the magnetic monopoles what would be the electric field produced by this magnetic monopoles current density?

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If we introduce a magnetic monopole density, $\rho_m$, and a magnetic current density ${\bf J_m} = \rho_m {\bf v}$, then Maxwell's equations can be written symmetrically (using a set of natural units where $c=1$) $$ \nabla \cdot {\bf E} = \rho\ \ \ \ \ \ \nabla \cdot {\bf B} = \rho_m$$ $$ \nabla \times {\bf E} = -\frac{\partial {\bf B}}{\partial t} - {\bf J_m}\ \ \ \ \ \ \nabla \times {\bf B} = \frac{\partial {\bf E}}{\partial t} + {\bf J}$$

Thus for magnetostatics, Faraday's law becomes $$\nabla \times {\bf E} = - {\bf J_m}\ $$ and a magnetic monopole current density acts as a source of curling electric field. In integral form, this could be written $$\oint {\bf E}\cdot d{\bf l} = -\int {\bf J_m} \cdot d{\bf S} = -I_m\ ,$$ where $I_m$ is the "magnetic current".

In terms of potentials you would have to introduce a new vector potential, ${\bf A_m}$ whose curl gave an electric field and a magnetostatic potential $\phi_m$, whose gradient gave a magnetic field: $${\bf E} = -\nabla \phi - \frac{\partial {\bf A}}{\partial t} - \nabla \times {\bf A_m}\ , $$ $${\bf B} = \nabla \times {\bf A} - \nabla \phi_m - \frac{\partial {\bf A_m}}{\partial t}\ . $$

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  • $\begingroup$ In an SR context, the scalar electric potential and vector magnetic potential are components of the four-potential $A^\mu$. If we add magnetic charge to the mix, then there is at least a new scalar magnetic potential. Where does this fit in the context of the four-potential? $\endgroup$ Nov 10, 2017 at 1:50

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