Electric field produced by magnetic monopoles current density 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?
 A: 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:
$${\bf E} = -\nabla \phi - \frac{\partial {\bf A}}{\partial t} - \nabla \times {\bf A_m}\ , $$
$${\bf B} = \nabla \times {\bf A} - \nabla \phi_m - \frac{\partial {\bf A_m}}{\partial t}\ . $$
