0  & \text{on } \Sigma  \, ,
\end{array}$$ where ${\boldsymbol{j}}$ is the applied current density, $\mu$ is the magnetic permeability and $\sigma$ is the electric conductivity. In what follows, we assume that $\mu$ and $\sigma$ are positive piecewise constant functions in $\Omega_{C}$ and that $\mu_{|\Omega_I}= \mu_0$ is the permeability constant of vacuum. It follows from the first equation that $$\label{lap}
0={\mathrm{div}}({\boldsymbol{h}}_{|\Omega_I})= {\mathrm{div}}(\nabla \psi + k \boldsymbol \rho)\, \hbox{ in } \Omega_I\, .$$ We point out here that the electric field ${\boldsymbol{e}}$ is not uniquely determined in $\Omega_{I}$. Nevertheless, the tangential components of the magnetic field and the tangential components of any admissible representation of the electric field should be continuous across the interface $\Gamma$, i.e., $$\label{magt}
{\boldsymbol{h}}|_{\Omega_{C}} \times {\boldsymbol{n}}_\Gamma = (\nabla \psi + k \boldsymbol \rho) \times {\boldsymbol{n}}_\Gamma \, .$$ and $$\label{mage}
{\boldsymbol{e}}_{|\Omega_C} \times {\boldsymbol{n}}_\Gamma = {\boldsymbol{e}}_{|\Omega_I} \times {\boldsymbol{n}}_\Gamma.$$ The electric field ${\boldsymbol{e}}$ is considered here as an auxiliary variable, it will be removed from the formulation. Hence, we should deduce from a transmission condition relating ${\boldsymbol{h}}$ and $\varphi$ on $\Gamma$. Applying the surface divergence operator ${\mathrm{div}}_\Gamma$ to both side of and recalling that ${\mathrm{div}}_\Gamma({\boldsymbol{e}}\times {\boldsymbol{n}}_\Gamma)= {\mathbf{curl}}\,{\boldsymbol{e}}\cdot {\boldsymbol{n}}_\Gamma$ we deduce that the field ${\mathbf{curl}}\,{\boldsymbol{e}}$ admits continuous normal components across $\Gamma$. 