As a consequence of the first equation of , $\mu{\boldsymbol{h}}$ should also have continuous normal components across $\Gamma$, i.e., $$\label{transe}
\mu {\boldsymbol{h}}\cdot {\boldsymbol{n}}_\Gamma = \mu_{0}(\nabla \psi + k \boldsymbol \rho) \cdot {\boldsymbol{n}}_\Gamma \,.$$ Finally, we deduce from and the property ${\mathbf{curl}}\, \boldsymbol \rho = {\bf 0}$ that $$\int_\Gamma  {\boldsymbol{e}}_{|\Omega_C}  \times {\boldsymbol{n}}_\Gamma \cdot \boldsymbol \rho =\int_\Gamma  {\boldsymbol{e}}_{|\Omega_I}  \times {\boldsymbol{n}}_\Gamma \cdot \boldsymbol \rho = \int_{\Omega_I} {\mathbf{curl}}\, {\boldsymbol{e}}\cdot \boldsymbol \rho,$$ thus $$\label{scal}
\int_\Gamma \sigma^{-1} ( {\mathbf{curl}}\, {\boldsymbol{h}}- {\boldsymbol{j}})  \cdot (\boldsymbol \rho \times {\boldsymbol{n}}_\Gamma)= \imath \, \omega \int_{\Omega_I} \mu_0  (\nabla \psi + k \boldsymbol \rho) \cdot \boldsymbol \rho \, .$$

From now on, for the sake of simplicity in notations, ${\boldsymbol{h}}$ will stand for ${\boldsymbol{h}}|_{\Omega_{C}}$. Taking into account , , and , we deduce that the eddy current problem can be formulated in terms of the magnetic field and its scalar potential representation in the insulator in the following form: Find ${\boldsymbol{h}}:\, \Omega_{C}\to \mathbb C^{3}$, $\psi:\, \Omega_{I}\to \mathbb C$ and $k\in \mathbb C$ such that, $$\begin{aligned}
\imath \omega \mu {\boldsymbol{h}}+ {\mathbf{curl}}\, [\sigma^{-1} ( {\mathbf{curl}}\, {\boldsymbol{h}}- {\boldsymbol{j}})] &= \boldsymbol 0          &\text{in $\Omega_C$}\label{ModelProblem1}\\[2ex]
{\boldsymbol{h}}\times {\boldsymbol{n}}_\Gamma & =   (\nabla \psi + k \boldsymbol \rho) \times {\boldsymbol{n}}_\Gamma  &\text{on $\Gamma$}\label{ModelProblem2}\\[2ex]
\mu\, {\boldsymbol{h}}\cdot {\boldsymbol{n}}_\Gamma & =   \mu_0(\nabla \psi + k 