The Maxwell's equations are

\begin{align} \frac{\partial \vec{B}}{\partial t} &= - \nabla \times \vec{E} \hspace{2cm} \text{Faraday equations}\\ \frac{\partial \vec{D}}{\partial t} &= - \nabla \times \vec{H} - \vec{J} - \vec{s} \hspace{2cm} \text{Ampere's law.} \end{align}

where $\vec{E}$ is the electric field, $\vec{H}$ is the magnetic field, $\vec{J}$ is the electric flux, $\vec{D}$ is the displacement flux, $\vec{B}$ is the magnetic flux and $\vec{s}$ is the source.

The constitutive realtations between the fields and flux are as follows.

\begin{align} \vec{J} &= \sigma \vec{E}\\ \vec{B} &= \mu \vec{H}\\ \vec{D} &= \epsilon \vec{E} \end{align}

Using the contitutive relations, the Maxwell's equations can written in terms of the electric field and the magnetic flux \begin{align} \mu \frac{\partial \vec{B}}{\partial t} &= - \nabla \times \vec{E} \\ \epsilon \frac{\partial \vec{E}}{\partial t} &= - \nabla \times \frac{1}{\mu} \vec{B} - \sigma \vec{E} - \vec{s} \end{align}

Futher by droping the $\epsilon$ term and using a $e^{-i\omega t}$ Fourier time relation, Maxwell's equations can be written in the frequency domain as

\begin{align} -i \omega \mu \hat{\vec{B}} &= - \nabla \times \hat{\vec{E}} \\ - \vec{s} &= - \nabla \times \frac{1}{\mu} \hat{\vec{B}} - \sigma \hat{\vec{E}} \end{align}

The weak form is \begin{align} -i \omega \mu (\hat{\vec{B}},\hat{\vec{F}}) + (\nabla \times \hat{\vec{E}},\hat{\vec{F}}) &= 0 \\ (\nabla \times \mu^{-1} \hat{\vec{B}},\hat{\vec{W}}) + (\sigma \hat{\vec{E}},\hat{\vec{W}}) &= (\hat{\vec{s}},\hat{\vec{W}}) \end{align}

$ (\mu^{-1} B, \nabla \times W) + (\sigma E, W) = (\mu^{-1}B W )|_0^{end} $

$ (i \omega B,F) + (\nabla \times E , f) = 0 $