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\subsection{}

\begin{equation}
\frac{\partial\EE}{\partial t}=\nab\times\BB-\JJ-\frac{\alpha}{f}\left(\dot{\phi}\BB+\nab\phi\times\EE\right)
\end{equation}
Use Ohm's law
\begin{equation}
\frac{\partial\EE}{\partial t}=\nab\times\BB-\sigma(\EE+\uu\times\BB)-\frac{\alpha}{f}\left(\dot{\phi}\BB+\nab\phi\times\EE\right)
\end{equation}
Omit displacement current and solve for $\EE$
\begin{equation}
\sigma\EE+\frac{\alpha}{f}\nab\phi\times\EE
=\nab\times\BB-\sigma\uu\times\BB-\frac{\alpha}{f}\dot{\phi}\BB
\end{equation}
Divide by $\sigma$ and define $\tilde\alpha=\alpha/(\sigma f)$, so
\begin{equation}
\EE+\tilde\alpha\nab\phi\times\EE
=\sigma^{-1}\nab\times\BB-\uu\times\BB-\tilde\alpha\dot{\phi}\BB
\end{equation}
or
\begin{equation}
\MMMM\EE=\sigma^{-1}\nab\times\BB-\uu\times\BB-\tilde\alpha\dot{\phi}\BB
\end{equation}
where
\begin{equation}
\mathsf{M}_{ij}=\delta_{ij}-\epsilon_{ijk}\,\tilde\alpha\phi_{,k}
\end{equation}
is a tensor, whose inverse is
\begin{equation}
\left(\MMMM^{-1}\right)_{ij}=\left[\delta_{ij}-\epsilon_{ijk}\,\tilde\alpha\phi_{,k}+\alpha^2\phi_{,i}\phi_{,j}\right]/N
\end{equation}
with the determinant $N=...$.

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\begin{figure}[h!]\begin{center}
%\includegraphics[width=\columnwidth]{2p5}
\end{center}\caption[]{
\url{2p5}
}\label{2p5}\end{figure}

%\begin{table}[htb]\caption{
%}\vspace{12pt}\centerline{\begin{tabular}{lccccccc}
%$R$ & $T$ & $\kf$ & $\omf$ & $\EEGW$ & $\hrms$ & $k_{\rm peak}$ \\
%\hline
%\label{Ttimescale}\end{tabular}}\end{table}

%r e f
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%\bibitem[Biskamp \& M\"uller(1999)]{BM99}
%Biskamp, D., \& M\"uller, W.-C.\yprl{1999}{83}{2195}

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