We’ve seen that the far field electric and magnetic fields associated with a magnetic vector potential were

\begin{equation}\label{eqn:dualFarField:40}

\BE = -j \omega \textrm{Proj}_\T \BA,

\end{equation}

\begin{equation}\label{eqn:dualFarField:60}

\BH = \inv{\eta} \kcap \cross \BE.

\end{equation}

It’s worth a quick note that the duality transformation for this, referring to [1] tab. 3.2, is

\begin{equation}\label{eqn:dualFarField:100}

\BH = -j \omega \textrm{Proj}_\T \BF

\end{equation}

\begin{equation}\label{eqn:dualFarField:120}

\BE = -\eta \kcap \cross \BH.

\end{equation}

What does \( \BH \) look like in terms of \( \BA \), and \( \BE \) look like in terms of \( \BH \)?

The first is

\begin{equation}\label{eqn:dualFarField:140}

\BH

= -\frac{j \omega}{\eta} \kcap \cross \lr{ \BA – \lr{\BA \cdot \kcap} \kcap },

\end{equation}

in which the \( \kcap \) crossed terms are killed, leaving

\begin{equation}\label{eqn:dualFarField:160}

\BH

= -\frac{j \omega}{\eta} \kcap \cross \BA.

\end{equation}

The electric field follows again using a duality transformation, so in terms of the electric vector potential, is

\begin{equation}\label{eqn:dualFarField:180}

\BE = j \omega \eta \kcap \cross \BF.

\end{equation}

These show explicitly that neither the electric or magnetic far field have any radial component, matching with intuition for transverse propagation of the fields.

# References

[1] Constantine A Balanis. *Antenna theory: analysis and design*. John Wiley & Sons, 3rd edition, 2005.