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.