Wait A Minute...

From previously,

$B=\cfrac { d }{ Rv } .\cfrac { 1 }{ 2\pi \varepsilon _{ o }r } E$.

since $x^{'}=i.2\pi r$,    $x^{'}$ is tangential to $r$ everywhere

$B=-i\cfrac { d }{ Rv } .\cfrac { 1 }{ \varepsilon _{ o }x^{'} } E$

where $i$ was mulitplied to both the denominator and numerator.

Comparing this expression with,

$B=-i\cfrac{\partial E}{\partial x^{'}}$

 which leads to the wave equation

 $\cfrac { \partial ^{ 2 }E }{ \partial t^{ 2 } } =\cfrac { \partial ^{ 2 }E }{ \partial x^{ 2 } } v^{ 2 }$,

with a typical solution of the form,

$E(x,t)=\cfrac{A}{x}.e^{i(\omega t\pm kx)}$

We find that, since $E$ can be replaced with $D.E$ where $D$ is a constant, and still satisfy the wave equation,

$-i\cfrac{\partial E}{\partial x^{'}}=-i\cfrac{E}{x^{'}}$

and

$D=\cfrac { d }{ Rv  \varepsilon _{ o }}$

where $-i$ rotates $x^{'}$ to the $x$ direction,

$\cfrac{\partial E}{\partial x}=\cfrac{E}{x}$

$\int{\cfrac{1}{E}}\partial E=\int{\cfrac{1}{x}}\partial x$

If we consider only the spatial component of $E$,

$E=Cx$    over the length of the wire.  Since $V=d.E$,

$C=\cfrac{V}{d^2}$

ie.,

$E=\cfrac{V}{d^2}.x$    where $d$ is the length of the wire.

$D.E=\cfrac { V }{ dRv  \varepsilon _{ o }}.x\\=\cfrac{J}{\varepsilon _{ o }}.\cfrac{x}{d}$

Since, $V=d.D.E$

$V=\cfrac{J}{\varepsilon _{ o }}.x$

where $V$ is the voltage along the length of the wire.

This is a very interesting result.  It is tempting to suggest that this $E$ field/$V$ voltage profile up the wire antenna is most efficient for radiation at the end of the wire.  Although it is true that this expression for $E$ is derived from $B=-i\cfrac{dE}{dx^{'}}$, that leads to the wave equation directly.  This seem like impedance matching, but

$V=IR=\cfrac{I}{\varepsilon _{ o }v}.d$

$R=\cfrac{d}{\varepsilon _{ o }v}$ where $v$ is the charge average velocity along the wire that changes with the driving voltage at the end of the wire.

Have a nice day.