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