Antenna equivalent radius

Derivation

The equivalent radius is derived by equating the average magnetic vector potential at the surface of a conductor of arbitrary cross-section with the potential on the surface of a cylinder.

Assume a conductor's cross-section dimensions are small compared to the wavelength, current only flows axially along the conductor, the current distribution slowly varies along the conductor's length, and current is approximately uniformly distributed along its circumference (owing to the skin effect). Furthermore, only the current in a neighborhood around any point on the conductor significantly contributes to the potential at that point. Time dependence is ignored, as it may be incorporated by multiplying the current distribution by a time-varying sinusoid. These conditions imply that a quasi-static condition exists and that the geometry is, effectively, one of an infinitely long conductor with a constant surface current density ${\displaystyle Q}$ (current per area), thereby reducing a three-dimensional problem to a two-dimensional one. Also implied is the magnetic vector potential is parallel to the conductor's axis.

First, consider the potential at a fixed point ${\displaystyle \mathit{y}}$ on the circumference of the arbitrary cross-section. With the circumference divided into differential segments ${\displaystyle dx}$, the current distribution may be approximated by placing a vertical line current within each segment, each with a linear density of ${\displaystyle Qdx}$ (current per length). It is well known that the potential of such a line current is ${\displaystyle -({\mu }_{0}Qdx/2\pi )\ln |\mathit{x}-\mathit{y}|}$, where ${\displaystyle {\mu }_0}$ is the permeability constant. The potential at ${\displaystyle \mathit{y}}$ is the sum of the potentials for all the strips, which is

${\displaystyle A(\mathit{y})=-\frac{{\mu }_{0}Q}{2\pi }{\oint }_{\ell }\ln |\mathit{x}-\mathit{y}|dx.}$

The average potential is then

${\displaystyle \overline{A}=\frac{1}{L}{\oint }_{\ell }A(\mathit{y})dy=-\frac{{\mu }_{0}Q}{2\pi L}{\oint }_{\ell }{\oint }_{\ell }\ln |\mathit{x}-\mathit{y}|dxdy.}$

Now consider the case of cylinder with the same linear current density as the conductor of arbitrary cross-section. It is also well known that the potential at any point on its surface, which is also equal its average potential, is

${\displaystyle A_c=-\frac{{\mu }_{0}QL}{2\pi }\ln \left(r_e\right).}$

Equating ${\displaystyle \overline{A}}$ and ${\displaystyle A_c}$ yields

${\displaystyle \ln \left(r_e\right)=\frac{1}{L^2}{\oint }_{\ell }{\oint }_{\ell }\ln |\mathit{x}-\mathit{y}|dxdy.}$

Exponentiation of both side leads to the formula for the equivalent radius.

The formula for the equivalent radius provides consistent results. If the conductor cross-section dimensions are scaled by a factor ${\displaystyle \alpha }$, the equivalent radius is scaled by ${\displaystyle |\alpha |}$. Also, the equivalent radius of a cylindrical conductor is equal to the radius of the conductor.