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# Área efectiva del dipolo:

${\displaystyle A_{dip}}$=${\displaystyle {\frac {P_{r}}{P_{0}}}={\frac {3\lambda ^{2}}{8\pi }}=0.1193\lambda ^{2}}$[0.3cm] ${\displaystyle P_{o}=E^{2}/120\pi }$
${\displaystyle Pr={\frac {E^{2}a^{2}}{4(80\pi ^{2}a^{2}/\lambda ^{2})}}}$ = ${\displaystyle {\frac {\lambda ^{2}E^{2}a^{2}}{320\pi a^{2}}}}$= ${\displaystyle {\frac {\lambda ^{2}E^{2}}{320\pi ^{2}}}}$
${\displaystyle A_{dip}={\frac {\frac {\lambda ^{2}E^{2}}{320\pi ^{2}}}{\frac {E^{2}}{120\pi }}}={\frac {120\pi \lambda ^{2}E^{2}}{320\pi ^{2}E^{2}}}={\frac {3\lambda ^{2}}{8\pi }}}$

Área efectiva de una circunferencia

# Small Dipole with No Heat Loss

For a small uniform current element the available output power is equal to the induced voltage squared divided by four times the radiation resistance. Thus

${\displaystyle P_{r}={\frac {E^{2}a^{2}}{4R_{rad}}}}$

where
E=effective value of the electric field of the wave.
a=length of the current element.
${\displaystyle R_{rad}=80\pi ^{2}a^{2}/\lambda ^{2}}$ Since the power flow per unit area is equal to the electric field squared divided by the impedance of free space,
i.e., ${\displaystyle P_{o}=E^{2}/120\pi }$, we have

${\displaystyle A_{dip}={\frac {P_{r}}{P_{0}}}={\frac {3\lambda ^{2}}{8\pi }}=0.1193\lambda ^{2}}$

The effective area of a half-wavelength dipole with no heat loss is only 9.4 per cent, 0.39 decibels,2 larger than the effective area of the small dipole. Therefore

${\displaystyle A_{0.5\lambda }=0.1305\lambda ^{2}}$

The area of a rectangle with one-half wavelength and one-quarter wavelength sides is ${\displaystyle 0.125\lambda ^{2}}$ and it is, therefore, a good approximation for the effective areas of small dipoles and half-wavelength dipoles.


${\displaystyle A_{0.5}=0.1305\lambda ^{2}}$
${\displaystyle 0.1305\lambda ^{2}=\pi r^{2}}$
${\displaystyle {\frac {0.1305}{\pi }}\lambda ^{2}=r^{2}}$
${\displaystyle 0.04154\lambda ^{2}=r^{2}}$
${\displaystyle {\sqrt {0.04152\lambda ^{2}}}={\sqrt {r^{2}}}}$
${\displaystyle r=0.2038\lambda }$