Chapter 9, Problem 1P

A center-fed Hertzian dipole is excited by a current I₀ = 20 A. If the dipole is λ/50 in length, determine the maximum radiated power density at a distance of 1 km.

To determine

The maximum radiated power density at the given distance.

Answer to Problem 1P

The maximum radiated power density at the given distance is $7.6\mu \text{W}/{\text{m}}^{\text{2}}$.

Explanation of Solution

Given data:

The amplitude of current $I_0$ flowing through the conductor is $20\text{A}$.

The distance $R$ is $1\text{km}$.

Calculation:

The dipole length $l$ is given as,

$l=\frac{\lambda }{50}$

Here,

$\lambda$ is the wavelength.

The conversion from kilometer $\left(\text{km}\right)$ to meter $\left(\text{m}\right)$ is given by,

$\text{1}\text{km}=\text{1000}\text{m}$

So, distance $R$ is,

$R=1000\text{m}$

The maximum radiated power density $S_{\max }$ is given by,

$S_{\max }=\frac{{\eta }_0{k}^2{I}_0{}^2{l}^2}{32{\pi }^2{R}^2}$ (1)

Here,

${\eta }_0$ is the intrinsic impedance of the air, which has the standard value as $\left({\eta }_0=377\Omega \right)$.

$k$ is the wave number.

The wave number $k$ is written as,

$k=\frac{2\pi }{\lambda }$

Substitute $377\Omega$ for ${\eta }_0$, $\frac{2\pi }{\lambda }$ for $k$, $20\text{A}$ for $I_0$, $\frac{\lambda }{50}$ for $l$ and $1000\text{m}$ for $R$ in equation (1).

$\begin{array}{c}{S}_{\max }=\frac{\left(377\Omega \right){\left(\frac{2\pi }{\lambda }\right)}^2{\left(20\text{A}\right)}^2{\left(\frac{\lambda }{50}\right)}^2}{32{\pi }^2{\left(1000\text{m}\right)}^2}\\ =\frac{377\times 4\times 400}{32\times 1000\times 1000\times 2500}\\ =7.6\times {10}^{-6}\text{W}/{\text{m}}^{\text{2}}\end{array}$

The conversion from $\text{W}/{\text{m}}^{\text{2}}$ into $\mu \text{W}/{\text{m}}^{\text{2}}$ is given by,

$1\text{W}/{\text{m}}^{\text{2}}={10}^6\mu \text{W}/{\text{m}}^{\text{2}}$

The conversion of $7.6\times {10}^{-6}\text{W}/{\text{m}}^{\text{2}}$ in $\mu \text{W}/{\text{m}}^{\text{2}}$ is given by,

$7.6\times {10}^{-6}\text{W}/{\text{m}}^{\text{2}}=7.6\mu \text{W}/{\text{m}}^{\text{2}}$

Hence, the maximum power density radiated by a half wave dipole is,

$S_{\max }=7.6\mu \text{W}/{\text{m}}^{\text{2}}$

Conclusion:

Therefore, the maximum radiated power density at the given distance is $7.6\mu \text{W}/{\text{m}}^{\text{2}}$.