Topic: Electromagnetics
The electric field measured in the far field of an antenna at a distance of 50 m is $1~\text{V/m}$. The average power densities at a distance of 500 m from the antenna is
- A. $26.6~\mu\text{W/m}^{2}$
- B. $0.1~\mu\text{W/m}^{2}$
- C. $10~\mu\text{W/m}^{2}$
- D. $13.3~\mu\text{W/m}^{2}$
Answer: D. $13.3~\mu\text{W/m}^{2}$
Explanation: In the far field, the electric field intensity $E \propto \frac{1}{r}$, so $E = \frac{K}{r}$.
At $r = 50~\text{m}$, $E_{1} = \frac{K}{50} \Rightarrow 1 = \frac{K}{50} \Rightarrow K = 50$.
At $r = 500~\text{m}$, $E_{2} = \frac{50}{500} = 0.1~\text{V/m}$.
The average power density is $P_{\text{avg}} = \frac{1}{2}\frac{E^{2}}{\eta}$ where $\eta = 120\pi~\Omega$.
$P_{\text{avg}} = \frac{(0.1)^{2}}{2 \times 120\pi} = \frac{0.01}{240\pi} \approx 13.3~\mu\text{W/m}^{2}$.