Question

An antenna located on the surface of a flat earth transmits an average power of 200 kW. Assuming that all the power is radiated uniformly over hte surface of a hemisphere with the antenna at the center, the time average Poynting vector at 50 km is
1. Zero
2. \(\frac{2}{\pi }\;{\bar a_r}W/{m^2}\)
3. \(\frac{40}{\pi }\;\mu W/m^2\)
4. \(\frac{2}{\pi }\;{\bar a_r}\mu \;W/{m^2}\)

Answer 1

Correct Answer - Option 4 : \(\frac{2}{\pi }\;{\bar a_r}\mu \;W/{m^2}\)

It is hemisphere, so area = 2πr²

Now transmitted power

P_T = P_avg × area

\({P_{avg}} = \frac{{{P_T}}}{{area}} = \frac{{200\;kW}}{{2\pi \left( {50\; \times \;{{10}^3}} \right){m^2}}}\)  

= \(\frac{{200\; \times \;{{10}^3}}}{{25\; \times\; 50 \;\times \;50\; \times\; {{10}^6}}}W/{m^2}\)

= \(\frac{{\frac{{2\; \times \;{{10}^5}}}{{2\pi \; \times \;25\; \times \;{{10}^8}}}W}}{{{m^2}}} = \frac{{40}}{\pi }\;\mu \;W/{m^2}\)

Poynting vector will be directed into a direction; So

= \({P_{avg}} = \frac{{40}}{\pi }\;{\bar a_r}\;\;\mu W/{m^2}\)