An 800-kHz radio signal is detected at a point 9.1 km distant from a transmitter tower.The electric field amplitude of the signal at that point is 0.440 V/m.Assume that the signal power is radiated uniformly in all directions and that radio waves incident upon the ground are completely absorbed.What is the average total power radiated by the transmitter? (c = 3.00 × 10⁸ m/s, μ₀ = 4π × 10^-7 T ∙ m/A,ε₀ = 8.85 × 10^-12 C²/N ∙ m²) A)0.27 MW B)0.32 MW C)0.38 MW D)0.45 MW E)0.50 MW

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Quizplus · Accepted Answer

The average total power radiated by the transmitter is given by:$$P_{avg}=\frac{1}{2}\varepsilon_0cE^2A$$where:* $\varepsilon_0$ is the permittivity of free space* $c$ is the speed of light* $E$ is the electric field amplitude* $A$ is the surface area of a sphere with radius equal to the distance from the transmitterIn this problem, we have:* $f = 800	ext{ kHz} = 8	imes10^5	ext{ Hz}$* $r = 9.1	ext{ km} = 9.1	imes10^3	ext{ m}$* $E = 0.440	ext{ V/m}$So, the average total power radiated by the transmitter is:$$P_{avg}=\frac{1}{2}(8.85	imes10^{-12}	ext{ C}^2/	ext{N}\cdot	ext{m}^2)(3	imes10^8	ext{ m/s})(0.440	ext{ V/m})^2(4\pi(9.1	imes10^3	ext{ m})^2)$$$$P_{avg}=0.27	ext{ MW}$$Therefore, the correct answer is A.

An 800-KHz Radio Signal Is Detected at a Point 9