An 800 kHz radio signal is detected at a point 2.1 km distant from a transmitter tower.The electric field amplitude of the signal at that point is 800 mV/m.Assume that the signal power is radiated uniformly in all directions and that radio waves incident upon the ground are completely absorbed.The intensity of the radio signal at that point is closest to: A)8.5 × 10-4 W/m² B)1.2 × 10-3 W/m² C)1.7 × 10-3 W/m² D)6.0 × 10-4 W/m² E)4.2 × 10-4 W/m²

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Quizplus · Accepted Answer

The intensity of the radio signal is given by:$$I = \frac{1}{2} \epsilon_0 c E_0^2$$where:* I is the intensity in W/m^2* $\epsilon_0$ is the permittivity of free space in F/m* c is the speed of light in m/s* $E_0$ is the electric field amplitude in V/mSubstituting the given values into the equation, we get:$$I = \frac{1}{2} (8.85 	imes 10^{-12} 	ext{ F/m})(3 	imes 10^8 	ext{ m/s})(0.8 	ext{ V/m})^2 = 8.5 	imes 10^{-4} 	ext{ W/m}^2$$Therefore, the intensity of the radio signal at that point is closest to 8.5 × 10-4 W/m^2.

An 800 KHz Radio Signal Is Detected at a Point