An 800-kHz radio signal is detected at a point 2.7 km distant from a transmitter tower.The electric field amplitude of the signal at that point is 0.36 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 intensity of the radio signal at that point? (c = 3.00 × 10⁸ m/s, μ₀ = 4π × 10^-7 T ∙ m/A,ε₀ = 8.85 × 10^-12 C²/N ∙ m²) A)170 µW/m² B)240 µW/m² C)340 µW/m² D)120 µW/m² E)86 µW/m²

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Quizplus · Accepted Answer

The intensity of the radio signal is given by:$$I = \frac{1}{2}ε_0 c E_0^2$$where:* I is the intensity in W/m^2* ε_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:$$I = \frac{1}{2}(8.85 × 10^{-12} C^2/N \cdot m^2)(3.00 × 10^8 m/s)(0.36 V/m)^2$$$$I = 170 µW/m^2$$Therefore, the intensity of the radio signal at that point is 170 µW/m^2.

An 800-KHz Radio Signal Is Detected at a Point 2