An 800-kHz sinusoidal 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 0.80 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? (ε₀ = 8.85 × 10^-12 C²/N ∙ m²,μ₀ = 4π × 10^-7 T ∙ m/A,c = 3.0 × 10⁸ m/s) 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 a radio signal can be calculated using the formula $I = \frac{1}{2} \epsilon_0 c E^2$, where $I$ is the intensity, $\epsilon_0$ is the permittivity of free space ($8.85 \times 10^{-12} \, C^2/N \cdot m^2$), $c$ is the speed of light ($3.0 \times 10^8 \, m/s$), and $E$ is the electric field amplitude ($0.80 \, V/m$). Plugging in the values, we get $I = \frac{1}{2} \times 8.85 \times 10^{-12} \times 3.0 \times 10^8 \times (0.80)^2 = 8.5 \times 10^{-4} \, W/m^2$.

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