An 800-kHz sinusoidal radio signal is detected at a point 2.1 km distant from a transmitter tower.The $\vec { E }$ 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

Question

Quizplus · Accepted Answer

The intensity of an electromagnetic wave is given by:$$I=\frac{1}{2}c\epsilon_0E^2$$where:* I is the intensity in W/m^2* c is the speed of light in m/s* $\epsilon_0$ is the permittivity of free space in F/m* E is the electric field amplitude in V/mPlugging in the given values, we get:$$I=\frac{1}{2}(3.0\times10^8\text{ m/s})(8.85\times10^{-12}\text{ F/m})(0.80\text{ V/m})^2=\boxed{8.5\times10^{-4}\text{ W/m}^2}$$

An 800-KHz Sinusoidal Radio Signal Is Detected at a Point E⃗\vec { E }E

An 800-KHz Sinusoidal Radio Signal Is Detected at a Point $\vec { E }$