Question

An AM radio transmitter broadcasts 63.2 kW of power uniformly in all directions. (a) Assuming all of the radio waves that strike the ground are completely absorbed, and that there is no absorption by the atmosphere or other objects, what is the intensity 30.5 km away

Answer 1

Answer:

Intensity of the transmitted radio wave is 5.406 x 10⁻⁶ W/m²

Explanation:

Given;

power of radio transmitter, P = 63.2 kW = 63200 W

distance of transmission, r = 30.5 km

Intensity of the transmitted radio wave is calculated as follows;

$I = \frac{P}{4\pi r^2}$

where;

I is the intensity of the transmitted radio wave

Substitute the given values and calculate the intensity of the transmitted radio wave;

$I = \frac{P}{4\pi r^2} = \frac{63200}{4\pi (30500)^2} = 5.406 *10^{-6} \ W/m^2$

Therefore, Intensity of the transmitted radio wave is 5.406 x 10⁻⁶ W/m²