Question

A 71 kW radio station broadcasts its signal uniformly in all directions. - What is the average intensity of its signal at a distance of 220 m from the antenna? -What is the average intensity of its signal at a distance of 2200 m from the antenna?

Answer 1

Answer:

Explanation:

Energy of signal being radiated per second on all sides = 71 x 10³ J .

At a distance of 220 m it is spread over an area of 4 π x (220)² because it is spreading uniformly on all sides.

So energy crossing per unit area

= $\frac{71\times10^3}{4 \times \pi\times(220)^2}$

= 11.67 x 10⁻² Wm⁻²s⁻¹.

This is the intensity of the signal.

At 2200 m this intensity will further reduce by 100 times

So there it becomes equal to

11.67 x 10⁻⁴ Wm⁻² s⁻¹.