Question

An antenna consumes 100W of electrical energy, all of which is converted to radio waves emitted equally in all directions. a) What is the magnitude of the Poynting vector 1km away from the source? b) A 1m by 1m panel that absorbs all radio waves going through is held facing the source,, 1km away from it, during 1 minute. How much energy does it receive from the source? c) Answer b) if the panel is tilted 45°.

Answer 1

a] It is given that 100 W of electrical energy is converted to 100W of electromagnetic waves energy in all direction.

So, using the inverse square law, the magnitude of Poynting vector 1 km (1000 m) away from the source will be:

b]

The energy absorbed will be:

E = Power x time = |S| x Area x time

=> E = 7.9577 x 10^-6 x 1 x 60 = 4.775 x 10^-4 J

c] If the angle is 45 degrees,

E' = Ecos45 = 3.376 x 10^-4 J.