How Does Distance Affect EM Wave Energy and Field Magnitude?

[leolaw]

Suppose a 50-kW radio station emits EM waves uniformly in all directions (a) How much energy per second crosses a 1.0m^2 area 100m from the transmitting antenna? (b) What is the rms magnitude of the E field at this point, assuming the station is operating at full power? (c) What is the voltage induced in a 10 m long vertical car antenna at this distance?

(a) if the definition of power is J/s , wouldn't be the energy acrosses that region also 50kW?

(b) I think this is simply \frac{\Delta U}{\Delta t} = (\epsilon _0)(E^2)(Ac\Delta t)}

Since that we have \frac{\Delta U}{\Delta t}, which is 50kW, already, we also have the area, speed of EM wave, and the time it takes to travel 100m, so we can just solve for E.

(c) I have no ideas how to start? any ideas?

 

[asrodan]

For part (a) energy is transferred as photons, basically "packets" of energy that have a particle-like quality to there behaviour. The move out in all directions so to find the energy across the region you have to consider the fact that the 5okW is distributed over a sphere of radius 100m.

Part (b) looks fine except for the power and that the left side shouldn't have a \Delta t

For part (c) you need to use Faraday's law which states that EA = \frac{-d\Phi_{B}}{dt} = V

This is only for constant E and A.

 

[jtbell]

Suppose a 50-kW radio station emits EM waves uniformly in all directions (a) How much energy per second crosses a 1.0m^2 area 100m from the transmitting antenna? [...]

(a) if the definition of power is J/s , wouldn't be the energy acrosses that region also 50kW?

Just for a moment, let's replace that radio station with a 50W light bulb. Imagine that your friend is holding the lit-up light bulb, and you are standing 100 m away, holding a 1-m^2 screen facing the light bulb. Do you really think that all of the 50W from the light bulb is landing on your screen?