Distance of hearing a radio station

  • Thread starter Thread starter gillyr2
  • Start date Start date
  • Tags Tags
    Hearing Radio
Click For Summary
SUMMARY

The discussion centers on calculating the distance from which a radio station broadcasting at an average power of 24 kW can be received, given an acceptable electric field amplitude of 2.6×10-2 V/m. The solution involves using the equation P/A = e0cE2 to derive the distance, resulting in an estimated range of 65.2 kilometers. Participants clarified the use of the average value of the Poynting vector and corrected the area calculation from a circle to a sphere's surface area.

PREREQUISITES
  • Understanding of electromagnetic wave propagation
  • Familiarity with the Poynting vector and its average value
  • Knowledge of the permittivity of free space (e0) and speed of light (c)
  • Basic algebra for rearranging equations
NEXT STEPS
  • Study the derivation of the Poynting vector in electromagnetic theory
  • Learn about the implications of electric field strength in radio wave propagation
  • Research the relationship between power, area, and distance in wave transmission
  • Explore practical applications of radio wave calculations in telecommunications
USEFUL FOR

Students in physics or engineering, radio frequency engineers, and anyone interested in understanding radio wave propagation and transmission distances.

gillyr2
Messages
45
Reaction score
0

Homework Statement


A radio station is allowed to broadcast at an average power not to exceed 24 kW.
If an electric field amplitude of 2.6×10−2 V/m is considered to be acceptable for receiving the radio transmission, estimate how many kilometers away you might be able to hear this station.


Homework Equations



P/A = e_0*c*E^2

The Attempt at a Solution



e_0*c*E^2/P = 1/pi r^2

sqrt(24000/(8.85 *10^-12*3*10^8*pi)) = 65.2 km

help!
 
Physics news on Phys.org
can anybody help?
 
I just had this problem and this is how I solved it

S=P/A where S=(.5)c(permittivity of free space)(E^2) - (this is the average value of S not an instantaneous value. Your equation for S was assuming the instantaneous value)

then a little rearranging:

A=P/S
(4)(pi)(r^2)=P/S - (4pir^2 is the surface area of a sphere, I think you were incorrectly using the area of a circle)

you should be able to solve from here!
 

Similar threads

Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
4K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 1 ·
Replies
1
Views
2K