Radio Broadcast Antenna Peak Intensity

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SUMMARY

The peak intensity of a radio broadcast signal from an antenna broadcasting at 104.3 FM with a power of 5.00 kilowatts can be calculated using the formula for intensity, I = P/(4πr²). For a receiving antenna located 25.0 km away, the correct calculation involves determining the surface area of a spherical shell, which is 4πr², rather than a circular area. The intensity is ultimately measured in watts per square meter, necessitating the conversion of distance to meters for accurate computation.

PREREQUISITES
  • Understanding of radio frequency concepts, specifically FM broadcasting.
  • Knowledge of the Poynting vector and its application in electromagnetic theory.
  • Familiarity with the formula for intensity in spherical coordinates.
  • Basic skills in unit conversion, particularly between kilometers and meters.
NEXT STEPS
  • Study the Poynting vector and its significance in electromagnetic wave propagation.
  • Learn about the relationship between power, intensity, and distance in radio broadcasting.
  • Explore the derivation and application of the formula for intensity in spherical coordinates.
  • Investigate the impact of distance on signal strength and quality in radio communications.
USEFUL FOR

Students in physics or engineering, radio frequency engineers, and anyone involved in the design and analysis of radio broadcasting systems.

Minescrushessouls
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Homework Statement


A radio broadcast antenna is located at the top of a steep tall mountain. The antenna is broadcasting 104.3 FM (in Megahertz) with a power of 5.00 kilowatts.

What is the peak intensity of the signal at a receiving antenna located 25.0 km away?

Homework Equations


Honestly, I don't really know

Intensity=P/d^2

Peak Intensity=some vector/μ0

The Attempt at a Solution


So I tried doing 5000/(25000^2)

The power divided by the distance squared. The system told me my units were incorrect...so I guess I don't even know where to start on this
 
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You should use equation of Poynting vector
 
Ugnius said:
You should use equation of Poynting vector

Ok so S=(ExB)/μ

But how would I find E and B from the information provided?
 
Intensity is measured in watts per square meter. Think of the radio antenna as putting out 5 kW in a spherical distribution. At some distance R from the transmitter, what's the size of the spherical "surface"?
 
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So the area would be the area of a circle, or pi*r^2

So would that mean it would be W/(pi*r^2)?
 
Minescrushessouls said:
So the area would be the area of a circle, or pi*r^2

So would that mean it would be W/(pi*r^2)?
No, the surface of a sphere is not a circle -- it's a spherical shell. What's the surface area of a spherical shell?
 
gneill said:
No, the surface of a sphere is not a circle -- it's a spherical shell. What's the surface area of a spherical shell?

4*pi*r^2?
 
Minescrushessouls said:
4*pi*r^2?
Yes.
 

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