Power and Intensity of Radio Waves

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SUMMARY

The discussion centers on calculating the intensity of radio waves from an AM radio station broadcasting at 21.0 kW and a frequency of 910 kHz, specifically at a distance of 18.0 km from the antenna. The intensity (I) is derived using the formula I = P/(4πr²), where P is the power and r is the distance from the source. Participants clarify that the frequency does not directly impact the intensity calculation in this context, focusing instead on the power and distance. The discussion emphasizes understanding the relationship between power, area, and intensity in free space.

PREREQUISITES
  • Understanding of radio wave propagation principles
  • Familiarity with the formula for intensity (I = P/A)
  • Knowledge of spherical wavefronts and their area calculations
  • Basic grasp of frequency and angular frequency (ω = 2πf)
NEXT STEPS
  • Research the derivation and application of the intensity formula I = P/(4πr²)
  • Explore the effects of frequency on radio wave propagation and intensity
  • Learn about the impact of environmental factors on radio wave transmission
  • Study advanced concepts in electromagnetic wave theory
USEFUL FOR

Students studying physics, engineers working with radio frequency technology, and anyone interested in understanding radio wave propagation and intensity calculations.

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Homework Statement



An AM radio station broadcasts with a power of 21.0 kW at a frequency of 910 kHz. Estimate the intensity of the radio wave at a point 18.0 km from the broadcast antenna.

Homework Equations



I= p/A I1/I2=r2^2/r1^2 w=2pi f

The Attempt at a Solution



From the given frequency, we can find omega (w). However, I am stuck on where to proceed afterwards. How would you find A? Thanks so much for helping.

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I cannot quite decode the relevant equation. Maybe its a fonts thing.
I= p/A I1/I2=r2^2/r1^2 w=2pi f
w (omega) = 2*pi*f is just expressing frequency in radians per second instead of cycles per second.

In free space, a radio wave intensity decreases as the area of the wavefront expanding sphere increases. In Earth atmosphere, it is also frequency-dependent because of propagation and absorption losses, ground losses, and ionospheric conditions. Not that we need be concerned..because
You have a given expression for the intensity, so we dodge all that hard stuff.

I can see I =\frac{p}{A} which makes sense. Not sure what the rest (I1/I2) is set equal to r_2^2/r_1^2 :confused:

I am unclear how \omega=2\pi f while true, actually affects anything here.

Sorry I don't quite get it yet. Have another try about what this intensity expression is.
 
just use the the amount of power and the distance. you don't need frequency.

I = P/(4*pi*r^2)
 

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