Sinusoidal electromagnetic wave help

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SUMMARY

The discussion centers on calculating the energy carried by a sinusoidal electromagnetic wave through a window with an area of 0.700 m², given an RMS electric field value of 2.30×10⁻² V/m over a duration of 30.0 seconds. The participant initially calculated the intensity using the formula I = E² / (2 * μ₀ * c), resulting in an intensity of 7.01 * 10⁻⁷ W/m². However, the participant's final energy calculation of 1.47 * 10⁻⁵ Joules was incorrect due to not converting the RMS value to a peak value before calculations. The correct approach involves recognizing the relationship between RMS and peak values for sinusoidal waves.

PREREQUISITES
  • Understanding of sinusoidal electromagnetic waves
  • Familiarity with the concepts of RMS and peak values
  • Knowledge of electromagnetic wave intensity calculations
  • Proficiency in using formulas involving μ₀ (permeability of free space) and speed of light
NEXT STEPS
  • Learn the conversion between RMS and peak values for sinusoidal waves
  • Study the derivation and application of the intensity formula I = E² / (2 * μ₀ * c)
  • Explore energy calculations in electromagnetic waves using different field values
  • Investigate the implications of using μ instead of μ₀ in various media
USEFUL FOR

Students and educators in physics, particularly those studying electromagnetic waves, as well as anyone involved in practical applications of wave energy calculations.

reed2100
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Homework Statement


A sinusoidal electromagnetic wave from a radio station passes perpendicularly through an open window that has area of 0.700m2 . At the window, the electric field of the wave has an rms value 2.30×10−2V/m .

How much energy does this wave carry through the window during a 30.0s commercial?

2. Homework Equations

Intensity = Power/area -> I=P/A

I = E^2 / (2 * μ-naught * speed of light)

Energy = power * time

The Attempt at a Solution



First find I since I am given all values needed ?

I = [ (2.3 * 10^-2 )^2 ] / [ 2 *4π * 10^-7 * 3 * 10^8 ]

I = 7.01 * 10^-7

Power = Intensity * area
Power = 7.01 * 10^-7 * (.7)
Power = 4.91 * 10^-7

Energy = power * time
Energy = 4.91*10^-7 * (30)
Energy = 1.47 * 10^ -5 Joules

That's incorrect and I'm having trouble figuring out where I went wrong, as I used the same approach for a similar problem and it worked. The difference between problems was that I wasn't given an rms value, instead I was just told the magnitude of the field if I remember correctly. I thought the rms was just root mean square and was just another way of expressing an "average" -ish value. Do I need to use μ instead of μ naught because it's in air and technically not vacuum? That would confuse me as well because I did a problem in this vein that involved high energy waves through flesh to eliminate cancer cells, and μ naught worked fine. I wouldn't think the program would suddenly switch it's expectations of what simplifications to use for the sake of the course level. Any and all help is greatly appreciated.
 
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Did you try converting E to a Peak value from an RMS value before doing your energy calculations? The relationship between RMS and Peak for a sinewave is straightforward.
 
gneill said:
Did you try converting E to a Peak value from an RMS value before doing your energy calculations? The relationship between RMS and Peak for a sinewave is straightforward.
Ah, that did it. I wasn't aware of the rms to peak necessity, I must have glazed over it in my reading or something. Thank you!
 

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