# Sinusoidal electromagnetic wave help

1. Nov 19, 2014

### reed2100

1. The problem statement, all variables and given/known data
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. Relevant equations

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

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

Energy = power * time

3. 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.

2. Nov 19, 2014

### Staff: Mentor

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.

3. Nov 19, 2014

### reed2100

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!