Calculating Electromagnetic Wave Intensity in a 30 sq m Room

[practicaleducator]

Hi,
If I build a machine that its sole purpose is to radiate xx Hz of electromagnetic wave, how do I calculate the intensity of the waves? Let's say I put it in the room of 30 sq meters.
Thank you.

 

[sophiecentaur]

Welcome to PF. That's a tough question to answer because you are not specifying enough. A "machine" is not normally the way we'd go about producing EM waves. I assume. you mean an electronic device of some sort - we could call that a Transmitter with an aerial and the room would have 'normal' walls; i.e. not metal, other than the odd bits of metal in the wiring and a couple of steel beams. If you're talking in terms of Radio Frequency waves, most of the Power that the transmitter produces would escape through the walls and some of it would be absorbed in the material.
You are jumping into the subject more than half way and it would be more usual to talk in terms of putting the aerial out in 'free space', where it would be radiating its power equally in all directions. (This is actually impossible to achieve and most aerials will have some significant directivity over just a range of directions but just let's assume it's omnidirectional) and that the wavelength is around 0.1m (so it's not large compared with the room size).

If you looked at the Power radiated through the walls of a spherical (well why not?) room with a total wall area of 100m² and, if the transmitter was radiating 100W, then the 'intensity' of the radiated power through the walls would be 100/100 = 1W per m². If you had a bigger room with 1000m² of wall area then the intensity would be 0.1W/m² as it spreads out over ten times the area etc. etc.

Now tell us what you were really wondering about. It sounds like the results of a conversation in a pub near closing time.(?) Is this a Health and Safety question?

 

[anorlunda]

Is it radio or light?

 

[sophiecentaur]

Is it radio or light?

Would it matter, in an ideal case? If the walls are reflective then, as in the RF case, the energy density would depend on the absorption coefficient. Iirc there was a similar thread not long ago in which the total energy stored in the room space would be equal to the power input divided by the fraction absorbed. This would be true whether or not there is a resonance but the transmitter matching might need to be re-tuned to obtain the same power input.

This may not be helping the OP, though.