I have a question about EMF's and the Inverse Square Law.
How does distance factor in?
For example, if you apply it to a scale of millmeters, it logically could not seem to apply, ie. 1r = 1mm, 2r = 2mm, etc.
In other words, how can there be 4 times less radiaition at 2mm than at 1mm.
I must be missing a relevant point, and someone is probaly laughing already.
Bary Latham
Probably not. We all have these moments of incomprehension.
In this particular case, you need to think that the same radiation is being spread over the surface of a 1mm diameter sphere or a 2mm diameter sphere.
The area of a sphere is 4.pi.r^2, so the density of radiation over a constant area - say 1mm^2 - has to fall with the square of the radius.
-- Bill Sloman, Sydney
It is all starting from a hypothetical point source at the origin, so yes a sphere radius r=1 has area 4pi.r^2 and for r=2 is 4x bigger.
This rule would break down somewhere near the Planck length but then that would not be classical electrodynamics any more.
-- Regards, Martin Brown
The wiki is fair. Not the clearest writing, but the diagram is fine.
It needs to be emphasized earlier in the write up that is is for a point source. A line source will exhibit a different roll off. This has some effect in real life speaker design if you mix driver technologies. That is some speaker have planar tweeters.
If you are mixing cones and ribbons, you do need to know the distance to the listener.
I do need to limit the audio speak, lest f*ck head Phil (the Aussie, not the Doctor) shows up.
Beside sound, there are real life application in lighting.
--- Imagine you have a point source of light such that at 1 meter away from the source a detector with an aperture of 1 square centimeter intercepts one watt of power.
Assuming the detector has an impedance of one ohm, then the voltage developed across it will be: E = sqrt (PZ) = sqrt 1W * 1R² = 1 volt
and the power radiated by the source will be:
4pi r² m W 12.57 * 1² m * W P = ------------ = -------------------- = 1257 watts 1² cm 1² * 1e-2 mAssume now that the detector is moved to a point 2 meters away from the source.
The 1257 watts from the source will now be spread over the surface of the new sphere, which will have an area of:
A = 4pi r² = 12.57 * 2m² = 50.28m² = 5028cm²
and, consequently, a new surface power density of:
1257W 0.25W P = --------- = ------- 5028cm² cm²Since the detector's impedance hasn't changed, its output voltage for 250 milliwatts in will be:
E = sqrt (PZ) = sqrt (0.25W * 1R) = 0.5V.
So, since surface power density decreases by a factor of four for a twofold increase in distance from the source, it follows a square-law relationship.
However, you were asking about EMF, and since it only halves for a twofold inrease in distance from the source, its response is linear, not square-law.
Electric field drops proportional to the inverse SQUARE of the distance. Magnetic field drops proportional to the inverse CUBE of the distance. Radiated fields drop proportional to ...etc
Which EMF drops differently?
Can you please explain why the inverse square law does not apply to EMF's which are a form of radiation from the point of origin?
For example, the 2.4GHz emitted by the antenna of a wireless router.
We have calibrated EMF meters here and observe the effect does not appear to be linear, but falls off disproportionately with distance.
On the other hand, if the inverse square law does apply across the board, one would imagine the effective radius of something like a cell phone tower would be a lot less than what it actually is.
Why would there be a difference between electric, magnetic and "EMF"?
Can this be expalined in practical terms?
Barry Latham
*Power density* (watts per square meter) falls off as the square of the distance. *Electric field intensity* (volts per meter) falls of as the first power of the distance. *Magnetic field intensity* (amps per meter) also falls off as the first power of the distance.
The three are related by a similar law to Ohm's Law, by virtue of the "Intrinsic Impedance of Free Space", 120*pi ~ 377 ohms.
The square root in the expression relating voltage, or current, to power, accounts for the first order fall off in field strength.
The EMF induced in an antenna depends upon the electric field strength, and the "aperture" of the antenna, which is a function of its dimensions, relative to a wavelength.
--
"Design is the reverse of analysis"
(R.D. Middlebrook)
If the radiating source were a point (infinitely small), you'd measure inversely proportional field strengths from the source. When the source is of realizable size, you have to start the measurement several wave- lengths from the source (wavelenth: 0.125m / 5 inches at 2.4 GHz).
The near field is pretty complicated and dependent of the antenna type and immediate surroundings.
-- Tauno Voipio
Light is just really really really high frequency radio waves. ;-)
Go back to my post with the wiki link. The inverse square law is for a perfect point source. Antennas are not perfect point sources, at least not if you are close to the antenna. Your monopole will radiate nothing directly above the antenna. If the antenna is colinear, it will have a radiation pattern this is like a doughnut.
Professional photographers often use lighting that is planar rather than a point source. This is to get more of a linear fall off of light, and also to reduce shadows.
The inverse square law is just conservation of energy--once the beam starts to spread out, i.e. somewhere past about R ~ d**2/lambda, where d is the beam diameter, the beam width in meters is proportional to the distance R from the source. Width times height is therefore proportional to R**2, and energy conservation requires that the average flux density go as 1/R**2.
Since the power in an electromagnetic field goes as E**2, the E field falls off as 1/R in a propagating wave.
Static fields fall off faster. For instance an electric monopole (e.g. a classical point charge) has an E field that falls off as 1/R**2. A dipole, e.g. a small magnet, has a B that falls off as 1/R**3.
The confusion arises from conflating the behaviour of the E field from a point charge with the power density of a propagating wave. Both go as
1/R**2, but only the second is the "inverse square law".Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 hobbs at electrooptical dot net http://electrooptical.net
Near field vs. far field effects? That would be your d**2/lambda, right?
Thanks to everyone for their precise explanations. The distinctions are clear to me now.
Barry Latham
Right. The inverse square law is a far-field phenomenon.
Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 USA +1 845 480 2058 hobbs at electrooptical dot net http://electrooptical.net
It is important to keep the near field and far fields separate.
In the near field the electric and magnetic field behaves as you described.
In the far field, both the electric and magnetic behave like the inverse of distance, while the power density is inversely proportional to the square of distance. In the far field, the constant relationship between the electric and magnetic field gives the characteristic impedance of space as 120pi or 377 ohms.
For a simple monopole/dipole style radiator the border between near and far field is somewhere between lamda/2pi and lambda, depending on who you ask, however for a parabolic dish, the border can be at hundreds of wavelengths.
At mains 50/60 Hz frequency, practically anything is in the near field and the fields behave as you described.
However at 10 GHz satellite TV band, practically everything is in the far field and hence both electric and magnetic fields drop by the inverse of distance.
At frequencies in between, you have to check if you are in the near or far field.
EMF is affectionately known as "voltage" in these parts.
How far away were you standing? Three feet? a hundred yards? You really have to be in the far field to reach the inverse square behaviour. A lot of the field you measure close to the antenna is evanescent, i.e. it never goes anyplace.
At the level you appear to be wanting, the answer is "it depends." (Sorry about that, but it does.)
I recommend getting an introductory fields textbook for electrical engineers.
Maxwell's equations are very practical indeed if you're asking this sort of question and want a crisp answer.
Cheers
Phil Hobbs
-- Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC Optics, Electro-optics, Photonics, Analog Electronics 160 North State Road #203 Briarcliff Manor NY 10510 hobbs at electrooptical dot net http://electrooptical.net
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