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In a previous question, I asked if having 2 coils of wire with electricity flowing from one to the other is RF, and they replied that it was not Radio Frequency. I was puzzled because I thought this was wireless transmission of a changing magnetic field, which thus was RF.

I thought the rate of the AC current oscillation was the frequency of the RF (so a 60 hertz AC input I thought would give me a 60 hertz RF signal). Well, I was told NOT.

I would like to know the difference between electromagnetic radiation and just a plain, changing electromagnetic field.

skyler
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4 Answers4

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There are two reasons why your earlier question wasn't about radio. The first is, that radio officially goes from 3kHz to 300GHz. The second is, that a transformer is based on a different principle than radio waves. That second reason is what's your question is about: a transformer is based on electromagnetism, radio waves are based on electromagnetic radiation.

Understanding on this topic is really hard, and exists for many people on a lot assumptions. I'll try to give an easy explanation for a layman, for which you'll have to accept some more assumptions than for the detailed explanation below.

Layman explanation

As you know, a magnetic field means that some materials like metals are attracted by others. One can generate a magnetic field by letting an alternating current flow through a wire or coil. That is what happens in the primary coil of a transformer. The other way around, a change in a magnetic field will generate a current in a coil - that's what happens in the secondary coil. These properties of magnetic fields and current are called electromagnetic induction.

Electromagnetic radiation is a particular form of the electromagnetic field. In electromagnetic radiation, the magnetic field will create an electric field (just assume that), but further away from the conductor that began with making the electromagnetic field. The electric field will create a magnetic field, even further away, and so on. It just goes on and on, due to specific properties of the field. That's the key to electromagnetic radiation.

When you are testing with a transformer, the secondary coil exists inside one wavelength of the wave that is produced. This means that the current in the secondary coil does not exist because of electromagnetic radiation, but because of electromagnetic induction: the fields don't create each other.

You can only prove the existence of electromagnetic radiation by transporting waves over more than one wavelength - only then, you can be sure the fields create each other.

Detailed explanation

There is some confusion here, and the cause of that is that the theoretical principle behind radio waves, and radio frequency, don't necessarily go together. Take a look at the Radio Wikipedia:

Radio is the wireless transmission of signals through free space by electromagnetic radiation of a frequency significantly below that of visible light, in the radio frequency range, from about 30 kHz to 300 GHz. These waves are called radio waves. Electromagnetic radiation travels by means of oscillating electromagnetic fields that pass through the air and the vacuum of space.

Note: I believe the 30kHz minimum should be 3kHz (reference: here and here)

You can see that there might be other waves, based on the same principle and working the same way, with a frequency <3kHz or >300GHz, that are just therefore not part of "Radio". Those waves aren't radio waves and they aren't in the RF spectrum, but they are just the same, when you forget about the frequency.

But there's more! Radio waves are electromagnetic radiation. Electromagnetic radiation contains of two components, one electrical and one magnetic. These components create each other, as said above. The red magnetic field creates a blue electric field, which creates the next magnetic field, and so on.

enter image description here

From the Electromagnetic radiation Wikipedia:

Electromagnetic radiation is a particular form of the more general electromagnetic field (EM field), which is produced by moving charges. Electromagnetic radiation is associated with EM fields that are far enough away from the moving charges that produced them that absorption of the EM radiation no longer affects the behaviour of these moving charges.

What we were trying to do in your earlier question was really just picking up the weak magnetic field, because that's what a secondary coil does.

I guess you're now wondering: but does a transformer do electromagnetic radiation, or is it just a magnetic field? Let's have a look, with the Electromagnetic radiation Wikipedia:

... the electric and magnetic fields in EMR1 exist in a constant ratio of strengths to each other, and also to be found in phase ...

1: electromagnetic radiation, compared to the electromagnetic field - note by author

Think about the transformer. A magnetic field is generated when the current changes. Let's say we have a pure sine as the current, \$I(t)=sin(t)\cdot{}c\$. We can get the change of the current on a specific moment by taking the derivative of that sinus, which is the cosine, so: \$B(t)=cos(t)\cdot{}c\$. Now have a look at the functions \$I(t)\$ and \$B(t)\$, which should exist in "a constant ratio of strengths to each other" and in phase.

Note: the constant \$c\$ is because the formulas depend on other things as well, that are irrelevant now and constant in a specific situation

enter image description here

You can already see those functions aren't in phase. They aren't in a constant ratio to each other either. You can see that by plotting \$f(t)=\frac{sin(t)}{cos(t)}=tan(t)\$:

enter image description here

So no, a transformer does not radiate electromagnetic radiation. The waves aren't in a constant ratio of strength to each other, neither are they in phase. The tests you did with a transformer in your earlier question, were just based on a magnetic field.

This difference between picking up a magnetic field and magnetic radiation is known as the difference between near and far field.

Summary

There are two main reasons why your experiments weren't about radio. The first is that it just was the wrong frequency. The second is that a coil with an AC current does not provide electromagnetic radiation.

Reference

  • This is all very sad. Every other time a layman explanation is needed the answer will contain a gazillion of graphs with trigonometric functions. I strongly believe it shouldn't be this way. – sharptooth Apr 05 '13 at 08:12
  • @sharptooth you're right. I think skyler is able to understand this, but I've added some layman explanation at the top of the answer. –  Apr 05 '13 at 08:47
  • @OlinLathrop are you sure? Wikipedia and my physics book show different. The two fields should have a fixed ratio, I believe, which cannot happen when out of phase. One field is horizontal and the other vertical, there's a 90 degrees angle - the diagram is an attempt to show three dimensions. –  Apr 05 '13 at 12:30
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    @OlinLathrop good point. I follow your logic, but cannot find any diagram with the two fields out of phase. I don't have the knowledge to answer this, so I asked a question about it: http://electronics.stackexchange.com/q/64519/17592 –  Apr 05 '13 at 12:45
  • @Kortuk hmm, and what about [Maxwell's equations](https://en.wikipedia.org/wiki/Maxwell%27s_equations#Vacuum_equations.2C_electromagnetic_waves_and_speed_of_light)? Also, you might consider answering http://electronics.stackexchange.com/q/64519/17592 :) –  Apr 05 '13 at 13:59
  • @Kortuk I actually don't have enough knowledge to participate in this discussion, I just know Wikipedia says something different than you are saying. Just one note: are you sure you aren't misunderstanding [this way](http://electronics.stackexchange.com/a/64528/17592)? –  Apr 05 '13 at 14:07
  • @Kortuk looking forward to it :) it will be a tough job to see what answer I should accept in the end... I'll have to rely on the votes :) –  Apr 05 '13 at 14:15
  • @Olin : very good point, but I think the magnetic induced _voltage_ is in quadrature with the electric field : which means the _rate_of_change_ of field is in quadrature, or the field itself is in phase. –  Apr 05 '13 at 14:24
  • camilstaps, I actually did the math and did realize I mixxed up the time domain and position graphs. Woops! FDTD is the time domain simulation, my bad! – Kortuk Apr 05 '13 at 14:54
  • @OlinLathrop This question came up today and I noticed that both Brian's answer and your version of the fields in an EM wave is wrong. I thought this had been resolved on another question but maybe it needs to be resolved here too. E and H fields rise and fall in amplitude together with a ratio that is the resistance of free space (E/H). If they weren't in phase there would be a reactance (as opposed to resistance) of free space and this is wrong. – Andy aka Jul 01 '15 at 13:53
  • http://electronics.stackexchange.com/q/64519/17592 –  Jul 01 '15 at 13:55
  • @Andy: This is the question I remember it being resolved in. I have deleted my erroneous comments. – Olin Lathrop Jul 01 '15 at 21:35
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50/60Hz transformer coupling is not RF because it works by reactive field coupling in what is known as the "near field". Here's a picture from wikipedia on near and far fields: -

enter image description here

At about the wavelength of the antenna (or the frequency you are using to couple energy) the near field becomes the far field. The far field is regarded as "proper RF" and is able to propogate with it's radiation decreasing as the square of the distance.

Now consider a transformer at 50Hz - what is the wavelength - 6,000 km - will near field magnetic coupling work at even a 1,000m - no. It's not rf

Andy aka
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  • Interesting discussion. I picture a tuned loop antenna at some lower frequency like the 472kHz ham band. Far field then occurs at a large physical distance. At 100m and closer, you're in the fresnel or reactive zones, and a coil other antenna here would act like the secondary of a very loosely coupled transformer, but an observer miles (or thousands of miles) away would say you have an antenna rather than a transformer. For high frequencies this transition occurs so close to the antenna in human terms that it mostly doesn't matter. –  Jun 13 '19 at 18:06
  • At that frequency and with a modestly sized tuned loop antenna it produces very little E field at all and it makes a terrible transmit antenna but, because it works well at converting a mag field to voltage, it can be an excellent receiver. As a receiver it is in the far field unless it’s a tag type application. – Andy aka Jun 13 '19 at 20:08
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I think you already largely get it from your last sentence. A changing magentic field is not the same as radio.

Real radio is propagating energy. You can think of the energy as tied up in a specific dance between the E (electric) field and the B (magnetic) field. The two together oscillating in the right way cause the energy to propagate at the speed of light thru free space. Visible light is one example of this. It is a tiny part of the larger spectrum that goes down to (but not at) DC and up past gamma rays and cosmic rays. Common AM radio is at around 1 MHz, which has 300 meter wavelength. Common FM about 100 times higher frequency and therefore 100 times shorter wavelength, so 100 MHz and 3 meters. WiFi operates at about 2.4 GHz, which is 125 mm wavelength. There are microwaves of a few 10s of mm wavelength, the "terahertz" radiation used at airports to look under your clothes, infrared, visible light (roughly 500 nm), ultraviolet, xrays, gamma rays, and on. All these are exactly the same thing except for the frequency of oscillation. Since they all travel at the same speed of light in free space, you can also characterize them by their wavelength.

The E and B fields can each support non-propagating fields too. Wrap some wire around a steel bolt or ferrite rod, turn on the current, and you have a magnetic field. Ferromagnetic materials, like steel, will be attracted to this electromagnet. However, note that the energy of this field isn't being sent anywhere. The field exists around the electromagnet, and falls off rapidly with distance. You can even vary the field over time by driving the electromagnet with AC current, and then have another nearby electromagnet work in reverse to make a electric signal in its wires from the changing magnetic field. In fact, this is the basis for how transformers work. Yes, you can transfer signals and even significant power this way, but it is not radio. For example, there is no way to arrange a bunch of electromagnets to cause a beam of B field disturbance to be radiated in a particular direcation. You can shape the field locally, and the field does in theory extend to inifinity at the speed of light, but its not the same as sending a radio wave (or light beam, or radar beam, etc).

Just like you can make a B field device, you can also create static electric fields. As with the magnetic field from the electromagnet, this electric field can be detected locally and significant power transferred at close distances. But again, the energy of that field isn't being "sent" anywhere. To have the energy really radiate on its own requires just the right interaction between the B and E fields that we call electromagnetic radiation. We often get a little sloppy and refer to anything radio as "RF". RF really stands for radio frequency, but often we use it to just mean radio of any kind.

Olin Lathrop
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From Wiki:

Radio frequency (RF) is a rate of oscillation in the range of about 3 kHz to 300 GHz, which corresponds to the frequency of radio waves, and the alternating currents which carry radio signals.

Why 3KHz and not, say, 2.9 KHz? convention!

The fact is that electromagnetic radiation can occur at any frequency, e.g., the ELF spectrum is from 3 Hz to 300 Hz, but EM radiation is not necessarily RF.

Alfred Centauri
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