The image is my visualization of drift velocity and electromagnetic (EM) propagation of charge wave in a closed circuit. The slow drift velocity of the electrons follows the path of the circuit (a circle wire). Does the the EM wave follow the same path of that of the drift velocity?

Since textbooks and online resources I found offer no understandable description/differentiation, I assume they take the same path (of the circuit wire).

But I cannot understand why:

(1) If the wave is induced by and propagation from the voltage source (battery), then it should take the vector path of the magnetic field created by the battery, instead of the circuit path.

(2) If the electromagnetic wave is caused by some ballistic effect (electron “pressuring” the next electron like water molecules in a tube), then shouldn’t the wave left tangent to the wire and shoot to outer space? (similar in sound wave, when enter image description hereI shout, the sound wave goes in all direction but not through a specific path of the target person). But we know there is magnetic field caused by current is wrapping around the wire; so what is confining the wave to go into wire path?


Perhaps I should elaborate that I am not asking about the radiation or antenna effect. I am curious on the actual "electricity/energy/signal" current (not the drift current by electrons) going in the path of the circuit wire instead of radiating outwards. I have amend the picture so it looks more like a current going through a bulb rather than looking like an antenna. (sorry for the bad drawing..I did my best job)

enter image description here


To rephrase my question with a better picture, when the battery apply a electric potential to an closed circuit wire, there are two currents - the very slow drift current from electrons, and the current in form of EM wave traveling near the speed of light. What is causing the EM wave the bend and turn along the wire?

enter image description here

  • $\begingroup$ Are the red balls electrons? If they are, then the circuit path is wrong here. $\endgroup$
    – user36790
    Commented Nov 28, 2015 at 13:09
  • $\begingroup$ yes the red balls are electrons. Thank you for pointing out the polarity I have change it. $\endgroup$
    – KMC
    Commented Nov 28, 2015 at 13:49

4 Answers 4


(1) If the wave is induced by and propagation from the voltage source (battery), then it should take the vector path of the magnetic field created by the battery, instead of the circuit path.

Yes, and the magnetic field follows the circuit path, but remains outside the wires. The electric current in a wire is causing a magnetic field which is located well outside the wire ...and the flow of electrical energy is all found in the field outside, and not inside the wire.

In other words, when electrons are flowing, there's no energy inside electrons, instead it's all in empty space; out in the electrons' surrounding fields.

Oh, also note that the battery pulls electrons in from the positive wire, which sends a wave out through that wire. In other words, the battery sends energy out along both wires, and not just along the negative wire as in your diagram. And, with hoses and water, you can send out waves by sucking, as well as by blowing. (Both wires were already full of charge-carriers, of course. The electrons are the "medium" for the fast waves.)

(2) If the electromagnetic wave is caused by some ballistic effect (electron “pressuring” the next electron like water molecules in a tube...

Not a ballistic effect, since this all works the same with zero-ohm perfectly conductive wires and no forces having to shove against each electron in the chain.

Yes, the forces in a water hose are passing through the water, but circuits are different. The "pressure" in circuits involves capacitance; it involves the e-field or "voltage-field" located in the space between the wires. Briefly, the "pressure" is created when the battery takes electrons from one wire, leaving positive protons behind. It forces the electrons into the other wire, charging it negative. The positive and negative wires are a capacitor, with the forces on the electrons only appearing in space, in the "dielectric." The forces don't affect the connecting wires, instead they drive the filament's electrons to flow against strong resistance there.

The full answer to your questions is in the "nearfield" part of the topic of EM theory, and also in antenna theory in the limit of low frequency and where the metal structures << wavelength. In other words, we can analyze a flashlight circuit as a loop-antenna, but then run it at zero or low frequency, where the radiation of EM waves goes to zero.

Just for your information, here's the AC magnetic field of a (tiny) coil at low frequency (no EM waves escaping):

MIT Teal: osc field v2

And here's the same coil running at higher frequency, where the magnetic field still expands and collapses, but also the EM waves are flying off:

MIT Teal: osc field intermediate v2

For DC battery circuits the EM radiation from the circuit is effectively zero. There are strong fields, but they aren't flying away as waves. The Poynting-vector energy-flow lines are out in space, but they must curve around to follow the circuit. The Poynting-vector lines are not allowed to point outwards into empty space as in your first diagram, since any changes in the fields occur too slowly, and the circuit is too small compared to the wavelength of any EM waves.

Rigorously we can view a flashlight circuit as a one-turn coil with its surrounding magnetic field. But in coils at DC, energy is stored in the magnetic field, but no energy is flowing...

b-field of flashlight circuit

To this model we include capacitance: the capacitor formed by the two charged wires connecting the battery and the bulb-filament. In a pure capacitor at DC, energy is stored, but no energy is flowing. Sketch in the e-field between these positive and negative wires:

e-field of flashlight circuit

Next: whenever electrical energy IS flowing, (when volts and amps are both present,) then both fields appear. Together these two fields show a cross-section of the EM-energy flowing from the battery to the resistor:

enter image description here

There's a common widespread misconception associated with this diagram above. Many students assume that it only applies to extremely high frequencies; to the RF wattage traveling along 2-wire transmission lines. Wrong. There's no lower limit to the frequencies where this applies. It works for speaker cables in your stereo, for telephone and telegraph lines, for 60Hz power lines, and works all the way down to DC with flashlights. (In other words, Maxwell applies to circuits regardless of frequency. Doh! Who'da thought!)

If we next look at these field-lines from another angle, and also sketch in the Poynting-vector energy-flow lines which connect between the "squares" of e and b flux, then the energy-flow looks something like this below:

energy flow of a DC circuit

It all connects together and makes sense: electrical energy spews out of the battery, runs roughly parallel to the connecting wires, then dives into the resistor/bulb-filament; entering the tungsten surface at right angles. At the same time the currents in the wires are connected to or "causing" the magnetic field, and the voltage across the wires is connected to or "causes" the electric field.

There's one big problem with all of this: it's only ever presented to 4th-year engineering students! Them, and at graduate level physics, but not to undergrads or to high school students. It's not found in technician course materials (except perhaps in radio tech courses, and then never applied to DC circuits, phone cables, or 60Hz power lines.) Probably this is done because the math homework would be far beyond a beginning student, and any textbook chapters about the above topics MUST ALWAYS HAVE math homework, right?, right? :) Forget that! We can just explain it all verbally, and draw pictures as above. No math allowed. Don't aim it at electrical engineers, instead do it for "Einstein's Audience," where Einstein doesn't truely understand a topic unless he can explain it to his grandmother.

  • $\begingroup$ Ha! You can see the answers here ;P $\endgroup$
    – user36790
    Commented Apr 7, 2016 at 2:34
  • $\begingroup$ This is one of the best answers on this site -- thank you! Is it correct to say that there are two different perspectives on all classical mechanics/electrodynamics problems: the 'fields' are the objects of consideration transferring energy, vs the 'particles' are the objects of consideration transferring energy? In which case there must be a generalization of the Poynting vector for classical mechanics too (e.g. energy flow through a 'strain gradient' of a gear train)? $\endgroup$
    – Matt
    Commented Aug 22, 2021 at 13:45
  • 1
    $\begingroup$ @Matt you're close: it's a waves-versus-medium duality, like sound traveling through wooden rods, or passing along "Newton's Cradle's" balls. With metal conductors, the electron-filled wires are the "medium" which guides the fast-moving electrical energy, where this energy is actually made of EM waves (and applies even at DC.) Quantum physics involves the fields and the energy-flow being quantized ...meaning that no electric companies are selling any electrons, instead they're selling us photons. Ours is a "radio-powered" civilization, with megawatt VLF waves racing along 3-wire waveguides. $\endgroup$
    – wbeaty
    Commented Aug 27, 2021 at 6:50
  • $\begingroup$ It's because it's not relevant to engineers and they just view those circuits as lumped circuits when they are all of course behaving like distributed circuits. A question I've always had is, let's say you have a short circuit, the EM waves apparently bounce off the open circuit as if it has a large lumped circuit impedance, back to the source again, but surely it would actually bounce off the characteristic impedance of free space. I'm not sure how the open circuit impedance ends up arising as the reflections settle. Similarly with a short circuit. $\endgroup$ Commented Dec 9, 2021 at 11:50
  • $\begingroup$ @LewisKelsey if wires are too close together, yet open circuit, they can't act as antennas, can't "see" the Z of free space. If the ends of the open circuit are an inch apart, they'll reflect AC 12KHz bursts differently than 12GHz bursts. So, the cut end of coax might look like a 200GHz antenna. Send it a perfect DC step, and it will shave off the corners before reflecting the DC pulse, because some of the picosecond energy didn't reflect, and flew out of the cut coax. Similar: in acoustics, use a horn to match the Z of empty air to the Z of a metal pipe. Antennas are z-match xfrmrs. $\endgroup$
    – wbeaty
    Commented Dec 9, 2021 at 14:37

The short answer to your question is that EM waves travel in the same direction as the wire and current, guided by two opposite conductors, and flow into any device that consumes power (has a voltage drop across it and current flow through it). So for your light bulb circuit, the wave flows from the battery to the lightbulb between the wires. Here's an image that illustrates this idea.

Power flow in a simple circuit

There is also the tendency for the wave to "escape" the circuit, especially at bends, or certain structures such as antennas, which are designed to maximize the escape of radiation from the circuit. This is all but negligible at low frequencies, however.

Long answer: read this page or this page to get a better view of how and why the wave travels the way it does. One thing to keep in mind is that current and charge descriptions of circuits are completely equivalent to EM wave descriptions, and it's thus possible to describe the same circuit either way. So when some of the author's describe the energy as "actually" travelling outside the circuit, that's somewhat misleading.

  • $\begingroup$ "current and charge descriptions of circuits are completely equivalent to EM wave descriptions" @David ...well, except that the circuit description usually doesn't explain energy location, and if they try, they get it wrong; ascribing it to currents rather than to [b-field X e-field] cross product. Very similar: in coils, where does circuit theory say the energy is? Inside the wire? Or outside, in the magnetic field? And is a capacitor's energy on the plate surfaces, or in the empty gap? Circuit explanations are wrong because they don't apply these descriptions to the connecting wires. $\endgroup$
    – wbeaty
    Commented Apr 6, 2016 at 1:01
  • 1
    $\begingroup$ @wbeaty You're making an argument of circuit descriptions vs field descriptions. I don't address this argument at all. What I'm saying is that current and charge descriptions are equivalent to EM wave descriptions. The main point I'm hinting at is the idea that energy transferred through work done on charges is equivalent to the Poynting vector description. These ideas don't address where the energy is stored, and are consistent with the idea of energy being stored in the fields. I agree that anyone who ascribes most of the energy to kinetic energy in the currents is very wrong. $\endgroup$
    – David
    Commented Apr 6, 2016 at 23:45
  • $\begingroup$ You say "So when some of the author's describe the energy as "actually" travelling outside the circuit, that's somewhat misleading." Actually they're being correct, not misleading. As with coax and 2-wire xmission lines, the energy really truely does travel in the space, and not inside the metal. The field-based explanations put the location of energy in the correct place: it's in the fields, not in the wires. Circuit-based explanations put the energy in the voltage and current ... I is actually amp-turns, so amps "are" magnetism, as voltage "is" the e-fields. $\endgroup$
    – wbeaty
    Commented Apr 7, 2016 at 5:34
  • $\begingroup$ @wbeaty I agree about the energy being transferred through the fields. What I think is misleading is that when the authors write that the energy actually travels through the fields, it implies that the previous description of energy transfer as charges moving through a potential is incorrect. Both explanations are correct, and mathematically equivalent, which is what I was trying to emphasize in my answer. $\endgroup$
    – David
    Commented Apr 7, 2016 at 15:17

Does the the EM wave follow the same path of that of the drift velocity?

I'll assume you're asking about a case where the wire loop is large enough to radiate effectively. Meanining you're asking about a resonant loop antenna, with a circumference approximately equal to the wavelength of the signal being applied.

In this case, no, the EM wave (or at least the E field) doesn't follow the path of the drift velocity. Why? The E field is what is causing the charge carriers to accelerate. So the locations with highest E-field are the locations with the greatest acceleration of the carriers. Since we are talking about a sinusoidal oscillation, we know the locations with the largest acceleration will actually be the places where the velocity is zero. And the carrier velocity must be greatest at the locations with zero acceleration (which must be the locations with zero E field).

The maximum of the H field, on the other hand, must be maximum near the locations where the carrier velocity is maximum, according to Ampere's Law.

If the wave is induced by and propagation from the voltage source (battery), then it should take the vector path of the magnetic field created by the battery, instead of the circuit path.

If you drive the antenna with a dc source like a battery, then no EM wave will be generated, because the battery only produces a dc current. The magnetic field will circle through the loop in the opposite direction to how you have drawn it, following the right-hand rule. The E-field will point from the higher-potential parts of the circuit to the lower-potential parts of the circuit, but would probably have to be calculated using an electrostatics simulation.

If the electromagnetic wave is caused by some ballistic effect (electron “pressuring” the next electron like water molecules in a tube), then shouldn’t the wave left tangent to the wire and shoot to outer space?

I'll go back to discussing an antenna with an ac stimulus.

Remember that the EM waves generated by each infinitesimal element of the antenna will interfere with each other.

The radiation we see in the far field will be the superposition of the waves generated by all the antenna elements working together.

As the voltage signal propagates around the loop, the E-field will at some times point from the the "top" of the antenna to the "bottom" and at other times from the "left side" to the "right side". As the signal oscillates, these oscillating fields will create an EM wave with an omnidirectional radiation pattern; that is, it will radiate nearly uniformly in all directions in the plane of the loop.

what is confining the wave to go into wire path?

The wave does not follow the wire path, it radiates outward. The current is confined to follow the wire path because air does not contain sufficient free charge carriers to carry the current.

  • $\begingroup$ Thank for the detail. But I am interested in the current of the wire, not the antenna radiation. The current is itself charges's EM wave not the drift current of the electron, and I do not understand how the "electricity" or signal or energy is delivered along the path of the circuit. $\endgroup$
    – KMC
    Commented Nov 29, 2015 at 4:13
  • $\begingroup$ @KMC, you should make your question more clear. If you ask about the "EM wave" you're going to get answers about electromagnetic waves. If you want to know how the carriers in the wire are moving, you need to ask about that instead. $\endgroup$
    – The Photon
    Commented Nov 29, 2015 at 4:15
  • $\begingroup$ but isn't the charge carrier propagate as wave (Electromagnetic wave) that how signal transfer 2/3 in speed of light? Please let me know if my description is incorrect and I will amend it further. Thank you. $\endgroup$
    – KMC
    Commented Nov 29, 2015 at 4:21
  • $\begingroup$ At a QM level, the electron is described by a wave. But this is a probability wave, not an EM wave. $\endgroup$
    – The Photon
    Commented Nov 29, 2015 at 4:22
  • $\begingroup$ But some wave is going in high speed along the wire from the battery and along the circuit wire to energize the light bulb, and certainly is not caused by the drift velocity of electron. If it's not EM wave and what is? The drift velocity goes along the path of the wire wire because this is where the electron resides, but what I do not understand is why the wave also falls along the same path. What is directing "this" wave (or current)? $\endgroup$
    – KMC
    Commented Nov 29, 2015 at 4:57

Current in the wire is because at each point there is a net Electric field along the wire. This E is because of the variation of surface charge density on the surface of wire as we move from Anode to Cathode. This E is continues and of same strength at a given instant of time at any point inside the wire under consideration. The variation of E(t) depends upon the type of source. For a dc source E(t) is constant and given by V/l(V=voltage and l=length of wire). For an AC source $E(t)=E_0 sin(\omega t)$ at any point inside the wire.

Now as per the Drude Model the electrons move randomly in all directions and collide with heavy nucleus'. On an average it takes $\tau$ time for every electron to collide again. In between this time the electron accelerates and hence must radiate EM wave. This EM wave may be trapped inside(absorbed) wire and appear as heat. Or it may appear as light coming out of the wire. What would happen mainly depends upon the nature of material. Some material like tungsten used in electric bulbs glow, some material like aluminum heat up.

That being said. There might be more details involved. E.g. I found a better model than Drude model here, Free Electron Model. Quantum Machenics is beyond my competence. So I could help only with Classical Physics. Feel free to ask in comments.


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