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If a charge travels in a circle it must accelerate, thereby producing EM. However, a wire in a circular loop is analogous to many charges moving in a circle. So, why don't circular currents produce EM? (I have not found any evidence that circular currents can produce EM).

A similar question is: Why doesn't alternating current produce light while a vibrating single particle with a charge will. However, my question asks about circular wires and direct current. Thanks.

Community
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    Don't forget that wires do produce EM radiation (though not through synchrotron radiation, as far as I'm aware) - that's why we have all those complicated rules for how to build electronic equipment so that it doesn't produce radiation (and interfere with other electronic equipment). – Luaan Jul 22 '16 at 12:31
  • Who changed the title? – OrangeDog Jul 22 '16 at 20:47

4 Answers4

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Circular currents do produce EM, and indeed this is exactly how X-rays are produced by synchotrons such as the (sadly now defunct) synchotron radiation source at Daresbury. In this case the current is flowing in a vacuum not in a wire, but the principle is the same.

Current flowing in loops of wire don't produce radiation in everyday life because the acceleration is so small. The electrons are moving at the drift velocity, which is only around a metre per second, so the amount of radiation released is immeasurably small. Synchotrons produce radiation because the electrons are moving at almost the speed of light.

John Rennie
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    I think it's worth adding that perfectly steady circular currents don't radiate - the (negligible) radiation only occurs because the current isn't steady at the microscopic level. – tparker Jul 22 '16 at 06:33
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    @tparker: Yes, I've upvoted your answer saying that :-) – John Rennie Jul 22 '16 at 06:34
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    "the amount of radiation released is immeasurably small" This is unfortunately completely untrue. The energy of most everyday drift velocity corresponds to radio waves. This is how typically make radio waves, using a wire with a current, we call these things antennas. – Aron Jul 22 '16 at 13:57
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    @Aron : Actually, antenna's work far, far better when you drive them with AC. As the question already points out, acceleration produces radiation, and the frequent flipping of the direction of AC current implies a high acceleration. – MSalters Jul 22 '16 at 14:38
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    @Aron: the emitted radiation is proportional to acceleration not velocity. However for a DC current flowing in a loop of radius $r$ the acceln is $a = v^2/r$, where $v$ is the drift velocity. Hence low drift velocity does mean low acceln and therefore low radiation. For an AC current with frequency $\omega$ the acceln is proportional to $\omega^2$, hence at RF frequencies $a$ and therefore the radiation can be very high regardless of the drift velocity. – John Rennie Jul 22 '16 at 15:06
  • @user104372: this is the sort of thing to discuss in the chat room – John Rennie Jul 23 '16 at 05:59
  • @user104372: see Why don't electrons crash into the nuclei they “orbit”? – John Rennie Jul 23 '16 at 06:36
  • @user104372 No, the reason that electron orbits are stable is a purely quantum-mechanical effect that cannot be explained by classical electromagnetism. See the question that John Rennie linked to. – tparker Jul 23 '16 at 07:22
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    This is wrong, I can't believe it has been upvoted so much. In circular solid, most material has crystal form which means there are electrons moving in 90 degree change of direction or at the very least 45. This is enough to cause a consistent pattern of detectable EM wave. They are right about drift velocity, but electron can produce burst of measurable EM each time they shift into different atom/molecule. Electron might not need to at light speed. In fact even in antenna, where it is not circular, you still have drift speed yet you get detectable EM. –  Jul 26 '16 at 08:38
  • Hi, John, I got a reply from Diamond House, if you are interested I can forward it to any email address, you might help me frame a request for relevant data. –  Aug 16 '16 at 05:32
  • @tparker I don't think one needs to appeal to quantum fields to explain the stability of electron orbit. Non-relativistic quantum mechanics for electrons in a classical EM field is enough. See https://physics.stackexchange.com/questions/70200/why-electrons-cant-radiate-in-their-atoms-orbits/70201#70201 – Diego Mazón Feb 25 '21 at 18:43
  • @DiegoMazón I agree, but I never did appeal to quantum fields or relativity! – tparker Feb 26 '21 at 00:23
  • @JohnRennie Is the radiation level given by the drift velocity or the Fermi velocity? I'm not sure of the answer, but my sense is that the Fermi "velocity" (actually a speed) gives the relevant (scalar) speed scale, while the drift velocity gives the relevant (vector) velocity scale, and the emitted radiation is determined by the instantaneous speed of the electrons, and not by their net velocity. I'm not at all sure about that though. – tparker Feb 26 '21 at 00:28
  • @tparker Ok. As you said that the lack of radiation (stability) could "not be explained by classical EM", I wanted to clarify that quantum particle in a classical EM field is enough (no need of QED). We agree then. – Diego Mazón Feb 26 '21 at 08:34
  • Since in electrodynamics permanent magnets are described as loops of current, would they also radiate? – Juan Perez Nov 25 '21 at 02:48
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If you consider that electric current is actually the flow of individual charged electrons, then as John Rennie pointed out, the radiation exists but is negligibly small. But if you were to imagine breaking the current into more and more point particles with less and less charge while holding the linear charge density $\lambda$ fixed, then the radiation would decrease more and more, because the radiation per particle decreases as $q^2$ while the number of charges only increases as $1/q$. So in the actual continuum limit where the current becomes perfectly steady, the radiation actually vanishes completely.

tparker
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    This puzzles me because it contradicts the central tenet of "accelerated charges radiate". Is it because there are no individual charges any more? Do we have a kind of resonance, like an oscillating electron in an "orbit" around a nucleus? Or is it just a theoretical reasoning assuming we had infinitesimally small charges (which we don't)? – Peter - Reinstate Monica Jul 22 '16 at 15:18
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    @PeterA.Schneider Accelerated point charges radiate. Since a steady current isn't made up of point charges, this rule doesn't apply to curved steady currents. – tparker Jul 22 '16 at 16:27
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    So is your scenario a hypothetical one without electrons, or is your scenario a (possibly idealized) real-world scenario in which the electrons (or other charged particles) "hybridize" or whatever and do not constitute single charges any longer? – Peter - Reinstate Monica Jul 22 '16 at 16:45
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    @PeterA.Schneider It's just a hypothetical scenario without electrons - a formal solution to Maxwell's equations. In the real world charge is discrete, so there's no such thing as a steady current, and curved currents will always radiate slightly. – tparker Jul 22 '16 at 18:32
  • @tparker In the real world, electrons don't follow trajectories either. An electron in an s orbital doesn't radiate. A quantum charged particle in an stationary state does not radiate https://physics.stackexchange.com/questions/70200/why-electrons-cant-radiate-in-their-atoms-orbits/70201#70201 – Diego Mazón Feb 25 '21 at 18:38
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    On its face, if seems like you could equally well argue that if you divide a single oscillating charge into $n$ superimposed sub-charges, then this would decrease the radiation by a factor of $n$. But this is of course incorrect; the radiation from $n$ charges is not necessarily $n$ times the radiation from one charge if they are oscillating coherently. So to make your argument more accurate, you'd have to address this effect for a ring of charge. – Michael Seifert Mar 17 '23 at 02:54
  • @MichaelSeifert That's a really great point. If you have $N$ point charges of charge $q$ moving on a ring of radius $R$ with a uniform angular velocity $\omega$, then the time-averaged current around the ring is $I = Nq\omega/(2\pi)$. So one way to take the continuum limit is to take $N \to \infty$ and $q \to 0$ while holding their product - the total charge of the ring - and $\omega$ constant, as I suggested. But then you get that tricky coherence question. – tparker Mar 17 '23 at 04:38
  • @MichaelSiefert Another approach would be to take $N \to \infty$ and $\omega \to 0$ while holding their product and $q$ constant. This corresponds to creating more and more point charges of charge $q$, so that the total charge of the ring grows unboundedly, while having them orbit slower and slower in order to keep the current the same. While less obviously equivalent, I think it does actually approach the same continuum limit, as long as we also increase a fixed opposite-sign background charge on the ring to keep the total charge constant. – tparker Mar 17 '23 at 04:42
  • @MichaelSeifert But in this case, the particles are moving arbitarily slowly, so they will emit EM waves of wavelength much longer than the ring's diameter. I want to wave my hands and say that in this nonrelativistic limit, things are basically quasistatic so radiation effects clearly go away. There's probably some more rigorous way to formalize that intuition. – tparker Mar 17 '23 at 04:46
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To back up John Rennie's answer, consider the Bremsstrahlung formula for velocity perpendicular to acceleration: $P= {{q^2a^2\gamma^4}\over{6\pi\epsilon_0c^3}}$. For all practical purposes $\gamma=1$, so we can simplify this to $P\approx ({q \over \mathrm{C}})^2 ({a\over \mathrm{m/s^2}})^2 {1\over{18.85\times 8.85\times 10^{-12}\times 2.7\times 10^{25}}}\mathrm{W}\approx ({q \over \mathrm{C}})^2 ({a\over \mathrm{m/s^2}})^2\times2.22\times 10^{-16}\mathrm{W}$.

leftaroundabout
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CuriousOne
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    I know what 'simplify' means in a mathematical sense, but as a layman it made me laugh... – Lyall Jul 22 '16 at 08:16
  • @Lyall: I was kind of cringing when I wrote it... it doesn't really do much, does it? :-) – CuriousOne Jul 22 '16 at 08:18
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    It doesn't help that you've got the units smeared out; there's a Watt at the end but a $m^2s^{-4}$ below the a. Considering that the Watt is a $kg m^2 s^{-3}$, that actually simplifies to $kg s$. I.e. $$(qa/C)^2 × 2.22×10^{-16}kg s$$ – MSalters Jul 22 '16 at 14:42
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    Can you give a typical result for typical everyday values (say, 10 amps, 1 meter radius)? – Peter - Reinstate Monica Jul 22 '16 at 15:15
  • @PeterA.Schneider: According to http://hyperphysics.phy-astr.gsu.edu/hbase/electric/miccur.html, at 10A in a 1mm wire the drift velocity is approx. 1mm/s, i.e. it would take the electrons over 6000s (almost 2 hours) to go around the loop even once. The moving charge would then be roughly 60,000C, but the acceleration would only be $1e-6/6m/s^2$, i.e. the first two factors give us a numerical value of $(60000*1e-6/6000)^2=10^{-4}$. Together with the numerical factor we end up with something on the order of $10^{-20W}$, which is utterly too small to be measured. – CuriousOne Jul 22 '16 at 21:07
  • @MSalters: Both the charge and the acceleration have units and they are being compensated in the first two terms. Having said this, please check the formula, I might have gotten it wrong. I was always poor at evaluating expressions. :-) – CuriousOne Jul 22 '16 at 21:08
  • @CuriousOne: Ow, wait a minute - that C doesn't happen to be a Coulomb? (i.e. an Ampere-second). – MSalters Jul 22 '16 at 21:19
  • @MSalters: Yes, it does. I am simply using a convention more often seen in engineering, I believe, where one takes the units out, first, then uses the unit free numerical value to do the calculation and then adds the unit of the result back at the end. Like I said, if you are concerned that it's not right, please check it, I don't have any illusions of grandeur about being correct all the time (I have plenty of experimental data that invalidates such a hypothesis about myself). :-) – CuriousOne Jul 22 '16 at 21:22
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According to Maxwell equations, steady currents and steady charge density won't produce EM waves. So if you have a steady current loop, you can calculate its magnetic field just by the Biot-Savart law, which gives an steady magnetic field in space. If a current loop would radiate energy, then it would be impossible to produce persistent currents in superconductor loops which are actually experimentally observed.(Actually there is also the quantum mechanics in play, and using "only" the classical Maxwell theory isn't completely correct, but it gives you the idea)

But to explain your point about a loop current being charges moving in circle, you are partially correct. If you break the current into tiny elements and calculate the electromagnetic field of each individual element, surely you see that the EM field of each individual element is a propagating magnetic field, but when you sum the EM fields of various elements together using superposition principle, the resulting "net electro magnetic field" would be the same as the one that is given by the Biot-Savart law. (Somehow the same that happen in interference experiments.)

But there is two points to make; Although the steady current and zero charge density is theoretically presumable, this could not be the case in the real world. Since the currents are moving electrons you expect that if you zoom in enough, you see single electrons with some gaps between them, So it seems that in that scale the charge density isn't steady in time and the continuous approximation of charge density breaks in that scale.(Although the currents are produced by electrons in the Bloch states which are not localized in space, So the moving single electrons picture isn't entirely correct either). And if you have a time-varying charge density, you would get EM radiation. Actually this is the situation in the synchrotrons. There you don't have an steady uniform current around the loop, but a bunch of individual electrons that are circulating the loop and you can see the time varying charge density easily. You just can't approximate the electrons running through the loop by a steady current.

Seyed
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