Timeline for If you run an electric current through a wire loop, do the accelerated charges radiate?
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| Oct 22, 2017 at 12:42 | history | edited | stafusa | CC BY-SA 3.0 |
Added link to relevant calculation.
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| Aug 15, 2011 at 20:47 | comment | added | user1631 | @endolith, yes even in that situation it will cancel. Keep in mind the cancellation is not instantaneous; the radiation from the left end emitted at time $t_1$ will cancel against the radiation from the right end emitted at a different time $t_2$, depending on the position where the field is being observed. | |
| Aug 15, 2011 at 19:55 | comment | added | endolith |
But for any loop of any shape where current is constant, the radiation will cancel? Even a very long narrow loop where the acceleration only happens at the ends? ⊂=============⊃
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| Aug 15, 2011 at 16:44 | comment | added | user1631 | @endolith, depends on whether the electrons are spaced closely enough that we may treat them as a continuum of charge and current. For a discrete set of electrons the cancellation won't be perfect. | |
| Aug 15, 2011 at 4:59 | comment | added | endolith | How can the radiation cancel perfectly if the charged particles are not in the same place? A ring of electrons traveling in a circle at relativistic speeds won't radiate? Really? | |
| Aug 10, 2011 at 20:56 | comment | added | user4552 | Without appealing to Green's functions and other esoterica, an easy way to see that it doesn't radiate is that since there is no time-variation in the currents, there is no parameter that would set the frequency of the waves. You might think that the period would equal the time required for an electron to go once around the circuit, but that would be different for different types of wire because of the different drift velocities. Maxwell's equations don't have velocity in them, only current. (The time to drift around the circuit is also ~1 s, so the frequency would be ~1 Hz!) | |
| Aug 10, 2011 at 20:35 | comment | added | drlemon | A "shut up and calculate" way to prove that a ring with current produces no electromagnetic waves is to solve Maxwell's equations for this configuration of electric currents. Thankfully, the solution for a loop with current in infinite space is readily available. It can be found in any physics textbook under "magnetostatics", along with a sketch of magnetic field lines curling around the wire. This solution is unique. No time varying magnetic or electric fields in this solution. Therefore, no electromagnetic waves. | |
| Aug 10, 2011 at 20:31 | comment | added | Revo | Yeah I found similar stuff on jackson's chapter 14 | |
| Aug 10, 2011 at 20:20 | comment | added | BebopButUnsteady | The waves that would be emitted by each electron destructively interfere with each other and produce no net radiation. If you want to learn about how to calculate this sort of thing the standard reference is Jackson, Classical Electrodynamics. To see immediately that there is no radiation, note that both the current and charge density do not depend on time and hence Maxwell's equation must have a solution that is time independent. No time dependence no radiation. You must have a time varying current and/or charge density to produce radiation. | |
| Aug 10, 2011 at 20:17 | comment | added | user1631 | I don't follow, why do you think a superconducting ring current would fade away? It cannot because magnetic flux through the ring cannot escape. There will be no Johnson noise for the superconducting ring if that's what you were after. | |
| Aug 10, 2011 at 20:11 | comment | added | Revo | Hmmm, that means a current in a superconducting ring will fade away by radiation? | |
| Aug 10, 2011 at 19:58 | comment | added | user1631 | Well, you can formally prove it using vector potentials and Green functions, if you are comfortable with those. However for a more intuitive take, consider a single electron going around in a loop. His total vector acceleration averaged over the loop must be zero. Now imagine many identical electrons spread around the loop; the total vector acceleration for all electrons at any instant must be the same as the loop-average for a single electron, i.e. zero. Since radiation is proportional the charge times acceleration, they cancel. | |
| Aug 10, 2011 at 19:20 | comment | added | Revo | I do not understand how individual radiations of many electrons would cancel each other. How can we show this using Maxwell's equations, is this a standard calculation or something? Pointing out some textbook in which this is explained (or even listed as an exercise) would be greatly appreciated and helpful. | |
| Aug 10, 2011 at 19:13 | history | answered | user1631 | CC BY-SA 3.0 |