Coil that conducts direct current

[south]

TL;DR

Does that coil radiate?

I assume that electric current is made up of moving electric charges. If this hypothesis is correct, then the current that flows through a coil is made up of accelerated charges.

I am not interested in evaluating the situation quantitatively. My question is conceptual. Does a coil that conducts direct current radiate electromagnetic waves?

 

[Dale]

I am not interested in evaluating the situation quantitatively. ... Does a coil that conducts direct current radiate electromagnetic waves?

No. I would explain further but you don’t want any math.

 

[south]

Hi Dave. Thanks for the reply.
My wording was poor.
I meant to point out that in practical terms, even if the coil radiated, it would emit a negligible amount of energy and we would not take the effect into account.
In purely theoretical terms it does not matter how much energy it emits, if any. I just want to ask, does it emit or not?

 

[renormalize]

I just want to ask, does it emit or not?

No it does not. A single charge traveling in a loop does radiate due to it's centripetal acceleration. But a vast number of closely-spaced charges moving in a loop (i.e., a macroscopic electric current) does not radiate, due to cancelations between the charges. (This follows from the math that you don't want to see!)

 

[south]

I understand what you have expressed regarding the set of charges.
Did I say that I do not want to see math?
Another flaw in my writing. I did not intend to say that.
I will gladly read the math.

 

[renormalize]

I will gladly read the math.

The answer to your question follows from solving this textbook problem from J.D. Jackson, Classical Electrodynamics (2nd ed.) involving Liénard-Wiechert potentials arising from $N$ equally-spaced charges:

You can likely find the full solution posted somewhere online.

 

[Dale]

OK, so if you look at Jefimenko's equations we see:
$$ \mathbf{E}(\mathbf{r}, t) = \frac{1}{4 \pi \varepsilon_0} \int \left[\frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^3}\rho(\mathbf{r}', t_r) + \frac{\mathbf{r}-\mathbf{r}'}{|\mathbf{r}-\mathbf{r}'|^2}\frac{1}{c}\frac{\partial \rho(\mathbf{r}', t_r)}{\partial t} - \frac{1}{|\mathbf{r}-\mathbf{r}'|}\frac{1}{c^2}\frac{\partial \mathbf{J}(\mathbf{r}', t_r)}{\partial t} \right] dV' $$

The first two terms decay as $r^{-3}$ and $r^{-2}$ respectively, so they represent the near field, not radiation. The radiation term is the third one, the one that decays as $r^{-1}$:
$$ - \frac{1}{|\mathbf{r}-\mathbf{r}'|}\frac{1}{c^2}\frac{\partial \mathbf{J}(\mathbf{r}', t_r)}{\partial t} $$

Notice that this term is proportional to $\frac{\partial \mathbf{J}(\mathbf{r}', t_r)}{\partial t}$. For a DC current this is, by definition, 0. So, the radiation is 0.

 

[south]

Thank you very much. Best regards.

 

[sophiecentaur]

A single charge traveling in a loop

The charge carriers are electrons. Electrons in a metal travel with an average speed of a couple of mm/s and their mass is very very small. In addition to the cancellation of induced fields, the basic amount of kinetic energy involved is very small so the field energy is not significant.
The classical instance of radiating em waves and the suggested radiation of orbiting electrons means that the first models of the atom (with electrons spiralling inwards) was a flawed model until quantum theory came to the rescue.