Coil that conducts direct current

[south]

TL;DR Summary

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.