2
$\begingroup$

The question of how a receiving antenna works has been asked on this site before, such as here How does a receiving antenna get an induced electric current? and here How does a receiving antenna work?. I understand the basic principle that the external EM field from the transmitting antenna causes electrons to move in the receiving antenna, creating a current.

My question though is in the title. How can an external electric field, as in the form of a radio or other EM wave, induce a current in a wire when the $E$ field is always zero in there? In terms of physical laws and math how can I calculate the current as a function of time if I know the external fields as a function of time?

I wanted to try to calculate the potential difference between two points of a wire from the external field using

$$\varepsilon=\int_{\text{start point}}^{\text{end point}}\mathbf{E}\cdot\text{d}\mathbf{l}$$

but that assumes that the electric field in the wire is as it would be if the wire weren't there and the waves were propagating through vacuum...

$\endgroup$
  • $\begingroup$ The electric field inside the conductor is not zero. $\endgroup$ – Farcher Feb 28 '18 at 8:31
7
$\begingroup$

When a metal antenna wire is put into the field of a propagating electromagnetic wave with time-varying fields, there will be an electric and magnetic field inside the wire and thus also a current but the penetration is exponentially damped. The penetration depth $\delta$ is called the skin depth. In treating boundary conditions with metals for electromagnetic waves often a "perfect conductor" is assumed with conductivity $\sigma \to \infty$. Then this penetration depth $\delta \to 0$ and the current can be assumed to be a surface current $J_s$. Thus, in this idealized situation, there will be no electrical or magnetic field inside the metal and the current is represented by a surface current $J_s= B_t/\mu_0$, which is normal to the tangential magnetic field at the surface (assuming the relative permeabilities are $1$). To get this current, you have to combine the incoming wave with the outgoing electromagnetic field of the wire and satisfy the boundary conditions at the wire surface for the total fields, similarly to a reflection at a plane perfect conductor metal surface.

$\endgroup$
  • $\begingroup$ Thanks, this is much more like what I was looking for. But your link is broken, is this what you meant to link to? $\endgroup$ – Ern Feb 28 '18 at 18:01
  • $\begingroup$ Sorry, I will correct this. You can also download older editions of this book for free on the internet. This is one of the best existing books on the topic of electromagnetic waves. $\endgroup$ – freecharly Feb 28 '18 at 18:14
  • $\begingroup$ Wonderful, I really appreciate the ref. I understand that this is a complex topic and thank you for pointing me towards a more thorough answer $\endgroup$ – Ern Feb 28 '18 at 18:19
4
$\begingroup$

There are multiple ways in which this is not a contradiction.

  1. By “inside a conductor”, we are referring not to the conductor as a whole but rather the interior volume as opposed to the surface. If the wire segment has a net charge, that charge will be found on the surface.

  2. The reason the field is said to be zero is that charges move as needed to bring it back to zero. Motion of charges is current. Current is what we hope to get from an antenna.

    (The field is exactly zero only in the electrostatic, equilibrium case, where the electrons have all settled down and stopped moving.)

A source for both claims

$\endgroup$
  • $\begingroup$ So how would I go about calculating the current knowing the external field as a function of time? $\endgroup$ – Ern Feb 28 '18 at 15:53
  • $\begingroup$ @Ern I can't help with that, sorry. $\endgroup$ – Kevin Reid Feb 28 '18 at 16:29
  • $\begingroup$ (And I didn't notice that that was part of your question. Sorry again.) $\endgroup$ – Kevin Reid Feb 28 '18 at 16:39
  • $\begingroup$ Oh that’s alright. I think this question may be much trickier than it seems at first glance. That’s electrodynamics I suppose, anyway thanks for answering $\endgroup$ – Ern Feb 28 '18 at 16:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.