As we know, a current carrying wire can create a magnetic field which is perpendicular to the direction of current (From relativity, we can understand why magnetic fields are produced ) and the net charge of the wire is zero. So, no electric field will be produced.

If the wire is moving, will the magnetic field move? I learned that a moving magnetic field can produce electric field. Where am I missing the concept clearly?

  • $\begingroup$ Actually it's precisely from relativity that an electric field is produced by an otherwise stationary neutral wire. The magnetic field is how the electrostatic field behaves when the charges are in motion. When they are still (relative to an "observing charge"), they produce no electric charge (relative to that particular "observing charge" in question) but if they start moving then space contracts from the perspective of the observing charge and now the wire is no longer neutral because of a perceived surplus of charge in the wire due to a perceived increase in density of charge in the wire. $\endgroup$ – Andrew Oct 7 '19 at 8:57
  • $\begingroup$ Pretty much every real wire carrying current does produce an electric field in its surroundings due to the electric charge induced on its surface. It does not have to move at all. I understand that you aim towards relativistic effects, this nevertheless answers your title question. $\endgroup$ – Tomáš Brauner Oct 7 '19 at 9:41
  • $\begingroup$ If there is a relative velocity between the wire and the charged particle (parallel to the wire), the charged particle will feel the magnetic force. If the relative velocity is zero, the particle will feel no force. Am I right? @andrew $\endgroup$ – Dumb_kaz Oct 7 '19 at 13:24
  • $\begingroup$ I am high school student. Sorry for my poor question. In this formula, F=qvB, here v is the velocity with respect to what? $\endgroup$ – Dumb_kaz Oct 7 '19 at 13:26

the net charge of the wire is zero. So, no electric field will be produced.

The charge of the wire being zero means that there is no electrostatic E-field. However there are other ways to produce E fields. Faraday’s law says $$\nabla \times E=-\frac{\partial}{\partial t}B$$ Which means that you can also get a circulating E field by having a B field which changes in time.

In your case, since the wire is moving the B field changes over time leading to an E field.


To explore the concept further, note that the overall charge density of the wire is in fact something that depends on your frame of reference.

To see this, imagine that you're on a starship next to an infinitely long, neutrally-charged but current-carrying wire. As you accelerate closer to the speed of light, you "catch up" to the electrons moving in the wire -- but conversely, the 'stationary' positive ions they're loosely attached to are now moving backwards.

Since they're moving relative to you, they now experience length contraction; meanwhile, the electrons, which were already moving and therefore already contracted relative to your original inertia frame, are spreading back out as you catch up. So the relative charge densities will themselves change as you move -- meaning that the wire appears to pick up a charge!

And, of course, whether you're moving relative to the wire or it's moving relative to you doesn't matter -- so you immediately see that yes, a moving neutrally-charged wire should general produce an electric field.

(This is somewhat separate to the Ampere's Law effect from Dale, though they turn out to be the same law when you move into the full spacetime formalism.)

  • $\begingroup$ From this, can I also explain why the magnetic force is perpendicular to the motion? $\endgroup$ – Dumb_kaz Oct 8 '19 at 9:26
  • $\begingroup$ Drift velocity is very very small than the speed of light. If I accelerate closer to the speed of light, shouldn't see that the electrons are moving backwards also? $\endgroup$ – Dumb_kaz Oct 8 '19 at 9:32
  • $\begingroup$ Yep -- but a whole Coulomb is huge, so even though the effect is tiny you can still get respectable field strengths out of it. (If you want to play around more with this, I suggest looking up four-vectors, specifically the fact that you can trade off vector with scalar potentials with the four-potential: en.wikipedia.org/wiki/Four-vector#Electromagnetism) $\endgroup$ – linkhyrule5 Oct 8 '19 at 18:10
  • $\begingroup$ Also, no matter how fast you go there'll be a difference in the relative velocities of the bulk and the electrons, and there's only one such relative velocity that causes the charge densities to 'cancel'. $\endgroup$ – linkhyrule5 Oct 8 '19 at 18:21

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