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I have a question about electromagnetism. Probably I understand something in a wrong way. So, I know that the drift velocity of electrons in conductors is very small ($\sim$ $0.1$-$1$ mm / s). Also I know that if I put the compass to the wire, the needle is deflected.

Suppose I've started moving compass along wire with drift velocity. Is there deflection of neadle in this case?

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Just for the record, and this is really just a reprise of Jonas' answer and the comments:

  1. a current flowing and the compass is stationary: in the rest frame of the compass there is a flow of negative charge and this generates the magnetic field.

  2. a current flowing and the compass is moving at the electron drift velocity: in the rest frame of the compass the (conduction) electrons are stationary but there is a flow of positive charge in the other direction and this generates the same magnetic field.

  3. a current flowing and the compass is moving at some velocity in between zero and the electron drift velocity: in the rest frame of the compass there is a flow of negative charge in one direction and a flow of positive charge in the other direction and the two flows generate the same magnetic field.

So whether the compass is stationary, moving at the drift velocity or any velocity in between, the deflection of the compass is the same.

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No, it won't. If you look at the situation from the rest frame of the compass, it looks as if the whole wire would be moving with the electron drift velocity. While there is a 'current' due to the moving electrons, there is also a current in the opposite direction but with the same magnitude due to the ions in the metal. The two cancel each other.

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  • $\begingroup$ I'm pretty sure that isn't true. If that were the case, then the magnetic field measured by a clamp ammeter would vary immensely if you slid the clamp up and down the wire at a speed on the order of the drift velocity in the interior wire (which is only 1 mm/s or so). Similarly, magnetic field strengths measured near a wire would be strongly dependent on how fast you were moving around parallel to it, and the material composition of the wire. Neither of these things happen in real life. $\endgroup$
    – DumpsterDoofus
    Commented Oct 15, 2013 at 13:41
  • $\begingroup$ Additionally, this is equivalent to saying that moving parallel to the wire at the drift velocity induces a secondary field which cancels the original field. By extension, from this you could conclude that moving at 10,001 times the drift velocity (which comes out to only a few mile/hr) would make the needle experience a magnetic field 10,000 times as strong as what the wire originally produced, and in the opposite direction. This is clearly not how the world works. $\endgroup$
    – DumpsterDoofus
    Commented Oct 15, 2013 at 13:45
  • $\begingroup$ Of course, let me know if you think I'm misunderstanding what you're saying. What I'm saying is that the magnetic field measured outside a wire by a moving device is independent of the velocity of the device (or at least until relativistic velocities are attained) or the drift velocity inside the wire, and only depends on the total current. As a result, the compass will experience the same field as in the stationary case, and will align itself as it normally does. $\endgroup$
    – DumpsterDoofus
    Commented Oct 15, 2013 at 14:09
  • $\begingroup$ The velocity at which you are moving is quite relevant here. This is a consequence of Einstein's theory of special relativity but it is important even at "non-relativistic" speeds. See van.physics.illinois.edu/qa/listing.php?id=20600 for a thought experiment on this issue. $\endgroup$
    – Jonas Greitemann
    Commented Oct 15, 2013 at 14:20
  • $\begingroup$ I think you are confusing relativistic effects with drift velocity effects. As I said before, the velocity of a measuring device moving parallel to a current-carrying wire does not affect the field it measures, except at (physically unrealistic) near-relativistic speeds. This is a basic fact of experimental physics known since the 1800s. The thought experiment you cited is not relevant here. $\endgroup$
    – DumpsterDoofus
    Commented Oct 15, 2013 at 14:40

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