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According to Einstein's relativistic interpretation magnetic field caused by charges moving inside a wire is caused only by the relative speed between charges and the observer. So, if electrons are moving inside a wire and observer is stationary, than observer will measure a magnetic field.

Now, electrons are moving relatively slowly inside a wire. Just few inches per second, or even less.

What would happen if somebody moved the instrument measuring the magnetic field at a same speed as the movements of electrons inside a wire and in the same direction?

Would that moving instrument measure no magnetic field at all?

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  • $\begingroup$ Keep in mind that in a real setup, not every charge moves at exactly the same speed, so you won't be able to perfectly match the speed of all of them exactly. $\endgroup$
    – Timaeus
    Oct 15 '15 at 13:52
  • $\begingroup$ Yeah, I knew that. But there would be a change and than you just use stats to work it out. As Floris said, problem is that most practical conductors have both positive and negative charges moving in the opposite directions. So far, we only identified electron beams as mono-polar current carrier. But electron beams move too fast for practical experiment. I looked into electrolytes, but they are bi-polar current carriers as well. $\endgroup$
    – COROVICD
    Oct 15 '15 at 14:54
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The problem is that in a wire, you have electrons moving one way and their positive counterparts, the atoms, being stationary. When you move along with the electrons, the lattice is no longer stationary, and you replace a "negative current in + direction" with a "positive current in - direction".

What you state would be true for electrons in vacuum. But in a wire, it doesn't work.

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  • $\begingroup$ Great answer. What about a plasma inside a neon tube? I am not sure, but I guess there we just have ions of one polarity moving in one direction. If we did the above experiment, but with neon tube instead of wire, would we get magnetic field to disappear. Or is plasma in neon tube moving at some really high speeds? $\endgroup$
    – COROVICD
    Oct 15 '15 at 13:17
  • $\begingroup$ Even a plasma is neutral: there are equal numbers of positive and negative charges. In that case, you have some positive and some negative current - but as you move, you will see more of one and less of the other. To see the effect you describe, you need an electron beam. And those typically have low space charge density (because they repel) and therefore any appreciable current requires very high speed. No free lunch. But really interesting question... I had to think about it for a minute! $\endgroup$
    – Floris
    Oct 15 '15 at 13:33
  • $\begingroup$ I agree with you about electrons in vacuum. But, how is this experimentally proofable? $\endgroup$ Oct 15 '15 at 14:51
  • $\begingroup$ I checked electrolytes, they are bi-polar current carriers. What about semiconductors? Is there a semiconductor that has only electrons as charge carriers? We need mono-polar current, with very slow charge carriers. Any ideas? $\endgroup$
    – COROVICD
    Oct 15 '15 at 14:58
  • $\begingroup$ I found a very relevant question and corresponding answer here. Two electron beams traveling in parallel are a great way to look at the problem more closely. $\endgroup$
    – Floris
    Oct 15 '15 at 19:13
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To answer this question you do not need to worry about the details of how the fields are produced, if you know the relevant Lorentz transformation of the E- and B-fields. If an observer in a particular frame of reference measures a magnetic field, but no electric field (e.g. the fields of a current carrying wire), then there must be a magnetic field in all other inertial frames.

The relevant Lorentz transforms for the B-field are that: $$\vec{B^{\prime}_{||}} = \vec{B_{||}}\ \ \ \ \ \vec{B^{\prime}_{\perp}} = \gamma (\vec{B_{\perp}} - \vec{v} \times \frac{\vec{E}}{c^2})$$ Thus if there is no E-field in one frame, but a non-zero B-field, there will always be a non-zero transformed B-field in any other frame of reference moving with velocity $\vec{v}$ (and Lorentz factor $\gamma$) with respect to the original frame.

The situation is different if your current was just caused by an electron beam, because then there would also be an E-field associated with them. If you move with the electrons in this case, the B-field will be zero in the moving frame, because the second term inside the bracket is exactly large enough to cancel with the first term.

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