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A current carrying conductor produces a magnetic field but what if there is an observer which has the same speed as current. Will there be magnetic field by the current carrying conductor for that observer?

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This depends on what kind of conductor and what kind of current: To give the most general answer, there will be no magnetic field in your frame of reference if there isn't a current in your frame of reference. The crucial question is: Is there a current in your frame of reference when you are moving at the speed of the current? Now it get's tricky, since you have to consider every charges present in the conductor. I will give a few cases to illustrate that:

Case 1: There is no conductor, you just have equaly distributed electrons moving in the same direction at the same velocity $\vec{v}$. In this case, moving with the same velocity as they do, what you observe in your frame of reference is equally distributed electrons, that don't move at all. You won't see any magnetic field here (but you will see an electric field).

Case 2: There is a conductor, which without the charges that the current consists of, is neutrally charged. The charges that the conductor consist of simply don't matter. This case is the same as case 1, you won't see a magnetic field. Note however, that your conductor is not neutrally charged, it contains the charges that the current is made of.

Case 3: Let the current be made up entirely by electrons, and the conductor be made up by protons. Then the conductor carrying the current is neutrally charged. But moving into the frame of reference with the same velocity as t he one of the electrons, you will experience the protons moving at $- \vec{v}$ in the other direction, and thus you will see the same magnetic field that you experienced in the first frame of reference.

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    $\begingroup$ Perhaps you should reword Case 2, as the description is convoluted. $\endgroup$
    – garyp
    Commented Dec 30, 2016 at 16:55
  • $\begingroup$ I agree with @garyp that you should reword Case 2 as well as Case 3. $\endgroup$
    – MrAP
    Commented Jan 15, 2017 at 0:33

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