Let there be a circular-shaped wire - a ring in which there exists a steady current. It creates magnetic field around the wire. This can be evaluated by Biot-Savart Law.

enter image description here

As shown in the fig., there is a magnetic field all around the wire.

But does current in one part of the wire create magnetic field on another part of the wire?

It can't create magnetic field just above or below it since $d\mathbf l$ and $\mathbf r$ are parallel to each other for part of the wire above or below it. But what about at other regions on the wire?

Is it possible that current at one part of the wire creates magnetic field on any other part of the wire carrying the same current?

At least I've seen no pic where there is magnetic field line on the wire which would imply there exists magnetic field on the wire created by current at other parts of the same wire.

Can anyone please clear me out this issue?

  • $\begingroup$ What do you mean "on the wire"? Do you mean does the magnetic field exist across the paired parallel segments of wire? $\endgroup$ Dec 24, 2015 at 15:03
  • $\begingroup$ @Daniel Griscom: Let you draw a circle & then draw a chord (not a diameter); name the ends of the chord on the circumference as $A$ and $B$. What I am saying by on the wire is whether there can be magnetic field at $B$ created by current at $A$ . Could I clear it now? $\endgroup$
    – user36790
    Dec 24, 2015 at 15:07
  • $\begingroup$ Is there some reason you think a moving charge doesn't affect the magnetic field everywhere? $\endgroup$
    – Asher
    Dec 24, 2015 at 21:23
  • $\begingroup$ @Asher: Actually I was a bit confused as to whether there can be force from the current on the other part of the same wire it is flowing into. That's why I asked it. Beside that, even Biot-Savart Law tells that there exists force on the other part of the wire. But still I wanted to confirm whether really current at one part of the wire exerts force on the other part of the wire. Also, the picture doesn't show field lines on the wire; all field lines are outside the wire. That's why I wanted to confirm it. $\endgroup$
    – user36790
    Dec 25, 2015 at 2:49

1 Answer 1


does current in one part of the wire create magnetic field on another part of the wire?

Yes, it does. But usual simplified models don't even have a magnetic field on the wire. Imagine the loop in the XY plane with the current going counter-clockwise, then the disk inside the loop has the magnetic field come out towards you in the Z direction. And the rest of the XY plane that isn't the disk inside the wire loop and also isn't the annulus of the wire loop has the magnetic field go away from you in the opposite of the Z direction. So on one side it is in and on the other side it is out. And in the limit where the wire loop is infinitely thin the magnetic field isn't defined.

See, there is an $I$ and a $J$ and if you have an infinitely thin wire with a nonzero $I$ then $J$ is infinite.

If you give your wire some thickness you can have a finite $\vec J$ throughout the wire and then use Jefimenko (or Biot-Savart, since it is steady) to compute the field at any point.

Is it possible that current at one part of the wire creates magnetic field on any other part of the wire carrying the same current?

Yes, the magnetic field anywhere is merely the buildup from the part of the field due to each little bit of current. And your idea that the wire contains "the same" current at different places is completely unfounded. It almost sounds mystical like you expect each piece of current to know that it is "the same" as some other current.

If you use Jefimenko instead of Biot-Savart you'll see that the field here and now is due to charges there and earlier. It is not related to current elsewhere right now.

It can happen that things right now are related to other things at the same moment. But that is an accident of having a symmetrically and common past, its an accident.

The fields now are causally related to the past charges and past currents (and their rates of change). The field now isn't caused by the current now, it is caused by the current in the past. But for a steady current those happen to be equal.

It's like if you and your friend look at each other while standing 30cm (or a foot) apart. Then you are seeing your friend as they were a nanosecond ago. And your friend is seeing you as you were a nanosecond ago. Neither of you is affected by the way the other one is right now. Sure. If you hold still then the way you are now and the way you were a nanosecond ago might be the same, but your friend doesn't see you now, they see your last self.

Same thing with a loop that is 30cm in diameter. The field at one point depends on the current on the other side as it was a nanosecond ago. And it depends on the parts closer depending on what the currents were more recently, but still in the past.

At some point you are getting closer to the point than the thickness of the wire. And by then your approximation of the wire as thin has failed you and you'd have to decide what to do.

  • $\begingroup$ Hi! Could you help me in this question? Would be grateful if you answer this. $\endgroup$
    – user36790
    Jan 30, 2016 at 7:38

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