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I have to calculate the pressure on a current carrying wire. Since there is a pressure on the wire, there must be a force on it, which is a magnetic force. Does the magnetic field produced by the wire, exert a magnetic force on the wire itself? If this is true, why?

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    $\begingroup$ Maybe you are looking for this: physics.stackexchange.com/a/394034/291127 $\endgroup$
    – Cluse
    Commented Mar 8, 2021 at 11:29
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    $\begingroup$ Is the wire straight or curved? $\endgroup$ Commented Mar 8, 2021 at 21:26
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    $\begingroup$ @JánLalinský It is a straight wire $\endgroup$
    – Pi314
    Commented Mar 9, 2021 at 7:18

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The answer is: no.

The 'magnetic force' on the wire is due (indirectly) to magnetic Lorentz forces acting on the moving electrons in it. It is true that there will be attractive magnetic forces between electrons moving in parallel paths at different points in the wire's cross-section (for example between electrons at opposite ends of a diameter). But these forces are equal and opposite, so there will be no resultant force on these electrons taken together, and no resultant force on the wire.

It's a different story when we apply an external magnetic field with a component at right angles to the wire. The electrons will then experience forces in a direction given by $\textbf F = -e \mathbf v \times \mathbf B$, (or by Fleming's left hand rule). [The moving electrons would be forced out of the wire, were it not for 'bonding forces' that stop them from leaving. Strictly, it is these 'bonding' forces (or their Newton's Third Law partners) that the rest of the wire experiences, rather than the magnetic Lorentz forces directly. However the magnitude of the force on the wire can be correctly calculated as the vector sum of the Lorentz forces, which is easily shown to be equal in magnitude to $F=BIL\ \sin \theta$ with the usual notation.]

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    $\begingroup$ Thank you for drawing attention to my sloppy wording. I've made a small correction. $\endgroup$ Commented Mar 8, 2021 at 13:45
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    $\begingroup$ Would this also be the case if it is a cylindrical conductor with thickness t and a radius of R, where t << R? $\endgroup$
    – Pi314
    Commented Mar 10, 2021 at 16:09
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    $\begingroup$ Yes indeed. The same argument applies. $\endgroup$ Commented Mar 10, 2021 at 16:15
  • $\begingroup$ @PhilipWood It might be helpful to stress in your answer that the forces are equal and opposite only in the case of a straight wire. For curved wire or wire made of non-parallel segments, this is not so and the wire can exert net macroscopic magnetic force on itself. $\endgroup$ Commented Nov 28, 2023 at 18:40
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If the wire is straight, then no, due to axial symmetry magnetic field is just compressing the wire a little but no net force is present.

However, if the wire isn't straight, then net magnetic force due to wire on itself may be non-zero. For example, consider wire in shape of upside-down letter J, in which current flows from bottom to top.

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The shorter oblique segment will experience magnetic force in north-west direction from the longer vertical segment. At the same time, the longer segment will experience force in the west direction. There forces add up to net non-zero magnetic force on the wire due to itself.

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  • $\begingroup$ From your description, I infer that north and east are towards the top and right of the diagram, respectively. Your analysis also concludes that the wire has a net force in the west-north-west direction. This is clearly incorrect, as an object cannot exert a force on itself. Another way of looking at it is to call the shorter oblique segment ‘A’ and the longer vertical segment ‘B’. Now the sum of the force exerted by A on B and the force exerted by B on A is nonzero, violating Newton’s third law of motion. $\endgroup$ Commented Mar 10, 2021 at 10:03
  • $\begingroup$ > "This is clearly incorrect, as an object cannot exert a force on itself." Magnetic forces in general do not obey Newton's third law of motion. $\endgroup$ Commented Mar 10, 2021 at 10:23
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I have to calculate the pressure on a current carrying wire. Since there is a pressure on the wire

This is not relevant to the main question, but what do you mean by “pressure”? Pressure normally means the ratio of force to area. What area would you be using here?

Does the magnetic field produced by the wire, exert a magnetic force on the wire itself?

No.

Perhaps the concept will be easier to understand in the context of gravity. For example, consider the Earth. Does the gravitational field exerted by the Earth exert a force on the Earth itself? Of course not; it is not possible for an object to exert a force on itself. This applies to all forces, including magnetic forces.

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  • $\begingroup$ Thank you for the answer! I would be using the outside surface of the wire (which is a cylinder) as the area. $\endgroup$
    – Pi314
    Commented Mar 9, 2021 at 7:19
  • $\begingroup$ @Pi314 If you find an answer useful, please upvote it. When you calculate pressure, you should be using the area that the force is spread over. Assuming there is some (not yet specified) force on the wire, does it make sense to say that this force is spread over the area of the wire’s surface? $\endgroup$ Commented Mar 10, 2021 at 9:57
  • $\begingroup$ Would this also be the case if it is a cylindrical conductor with thickness t and a radius of R, where t << R? $\endgroup$
    – Pi314
    Commented Mar 10, 2021 at 16:09
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The question asks about the force that produces "the pressure on a current carrying wire", and whether this force "exert(s) a magnetic force on the wire itself". There is most definitely a radial magnetic force that creates the pressure on the wire, as discussed in these similar questions:

I would flag this question as a duplicate except that the currently accepted answer correctly states that there is no net force on the wire that would cause the wire's axis to move laterally. The fact that this answer was accepted by the Original Poster suggests that the question is unclear and that the OP might actually have been asking whether there is a net transverse magnetic force on the wire.

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