Sonia Rowland
2022-10-02
Answered

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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vakleraarrc

Answered 2022-10-03
Author has **6** answers

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 $\mathbf{\text{F}}=-e\mathbf{v}\times \mathbf{B}$, (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\text{}\mathrm{sin}\theta $ with the usual notation.]

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 $\mathbf{\text{F}}=-e\mathbf{v}\times \mathbf{B}$, (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\text{}\mathrm{sin}\theta $ with the usual notation.]

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