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I'm trying to reconcile two seemingly conflicting sources of information about Lorenz force.

Let's consider a solenoid in proximity to a permanent magnet with the field pointing in the same direction as the axis of the solenoid.

From the right hand rule I conclude that the wire should either be pushed towards the center of the solenoid or outside.

However, the solenoid also acts as a magnet with poles on each sides and will attract/repel other magnets. How does one reconcile the two explanations ? Is the solenoid both contracting on itself and moving towards the other magnet at the same time ?

If yes which force/equations describe the movement of the electromagnet towards a permanent magnet.

Thank you!

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  • $\begingroup$ The correct spelling is Lorentz force. Lorentz force is force acting on single charged particle due to external EM field. However, what you're asking about is not the Lorentz force, but magnetic force on current-carrying wire, sometimes called the Laplace force, or motor-effect force, or Amperian force, or ponderomotive force. physics.stackexchange.com/questions/540496/… $\endgroup$ Commented Jun 2, 2023 at 13:48
  • $\begingroup$ @JánLalinský silly engineering comment. It’s the same force $\endgroup$
    – mr chap
    Commented Jul 23, 2023 at 8:20
  • $\begingroup$ @mrchap It's not. Lorentz force acts on microscopic charged particle, ponderomotive force acts on macroscopic body. The latter is due to large number of microscopic forces, not all of which are necessarily Lorentz forces. $\endgroup$ Commented Jul 23, 2023 at 12:35

2 Answers 2

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Assume the solenoid to be rigid. Since the magnet is (supposedly) aligned with the axis of the solenoid, which is the center, the Force by the magnet experienced by the wire isn't exactly orthogonal to the wire. Indeed, if you draw a line from the magnet to the wire, the force is perpendicular to this line. Depending on if the force is inwards or outwards, it is tilted towards or away from the magnet relative to the normal (radial) lines. But since the solenoid is rigid, we assume it doesn't get deformed so the perpendicular component gets contrasted by the structural stability and therefore vanishes. What remains is a parallel component caused by this tilting of the force. Thus the solenoid experiences a force toward or away from the magnet along the axis.

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  • $\begingroup$ So the attraction is solely from the curvature of the field created by the permanent magnet and not because "opposite Poles attract" $\endgroup$
    – DARK_DUCK
    Commented Jun 2, 2023 at 6:34
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    $\begingroup$ @emiliograndinetti - So what forces does the solenoid feel when its axis is placed parallel to a uniform magnetic field? $\endgroup$
    – Farcher
    Commented Jun 2, 2023 at 6:59
  • $\begingroup$ @Farcher good question. No axis-parallel forces in such an experiment. If you don’t believe it then Imagine a large fixed solenoid attracting a small solenoid. As the small passes through the halfway through the large, the north and south attracting side flips, so the forces act opposite. Therefore the equilibrium exists on the center of the large solenoid, where the magnetic field is roughly uniform $\endgroup$
    – mr chap
    Commented Jul 23, 2023 at 8:22
  • $\begingroup$ @Farcher positive centripetal force on charge when solenoid and uniform fields oppose, negative centripetal when they point in the same direction. Imagine again a small solenoid wants to attract into a large solenoid, when it is inside, equilibrium will be reached due to expansion as that will form a single solenoid again. These axial forces fall to zero if two solenoids rings of the same size get closer To each other because the axial component of magnetic field falls to zero at the radius of the solenoid ring $\endgroup$
    – mr chap
    Commented Jul 23, 2023 at 8:26
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For a different perspective from emilio grandinetti's answer, note that the solenoid could be approximated as a magnetic dipole with dipole moment $\vec m$ (or could be viewed as a distribution of such dipoles). The force on such a dipole due to an external magnetic field $\vec B$ is $$\vec F = (\vec m \cdot \vec\nabla)\vec B = m_x \frac{\partial \vec B}{\partial x} + m_y \frac{\partial \vec B}{\partial y} + m_z \frac{\partial \vec B}{\partial z}. $$

So in response to your comment on emilio's answer, yes, the net attractive or repulsive force on the solenoid can be described as being due to the gradient of the external magnetic field, and the observation that "opposite poles attract" can be seen as a consequence of this.

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  • $\begingroup$ So what forces does the solenoid feel when its axis is placed parallel to a uniform magnetic field? Does the dipole approximation fail? $\endgroup$
    – Farcher
    Commented Jun 2, 2023 at 7:00
  • $\begingroup$ @Farcher The "compressive" forces on the solenoid are still there, but I believe this is hidden in the dipole picture because the compressive forces on each dipole cancel out. $\endgroup$
    – Puk
    Commented Jun 2, 2023 at 7:09
  • $\begingroup$ But with a uniform field there is no gradient and hence no force due to the uniform external field. Yes, within the solenoid itself there are inward axial forces and outward radial forces. $\endgroup$
    – Farcher
    Commented Jun 2, 2023 at 9:13

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