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I'm misunderstanding something very important regarding magnetic force and its relation to velocity.

According to the Lorentz force, the magnetic force ${\mathbf {F_B}}=q{\mathbf {v}}\times {\mathbf {B}}$. Assuming the charge is not moving, then ${\mathbf {v} = \begin{pmatrix}0&0&0\end{pmatrix}}$. Therefore, ${\mathbf {F_B}=0}$.

So why when I hold a magnet close to a piece of metal, I can feel the magnet applying a force on the piece of metal? Since the piece of metal and the magnet are not moving, shouldn't the net magnetic force be 0? I assume the force I am feeling is from the magnetic field from the magnet and the charges in the piece of metal.

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The magnet and the metal are both made of magnetic dipoles. These can be thought of as tiny loops of electric current that generate the kind of magnetic field that you would expect out of the ends of a macroscopic bar magnet. Unlike electric charges, magnetic dipoles can feel a force when not moving, as long as they are exposed to a field that changes across space. Namely, the force on a dipole with moment $\vec{m}$ by an external field $\vec{B}$ is:

$$\vec{F}=\vec{\nabla}(\vec{m}\cdot\vec{B})$$

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  • $\begingroup$ I understand I can model the metal as a magnetic dipole, but isn't the force I'm observing ultimately a result of the electrons and protons in the metal interacting with the magnetic field? I think my false assumption may be that the electrons and protons are not moving, when in reality they are. If the metal was at absolute zero and there was negligible movement on the atomic level, I would not expect to feel the force. $\endgroup$
    – hetelek
    Sep 25, 2020 at 7:26
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    $\begingroup$ @hetelek Ferromagnetism doesn't really come from the motion of electrons (and nuclei are basically stationary in a metal anyway). It comes from the fact that electrons have an intrinsic magnetic dipole moment: they act as tiny bar magnets even when stationary. Due to a historical misconception, we usually call this property "spin", though the electron itself is not spinning. Due to the Pauli exclusion principle, under certain conditions it becomes favorable for unpaired outer electrons in different atoms to arrange themselves such that their spins are all aligned with each other. $\endgroup$ Sep 25, 2020 at 16:08
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    $\begingroup$ @hetelek When you have a bunch of atoms' magnetic dipole moments aligned in the same direction like that, you get a macroscopic ferromagnet. There is a contribution from the electrons' orbital angular momentum, but it is a minor one compared to the effect of spin alignment. The point is: ferromagnetism is an intrinsically quantum-mechanical phenomenon. It is not possible to explain it using classical electromagnetism; this was proven by the Bohr-van Leeuwen theorem. $\endgroup$ Sep 25, 2020 at 16:12
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    $\begingroup$ @hetelek Due to this spontaneous alignment of atomic magnetic dipoles, you would expect to feel a magnetic force even at absolute zero. A semi-classical model of this behavior is provided by the Ising model (I say "semi-classical" here because, although the model itself is entirely classical, the interaction between atomic magnetic dipoles that it prescribes is not the usual one that you would get from classical electromagnetism, but is much closer to the interaction you would expect due to the Pauli exclusion principle). $\endgroup$ Sep 25, 2020 at 16:20
  • $\begingroup$ Thank you, this is starting to make more sense. So is Lorentz force incorrect at very low zero temperatures? It seems it would incorrectly predict a magnetic force of 0 if the electrons/protons are not moving. $\endgroup$
    – hetelek
    Sep 25, 2020 at 21:53

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