I don't understand why the magnetic force only acts on moving charges. When I have a permanent magnet and place another magnet inside its field, they clearly act as forces onto one another with them both being stationary. Also, I am clearly misunderstanding something.

  • 1
    $\begingroup$ "and place another magnet inside its field they clearly act a forces onto one other with them both being stationary" -> How exactly are you able to place something somewhere while at the same time not moving it at all? $\endgroup$
    – walen
    Commented Feb 23, 2023 at 9:28
  • 6
    $\begingroup$ With hindsight, this was the biggest clue in the 19th century that something was wrong with classical physics: this special force between moving charges means the force between charges changes when one transforms between different inertial frames using the Galilean transforms. It's sorted out by the special-relativistic idea of using the Lorentz transforms instead of the Galilean transforms. $\endgroup$ Commented Feb 23, 2023 at 21:15
  • 1
    $\begingroup$ @DanielHatton To your 'with hindsight' comment: If the magnetic force would manifest itself as an effect that is proportional to the relative velocity between two charges then Galilean transforms will be the correct transforms. As we know: the Faraday/Maxwell concept of a field is one of a state of stress of some medium, in the case of Maxwell's electromagnetism usually referred to as Luminiferous Aether. Magnetic effect is then described in terms of velocity with respect to this Aether. As we know: any experiment to obtain a value for that velocity produces a null result. $\endgroup$
    – Cleonis
    Commented Feb 24, 2023 at 21:29
  • $\begingroup$ What led to believe that? Were you never, as a child, given a magnet to be used as either a toy or a scientific experiment? Did that magnet not act on anything it came across, moving charges or not? $\endgroup$ Commented Feb 26, 2023 at 20:19
  • $\begingroup$ Yes I did that's why I didn't understand why it only acted on moving charges $\endgroup$ Commented Feb 26, 2023 at 20:47

7 Answers 7


This is basically just a matter of definition. The magnetic force is not something independent or complete on its own. Instead, it is the electromagnetic force that is the underlying concept. The electromagnetic force can be split into a portion that acts on stationary charges and a portion that acts on moving charges. We call the first portion the electric force and the second portion we call the magnetic force. But they are not independent. They exist together and it is just an essentially arbitrary division to split the overall electromagnetic force into an electric force and a magnetic force.

Regarding a permanent magnet: a permanent magnet is typically uncharged, so you would not consider a permanent magnet as though it were an electric charge. It would typically be better to consider a permanent magnet as an uncharged magnetic dipole. Inhomogeneous magnetic fields can exert forces on stationary magnetic dipoles. The formulas are not at all the same for magnetic dipoles as for electric charges.

  • $\begingroup$ Regarding the last paragraph, how would you respond to the (likely false) claim that this demonstrates that the Lorentz force law is incomplete? I had a question regarding this, but honestly I'm still not sure if I understand how to think about it. $\endgroup$ Commented Feb 22, 2023 at 17:13
  • 3
    $\begingroup$ @MaximalIdeal I would need to get more information about what the claimant means by "incomplete". The Lorentz force law is a law for calculating the force on a classical point charge. So it is complete in the sense that it describes the complete force on a classical point charge. But it does not describe the force on classical charge distributions, macroscopic polarized or magnetized materials, or quantum mechanical particles. So if any of those are required for the sense of "complete" mentioned then indeed it would be incomplete. $\endgroup$
    – Dale
    Commented Feb 22, 2023 at 17:17
  • 4
    $\begingroup$ +1. Note that experimentally they are very different: you cannot measure a magnetic field with an electrostatic probe, nor can you measure an electric field with a magnetometer. Unless you move... $\endgroup$
    – John Doty
    Commented Feb 22, 2023 at 17:24
  • 7
    $\begingroup$ Importantly: if one observer sees a pure electric or magnetic field, an observer moving with respect to the first sees a mixture. The division of "EM field" into "electric field" (force on particles you see at rest) and "magnetic field" (force on particles you see as moving) is subjective. $\endgroup$
    – HTNW
    Commented Feb 23, 2023 at 0:33
  • $\begingroup$ @HTNW: Don't you mean "relative" rather than "subjective". That is a far better description of the relationship, as the phenomenon is quite objective. $\endgroup$ Commented Feb 25, 2023 at 20:11

At an effective classical level, the atoms in permanent magnets do contain moving electric charges at the microscopic level: the orbiting electrons. These moving charges correspond to microscopic electric currents, and the magnetic fields act on these microscopic currents.

This picture is certainly a simplification of the underlying quantum effects, but I think it's accurate enough to clear up your confusion.

  • 3
    $\begingroup$ The magnetic moment of a permanent magnet comes mostly from the spin of the electrons, with only a much smaller contribution from their orbital motion. $\endgroup$ Commented Feb 23, 2023 at 20:35

To be more precise, the magnetic force acts on currents. A moving particle implies a current, although not all currents correspond to a classical, visible movement. This is the case with the microscopic currents in permanent magnets.

  • 2
    $\begingroup$ your answer begs the question: in what sense is the magnetic moment of an electron related causally to a microscopic current? $\endgroup$
    – hyportnex
    Commented Feb 22, 2023 at 14:26
  • $\begingroup$ @hyportnex it is not. The electron magnetic moment is intrinsic. Some atomic magnetic moments are tied to electrons orbitals where electrons do not classically move per se. $\endgroup$
    – Mauricio
    Commented Feb 22, 2023 at 17:23
  • 1
    $\begingroup$ @hyportnex I guess you mean the intrinsic magnetic moment. That magnetic moment too can be interpreted as the effect of the current density in the rest frame due to the multi-component nature of spinors. See, for instance aapt.scitation.org/doi/abs/10.1119/1.19421 $\endgroup$ Commented Feb 22, 2023 at 18:00

As mentioned in a comment to the question: in terms of relativistic physics there is a way to address your question.

In terms of pre-relativistic physics it is something that must be granted in order to formulate a theory of electromagnetism. Phrased differently: in terms of the physics prior to the introduction of relavistic physics the fact that magnetism is proportional to velocity has to be accepted as is.

There is a set of notes by Daniel V. Schroeder titled:
Magnetism, radiation, and relativity

Preliminary remark:
The electric field is extremely strong. It is far stronger than we would tend to expect, since in daily life we never experience that. The effects of the kind of static electric charge we can feel in daily life are minute because there is never more than the tiniest difference of charge.

In these set of notes by Daniel Schroeder the following is pointed out: a current flowing through a wire is a population of electrons that is moving with respect to the wire.

In terms of pre-relativistic theory of electricity and magnetism: the wire is neutral everywhere because the density of positive charges and negative charges is the same everywhere.

Now in terms of relativistic physics:
We take the wire as a whole as stationary reference. The population of free-to-move-around electrons in the wire then has (on average) a velocity with respect to that stationary reference. For that velocity there is a corresponding length contraction. So in that sense the wire and the electron population do not have the same charge density.

Schroeder shows that it is possible to account for magnetic effects with this difference in charge density.

That is counter-intuitive on several levels, of course. For one thing: it is always asserted that relativistic effects require relativistic velocities in order to be measurable at all.

The (average) velocity of the electron population is very, very small. The thing is: the Coulomb force is so mindbogglingly strong that magnetism can indeed be accounted for in this way.

(One thing that concerns me though is that in those Schroeder notes relativity of simultaneity is not discussed. I suspect the Schroeder notes present an oversimplification. The numbers check out, so Schroeder is doing something right, but we can't be sure of what it is that is done right.)


A magnetic field does have an influence on a stationary electric charge.

"when i have a permanent magnet and place another magnet inside its field they clearly act a forces onto one other with them both being stationary"

is exactly the case that is also true for an electron.

An electron is both an electric charge and a magnetic dipole. Any description of internal currents in the electron to create the magnetic dipole is superfluous. It is quite sufficient to consider the magnetic field of the electron to be just as fundamental as its electric field.

For a detailed description of why an electron moving through a magnetic field is deflected laterally (Lorentz force, Hall effect(s) ), see here.

  • 3
    $\begingroup$ One might ask why, given that all materials are full of electrons, not all materials are attracted to magnets. It's because in most materials, the electrons are aligned randomly so that attraction and repulsion are equal. Only in certain materials do they align with some order to set up an overall magnetic field that causes a net attraction or repulsion. $\endgroup$
    – Rich006
    Commented Feb 23, 2023 at 13:15
  • $\begingroup$ An electron moving is an electric current. Calling it an internal current is confusing and misleading, but not superfluous. $\endgroup$
    – david
    Commented Feb 23, 2023 at 20:35

After reading some of the other answers I think there is a simpler more fundamental explanation.

According to Field Theory, rather than considering the interactions between particles and field, we can consider the interactions between the fields associated to the particles.

Under this understanding, Classical gravitation $F = G \frac{m_1 m_2}{r^2}$ is the interaction between 2 objects, each with it's own gravitational field. Similar Coulomb force $F = k\frac{q_1 q_2}{r^2}$ is the interaction between 2 charged particles, each with it's own electrical field.

Magnetic interactions are a bit more complicated, but we can also understand magnetism as interaction between objects that each have their own magnetic field.

Now to your question.

A static electric charge generates an electric field, but not a magnetic field, as per the relevant Maxwell equation. $\vec\nabla\cdot \vec E = \frac{\rho}{\epsilon_0}$.

A moving charge has a magnetic field, given by $\vec\nabla\times \vec B = \mu_0 ( \vec J + \epsilon_0 \frac{\partial \vec E}{\partial t})$. in which the current density $J = q\cdot \vec v$

Therefore a magnetic field does interact with a moving charge, because the moving charge also has a magnetic field, but not with a static charge, because the static charge does not have a magnetic field.


As a simple(ish) special case of the principle set out by @Cleonis, imagine two initially stationary charges, starting a short distance apart. Under the Coulomb (electrostatic) force, they will accelerate apart (if they're like charges) or together (if they're opposite charges).

Now think about this same pair of charges from the point of view of an observer who is moving perpendicular to the displacement between the two charges, so that, to this observer, the charges appear to be initially moving parallel to each other and perpendicular to the spacing between them.

Because the displacement between the two charges is perpendicular to the frame-of-reference motion, it will not appear Fitzgerald-contracted. However, there will be time dilation, so that to this observer, it will appear to take longer for the two charges to reach any given distance apart than it did to the stationary observer; i.e. the acceleration of the charges will appear to be smaller to the moving observer than to the stationary observer; i.e. (again) the force between the charges will appear to be smaller to the moving observer than to the stationary observer. That is, an observer who encounters moving (in the same direction, perpendicular to the spacing between them) charges will think there is an extra force between them, acting in the opposite direction to the Coulomb force, compared with an observer who encounters stationary charges. That extra force is what we call the magnetic force.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.