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I know the gravitational force between two particles with masses $m_1$, $m_2$:

$$ \vec{F} = \frac{Gm_1m_2}{|\vec{r}|^2} \hat{r} $$

And I know the electrical force between two particles with charges $q_1$, $q_2$:

$$ \vec{F} = -\frac{Kq_1q_2}{|\vec{r}|^2} \hat{r} $$

($\vec{r}$ is the position vector of particle 2 from the referential of particle 1 and $\hat{r} = \frac{\vec{r}}{|\vec{r}|}$)

I've been looking for an expression like these for the magnetic force since 2012... Then I found Physics Stack Exchange. Does anybody know if such formula exists?

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    $\begingroup$ Magnetic fields don't follow the simple inverse-square law, if that's what you're getting at. $\endgroup$
    – HDE 226868
    Feb 21 '15 at 17:33
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    $\begingroup$ I don't want an inverse square law, I just want a law. I just want a magnetic force vector for particle 1 describing how it interacts with particle 2. $\endgroup$ Feb 21 '15 at 17:35
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    $\begingroup$ @Rococo: So we have magnetic monopoles? $\endgroup$
    – Kyle Kanos
    Feb 21 '15 at 19:18
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    $\begingroup$ @KyleKanos obviously we have a different standard for what the phrase "inverse square law" should mean. I don't really care to get into an argument over wording, but since I just read it anyway I'll say that Jackson shares my usage: "we see that (5.4) [Biot-Savart] is an inverse square law, just as is Coulomb's Law." $\endgroup$
    – Rococo
    Feb 24 '15 at 5:35
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    $\begingroup$ @Rococo: The Biot-Savart law is for electric currents and not for point charges. For point sources, an inverse square law follows from $\nabla\cdot\mathbf N\neq0$ for some field $\mathbf N$ (i.e., a monopole is needed). $\endgroup$
    – Kyle Kanos
    Feb 24 '15 at 13:42
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What you want is essentially the Biot-Savart Law.

For a point charge that is moving slowly compared to the speed of light (which is also a condition for the Couloumb law that you give to be true, by the way), Biot-Savart says that a point charge makes a magnetic field like:

$\vec{B}=\frac{\mu_0}{4\pi}q_{1}\vec{v_1}\times\frac{\hat{r}}{r^2}$,

where $\vec{v_1}$ is the velocity of particle 1 and $q_{1}$ is its charge.

Then, the force particle two feels from it is the Lorentz force,

$\vec{F_2}=q_{2}\vec{v_2}\times\vec{B}$,

where $\vec{v_2}$ is its own velocity and $q_{2}$ its charge.

Put them together and you get the magnetic force one particle feels from the other,

$\vec{F_{1 \rightarrow 2}}=\frac{\mu_0 q_{1}q_{2}}{4\pi r^2}\vec{v_2}\times\{\vec{v_1}\times\hat{r}\}$

So it is a force that is very direction-dependent, unlike the other two formula you give: it depends on the velocities of each particle, both directions and magnitudes, as well as how these directions compare to the direction of the line that separates the two particles. For a given combination of these directions and speeds, it falls off as r^2 just like the other two forces.

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    $\begingroup$ Brackets missing in the final equation. $\endgroup$
    – SAKhan
    Feb 22 '15 at 5:10
  • $\begingroup$ The Biot Savart law is for a closed circuit current, and extending this to an isolated moving charge isn't trivial. $\endgroup$ Feb 23 '15 at 0:18
  • $\begingroup$ @LarryHarson Fair point. I must confess that I've read this disclaimer plenty of times in discussion of B-S, but always regarded it as slightly fussy. Maybe it's a lack of imagination on my part: I can't imagine how an infinitesimal source of a linearly superposed field could possibly be different than a point charge under suitable limits, even if the formal equivalence is delicate. $\endgroup$
    – Rococo
    Feb 24 '15 at 5:40
  • $\begingroup$ (your point to g3n1uss about non-reciprocity, on the other hand, requires an edit, thanks for implicitly drawing my attention to this.) $\endgroup$
    – Rococo
    Feb 24 '15 at 5:41
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Probably you are interested in the magnetic force between two moving charges which is

$$\vec{F}=\frac{\mu_0}{4\pi}\frac{q_1 q_2}{r^2}\vec{v}_1\times (\vec{v}_2\times\hat{r})$$

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    $\begingroup$ Whilst this may theoretically answer the question, it would be preferable to include the essential parts of the answer here, and provide the link for reference. $\endgroup$
    – Kyle Kanos
    Feb 21 '15 at 19:18
  • $\begingroup$ I've already seen this formula before, but what I don't understand about it is: what if only one of the particles is not moving? Then F = 0????? $\endgroup$ Feb 21 '15 at 22:50
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    $\begingroup$ For example: v1 measured in the referential of particle 1 is zero. So the magnetic force depends on the referential?? What a hell... $\endgroup$ Feb 21 '15 at 22:52
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    $\begingroup$ @matheuscscp: yes, both the magnetic force and the electric force depend on your frame of reference. Only when you consider both together do you get a frame-independent theory. (And even then, they are only perfectly frame-independent when you use the full equations rather than the low-speed approximations.) $\endgroup$ Feb 22 '15 at 0:33
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    $\begingroup$ For a game simulation, and assuming that the distances and speeds involved are low enough, the most reasonable option is to simply ignore the magnetic forces. They'll be negligible anyway. For a more accurate simulation you would need to include the electromagnetic field (and the gravitational field too, I suppose) but unless you've got a supercomputer to play it on it isn't going to run in anything like real time. $\endgroup$ Feb 22 '15 at 8:43
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If there where magnetic monopoles, the force between them in static conditions would be exactly the same as that described by Coulomb. You'd need to replace the electric charges with the magnetic charges, and possibly the universal constant as well. All of this is just a consequence of the symmetry of Maxwell's equations with a magnetic source.

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  • $\begingroup$ I didn't say that the particles are stopped. They may have a non-zero relative velocity. But, anyway... Is there a magnetic force between two charged particles???? $\endgroup$ Feb 21 '15 at 17:41
  • $\begingroup$ You should probably add that magnetic fields are usually caused by currents, so that without monopoles, there is no law for a magnetic force between particles. $\endgroup$
    – ACuriousMind
    Feb 21 '15 at 17:41
  • $\begingroup$ @ACuriousMind Okay, we have no law... But, is the magnetic force there?? Between the two charged particles? $\endgroup$ Feb 21 '15 at 17:43
  • $\begingroup$ @matheuscscp: Not between the particles as such. If they move, they behave as currents, and then their current will generate a field by the usual laws, but you cannot give a law for the particles, because the magnetic field is not caused by them, but by currents. $\endgroup$
    – ACuriousMind
    Feb 21 '15 at 17:45
  • $\begingroup$ @ACuriousMind Okay... Is it possible to describe the currents through the relative velocity between the two particles? Can you write an answer with that? $\endgroup$ Feb 21 '15 at 17:49
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To clarify upon the other answers: There is no magnetic force between non-moving charged particles. Other answers posted here have shown that if there is motion between two charged particles, there will be a magnetic force between them given by the Biot-Savart law.

Actually, the Biot-Savart law covers both moving and non-moving cases. If the velocities are equal (i.e. same speed, same direction), the force will be zero. Two non-moving particles are considered to have equal velocity in some frame of reference.

Additionally, our understanding of electrodynamics tells us that two non-moving magnetic monopoles should experience a force analogous to the electric force between two electric charges. These magnetic monopoles would possess some kind of "magnetic charge" analogous to electric charge. However, magnetic monopoles are not observed in nature; only dipoles are found.

Some of the comments have correctly stated what we observe empirically: There is an electric force between charges, and a magnetic force between moving charges.

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  • $\begingroup$ Thanks for the answer! You seem to understand quite a lot about this subject... Have you seen the comments in g3n1uss' answer? Someone stated that is possible to get a frame-independent electromagnetic force vector, if we use Maxwell's equations. I have tried to do that, but I'm having much trouble. I'd be very grateful if you could manage to do that and update this answer! $\endgroup$ Feb 25 '16 at 4:39

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