As far as I know, if magnetic monopoles existed, the force between them would be very similar to the force between electric charges, which reads $\mathbf {F}=\frac{k Q_1 Q_2}{r^2}\hat{\mathbf{r}}$ where $k\approx 9\cdot10^{11}\:\mathrm{N\:m^2/C^2}$.

  • What would this constant $k$ be for magnetic monopoles?
  • What would be the unit of the magnetic charge?

2 Answers 2


This is a good question, but at first blush it is hard to answer. This is because there is always an ambiguity in where you put the meaningful definitions - you can put it in the force constant, or you can put it in the units for the magnetic charges.

For magnetic monopoles, the first place one naturally turns to is to the magnetic Gauss law, which would be modified to the form $\nabla\cdot\mathbf B\propto \rho_\mathrm{m}$ ... except that you don't really have a way to fix that proportionality constant. By symmetry with respect to the electric Gauss law ($\nabla\cdot\mathbf E=\rho_\mathrm e/\varepsilon_0$) you'd hope that would be $\mu_0$, but that's some hardy guessing.

What you do want, very much, is for magnetic monopoles to have the same relationship to magnetic dipoles as electric monopoles have with electric dipoles - and magnetic dipoles we know how to handle.

If you take this as your basis, then, you can postulate* a magnetic field of the form $$ \mathbf B=\frac{k_1q_\mathrm{m}}{r^2}\hat{\mathbf r}, \tag 1 $$ with some as-yet-to-be-determined constant $k_1$, for a magnetic monopole of magnetic charge $q_\mathrm{m}$. The key physical input is requiring a pair of opposite magnetic charges a distance $d$ apart to have a magnetic dipole moment of $$ m=q_\mathrm{m}d, $$ as in the dipole case, and this fixes the units of the magnetic charge: you know that this magnetic dipole (made from two opposite magnetic monopoles) must be exactly equivalent to a standard magnetic dipole: a current $I$ in a loop of area $A$, with magnetic dipole moment $m=IA$. This requires you to have $[q_\mathrm{m}]=\mathrm{A\,m}$.

From here you can get the dimensionality of the magnetic field constant: $$ [k_1] =[Br^2/q_\mathrm{m}] =[BL/I] =\left[\frac{\mu_0I}{L}\frac{L}{I}\right] =[\mu_0] =\mathrm{N/A^2}. $$ However, it doesn't tell you the numeric value of this constant, for which you need to go back to your initial physical input - the equivalence of opposite pairs of magnetic monopoles with current loops. In particular, if each magnetic monopole has a magnetic field as in $(1)$, then it follows that a pair of them, with opposite magnetic charges and separated by a distance $d$ along an axis $\hat{\mathbf u}$, must have the magnetic field $$ \mathbf B=\frac{k_1q_\mathrm{m}d}{r^3}\left(3(\hat{\mathbf u}\cdot\hat{\mathbf r})\hat{\mathbf r}-\hat{\mathbf u}\right). $$ This contrasts with the magnetic field of a current-loop magnetic dipole with magnetic dipole moment $\mathbf m$ $$ \mathbf B=\frac{\mu_0/4\pi}{r^3}\left(3({\mathbf m}\cdot\hat{\mathbf r})\hat{\mathbf r}-{\mathbf m}\right), $$ which needs to be equivalent under the identification $\mathbf m=q_\mathrm{m} d\:\hat{\mathbf u}$, and this completely fixes the magnetic-field constant at $$ k_1=\frac{\mu_0}{4\pi}, $$ i.e. $$ \mathbf B=\frac{\mu_0}{4\pi}\frac{q_\mathrm{m}}{r^2}\hat{\mathbf r} $$ for a point magnetic monopole of magnetic charge $q_\mathrm{m}$. Similarly, this fixes the magnetic Gauss law to the naive $$ \nabla\cdot\mathbf B=\mu_0\rho_\mathrm m $$ (where $\rho_\mathrm m$ is of course the volumetric density of magnetic charge).

OK, so that's a lot of work - and we're nowhere near the force that you asked about. The reason for this is that we can only get out of the formalism what we put in, and we have only specified how magnetic monopoles should produce magnetic fields, but not how they should react to them. To get that from the formalism, we need to give it more information, and here again we are constrained in that a opposed-point-monopoles magnetic dipole needs to feel exactly the same force that a current-loop magnetic dipole of the same magnetic moment does.

Similarly to the above, we can postulate* that an external magnetic field $\mathbf B$ will produce a force $$ \mathbf F=k_2q_\mathrm m\mathbf B $$ on a point magnetic monopole of magnetic charge $q_\mathrm m$, and see what happens. Given this postulate, the same algebra that worked for electric dipoles implies that if you have two magnetic monopoles of opposite magnetic charges $q_\mathrm m$ a distance $d$ apart along the unit vector $\hat{\mathbf u}$, then the torque on them exerted by an external uniform magnetic field $\mathbf B$ will be $$ \boldsymbol{\tau} = (k_2q_\mathrm md\:\hat{\mathbf u})\times\mathbf B, $$ which contrasts with $\boldsymbol{\tau} = \mathbf m\times\mathbf B$ for a usual current-loop magnetic dipole of magnetic dipole moment $\mathbf m$. This equivalence then forces our second constant to be $$ k_2=1. $$ Similarly, you can check that this force will give identical expressions for the force on a magnetic dipole in a non-uniform external magnetic field $\mathbf B(\mathbf r)$, regardless of whether it is made up from opposing magnetic monopoles or from a small current loop.

With this, then, we can just connect the two main laws - how magnetic monopoles produce magnetic fields and how they react to them - to get the answer we're after. If you have two magnetic monopoles of magnetic charges $q_{\mathrm m,1}$ and $q_{\mathrm m,2}$, separated by a distance $r$ along the separation vector $\hat{\mathbf{r}}_{1\to2}$ pointing from $1$ to $2$, then the magnetic force exerted by magnetic monopole $1$ on magnetic monopole $2$ is $$ \mathbf F=\frac{\mu_0}{4\pi} \frac{q_{\mathrm{m},1}q_{\mathrm{m},2}}{r^2}\hat{\mathbf{r}}_{1\to2}. $$ As expected, like poles repel each other (e.g. two point north poles repel).

Here the magnetic charges are measured in ampere meters, and the magnetic charges can be calibrated by observing the interaction with a moving electric charge: if you have a point magnetic monopole of magnetic charge $q_\mathrm{m}$ and an electric charge $q_\mathrm{e}$ separated by a distance $r$ along the unit separation vector $\hat{\mathbf{r}}_\mathrm{m\to e}$, with the electric charge moving at velocity $\mathbf{v}_\mathrm{e}$, then the electric charge will be subject to a force $$ \mathbf F=\frac{\mu_0}{4\pi} \frac{q_{\mathrm{m}}q_\mathrm{e}}{r^2} \mathbf{v}_\mathrm{e} \times \hat{\mathbf{r}}_\mathrm{m\to e}. $$ This gives you an 'anchor' for the unit of magnetic charge - it's not free-floating, and it's completely tied to the unit of electric charge.

* Note that these are also additional impositions on the form of the produced magnetic field and the response to external fields. However, both postulates are reasonable things to assume: if the relations were substantially different, then we wouldn't want to call those objects magnetic monopoles. Moreover, it should be essentially possible to get to those forms using very little more than symmetry considerations.


Your new constant would probably be $k/c^2.$

$k$ is actually a simplified expression for $1/(4 \pi \varepsilon)$ which is the permittivity of free space. For magnetic fields, you use the permeability of free space, which is the constant $\mu$. $\mu$ and $\varepsilon$ are connected by their relationship with the speed of light $c = 1/\sqrt{(\mu \varepsilon)}.$

I think magnetic charge is measured in Ampere-meters. You can find a good derivation here.

  • $\begingroup$ While nice, the linked document is not really a derivation as such - it's an exposition of the results, with some internal checks to show that they make sense. $\endgroup$ Oct 4, 2016 at 16:05

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