We define electric potential and gravitational potential and use them quite often to solve problems and explain stuff. But I have never encountered magnetic potential, neither during my study (I am a high-schooler), nor during any discussion on physics.

So, does magnetic potential even exist? According to me, it should because a magnetic field is a conservative one and so we can associate a potential with it? Also if it's defined, then why don't we encounter it as often as we do the others (electric potential, gravitational potential, etc.)?

I have encountered magnetic potential energy only in the cases where a dipole is subjected to a magnetic field. Is magnetic potential limited only to this scenario, or is there a general expression for the magnetic potential?

I should have said this earlier, but don't restrain your answers due to my scope. You can use vector calculus as I am quite familiar with it. Also, this question is meant for everyone, so even the answers which are out of my scope are appreciated.

  • 2
    $\begingroup$ FWIW, there is a magnetic vector potential. $\endgroup$
    – Qmechanic
    Dec 26, 2019 at 15:28
  • 6
    $\begingroup$ Why do you think that magnetic field is a conservative one? It is not. That's why you don't see magnetic scalar potentials (except in more advanced books) $\endgroup$
    – FGSUZ
    Dec 26, 2019 at 15:46
  • 3
    $\begingroup$ @FakeMod you don't need to take the curl of the force, but rather the field (as you started with, but switched midstream). Just consider an infinite current, and it's pretty easy to take the curl around a loop concentrric with the current. Keep it up :) $\endgroup$
    – DeltaG
    Dec 27, 2019 at 0:30
  • 2
    $\begingroup$ @FakeMod yes, that's right, which is why you don't learn about one in HS or use it with cons. of E. There's some more info about issues with trying to define a scalar mag. potential and with what you really need to do (vector potential) in the answers below. It's a different beast from the electric field, which has a well-defined scalar potential, due to its irrotational nature. $\endgroup$
    – DeltaG
    Dec 27, 2019 at 5:42
  • 2
    $\begingroup$ You can define magnetic scalar potentials. but usually such a thing is useless. $\endgroup$
    – Shing
    Dec 28, 2019 at 11:31

5 Answers 5


Actually, we do!

It's just that it's not the same "kind" of potential - and the reason for this is that magnetic forces work differently than electric forces.

Magnetic fields, if you know, do not directly exert a force on charged particles, simply for being charged. (They would exert such on hypothetical "magnetically charged particles", but we've never found any to exist.) Rather, the force they exert does one thing only: to change the direction of motion of moving charged particles.

The usual thing we call "potential" is what you can think of as a kind of "specific potential energy": it is the potential energy that a unit quantity of charge has from sitting at a certain place in an electric field, and if the particle moves between two areas of different potential, it gains or loses energy as a result of the shifting - but ever-present - pull of the electric force upon it.

Magnetic fields, though, cause no changes in energy - changing something's direction of motion takes no energy, only speeding or slowing its motion does. Think about how that a bullet fired from a gun doesn't hurt more or less depending on the direction from which it comes, only from how powerful the gun is. In theory, to deflect the bullet from one to the other direction in flight, likewise, would not take any energy (though you would need a rather strong source of deflecting force).

But nonetheless, that doesn't mean you can't still describe them using something like a potential, but it doesn't have quite the same meaning any more. As I just see you mentioned you've tried some vector calculus, I'll give this a shot. You see, there's a kind of "duality", if one will, between two operations one can do with at least three-dimensional vectors: the dot product and the cross product, which gives rise to related differential notions of the divergence and gradient, vs. curl, respectively.

The usual ideal of a "potential", that is, for a field like the electric field (and also the Newtonian gravitational field), is based on the following result. "Under certain reasonable conditions", the following implication holds true. If $\mathbf{F}$ is some sort of force field, and

$$\oint_C \mathbf{F} \cdot d\mathbf{l} = 0$$

for all closed paths $C$, then there exists a scalar function $V$ (i.e. with 3 real spatial coordinate arguments and outputting one real number) such that

$$\mathbf{F} = -\nabla V$$

Intuitively, the first equation is a kind of "conservation of energy": the left hand integral is, in effect, a work integral if $\mathbf{F}$ is serving as a force field, describing the amount of energy gained or lost (positive is gain, negative is loss) by a particle moving in a closed circuit through that field as it is pushed and pulled by the exerted force. The above implication, then, says that "if the force field conserves energy, we can describe it by a potential energy". This is how you get the usual electric potential, which is the "specific" potential energy: energy per unit of charge, which in SI units comes to be joules per coulomb, which we call as "volts". Moreover, the first equation, "under certain reasonable conditions", corresponds to the one

$$\mathbf{\nabla} \times \mathbf{F} = \mathbf{0}$$

where the left-hand side is a differential operation called "curl", and intuitively represents the amount by which a vector field, thought of as force, locally *fails* to conserve energy.

Now, it turns out that, however, there is another, analogous form but involving this integral: if

$$\mathop{\vcenter{\huge\unicode{x222F}}}_S \mathbf{F} \cdot d\mathbf{S} = 0,$$

a surface integral over a closed surface $S$, then it follows that another, vector field $\mathbf{A}$ exists such that

$$\nabla \times \mathbf{A} = \mathbf{F}$$

which is very much like the relationship to potential and, indeed, we call this $\mathbf{A}$ a vector potential.

Again, we should think about the intuitive meaning of the first integral: this integral now is a flux integral - in effect, if you imagine the field as representing the stream lines of some fluid, i.e. if the vectors returned by $\mathbf{F}$ are mass-flow, i.e. mass per time, with direction of flow, vectors, the flux integral would represent the net amount of fluid flowing into or out of that space - and to set it to zero says that, in effect the field "conserves fluid": no new fluid is destroyed or created at any point. Analogously, this corresponds to a similar "local" statement per the divergence:

$$\nabla \cdot \mathbf{F} = 0$$

which, you may recognize, is exactly the equation satisfied by the magnetic field, $\mathbf{B}$:

$$\nabla \cdot \mathbf{B} = 0$$

and says there are "no magnetic sources", i.e. no magnetic charges. In a sense, magnetic "flux", which can be thought of as a kind of "fluid", swirls around magnetic objects but none is created or destroyed, and this conservation of flux gives rise to a magnetic vector potential, also typically denoted $\mathbf{A}$. This "potential" is a vector, not scalar, quantity - and this is the answer to your question. It doesn't represent energy, but more like "specific flux", I'd suppose, though it's hard to answer and, moreover, interestingly, is much less unique.

  • 5
    $\begingroup$ "Magnetic fields, though, cause no changes in energy" Wouldn't magnets cause ferromagnetic objects to have "magnetic potential energy" the same way that gravity wells cause objects with mass to have gravitational potential energy, due to the force attracting them to the magnet, though? $\endgroup$
    – nick012000
    Dec 28, 2019 at 5:50
  • 3
    $\begingroup$ @nick012000 : Good question. This is essentially the common question of "given that magnetic forces do no work, how then can you attract metal with a magnet, which clearly is doing work?" And the answer is that actually, it is electric forces that do the work in those cases. It would be complicated to describe exactly how in the comments - I believe though that there are answers given to this question elsewhere on this site, though, if you search for something like "magnetic forces do no work" etc. $\endgroup$ Dec 28, 2019 at 14:22

The potential is a kind of primitive function of a vector field, primitive in the sense of being the reverse of a differentiation, ie., an integral with a variable upper limit. The derivatives of the potential in all directions represent the vector field; not all vector fields can be represented that way but some do, for example the electrostatic field. Some other vector fields do not have such representation but one may define another type of differentiation with a corresponding primitive function but it is not a scalar potential and the magnetic field is such an entity. The field usually denoted by B has a primitive function called the vector potential. The mathematical difference between the electric and magnetic fields is that the electric field is an "along a line thing" while the magnetic fields is a "surface thing".

In the electrostatic case the total work done in any loop is zero from which it follows the existence of a potential function. In the case of the B field the total flux passing through any closed surface being zero is the reason for the existence of a vector potential such that the flux through any simple surface spanned by an arbitrary loop is the same as the loop integral of the primitive function, here vector potential integrated around the loop.

  • $\begingroup$ And how do we use this potential? Like, we often use electric and gravitational potential along with energy conservation to solve problems. Similarly, how do we use magnetic potential? Because the way it is defined I don't think it would be any useful while conserving energy. Please add the answers to these questions in your answer rather than posting a comment. It would help us avoid having a long discussion in commentsa and also enrich your answer. $\endgroup$
    – user243267
    Dec 26, 2019 at 14:07
  • 3
    $\begingroup$ Maxwell called it electrokinetic momentum and gave the letter A to show its primacy over other quantities... The vector potential A (not the magnetic field B) shows up directly in the Hamiltonian and is fundamental in QM and being a line thing not a surface thing analysis/calculation is easier with it. $\endgroup$
    – hyportnex
    Dec 26, 2019 at 14:14

There is a "magnetic potential" that appears in more advanced books and is defined as ${\bf B}= \nabla \phi$, just as ${\bf E}= -\nabla V$. (please excuse my equations if you are not familar with the $\nabla$ symbol) It's useful in regions where there is no current or no time-changing electric field, but it is "multivalued." Because of Ampere's law, if you encircle a current carrying wire, the potential decreases all the way round so when you get back to where you are started you have a different potential -- rather like the Escher prints of water-driven perpetual motion machine in https://en.wikipedia.org/wiki/Waterfall_(M._C._Escher). In many cases we can live with this multivaluedness because $\phi$ is not a potential energy (as is $V$), but is just a mathematical convenience.

This scalar $\phi$ is a simpler thing than the vector potential ${\bf A}$ (defined so that ${\bf B}=\nabla \times {\bf A}$) mentioned by hyportnex, but it is less useful.


A charged particle in a magnetic field maintains its kinetic energy. You can see this from the fact that $\vec{F}_B \propto \vec{v} \times \vec{B}$ which means that $\vec{v}\cdot\vec{F}_B=0$ and therefore the force by the magnetic field cannot preform any work on the particle $$ W_B = \int \vec{F}_B \cdot \vec{dx} = \int dt \vec{F}_B\cdot\vec{v} = 0$$ This means that there cannot be a corresponding conserving potential from which it will be derived.

  • $\begingroup$ But what about two magnets kept at a small distance without touching each other? They would either repel or attract, and in either of these cases, there would be an increase in the kinetic energy. Where does that kinetic energy come from? I suppose it is the energy store in the magnetic field which gets converted to kinetic energy. So , there is energy stored in magnetic fields and thus we even define magnetic field energy density, then why not define magnetic potential? $\endgroup$
    – user243267
    Dec 26, 2019 at 13:57
  • 1
    $\begingroup$ That is true, but a bit more complicated than that. In order to fully define this problem one has to consider both electric and magnetic fields, and the corresponding potential that describes both of them is called a vector potential. See the excellent answer of @hyportnex below $\endgroup$
    – user245141
    Dec 26, 2019 at 14:03
  • $\begingroup$ I see! You are talking about something like "Lorentz force potential", right? $\endgroup$
    – user243267
    Dec 26, 2019 at 14:08

There is the vector potential $\bf A$ for which ${\bf \nabla} \times {\bf A} = {\bf B}$. So $B_z = dA_x/dy - dA_y/dx$ and similar for other components.

The classical field theory of electromagnetism is based on the four potential $A^\mu = (\phi,{\bf A})$ . $\phi$ is the Coulomb potential .


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy