# If magnetic force can't do any work, then how can we define a potential?

I am confused about the idea that magnetic forces do no work. If something has a potential, then it has energy and hence can do work. I am using the logic of electric fields for this reasoning. I have seen the mathematical proof of why magnetic fields can't do work but I am not convinced since I can't get the intuition.

• Are you thinking of magnetic scalar or vector potentials? – Emilio Pisanty Sep 4 at 11:00
• Magnetic vector that is B(vector) = Gradient X A(vector) where A is the magnetic potential. – Megha Sep 4 at 11:09
• Valid question since magnetic force is non conservative force – yuvraj singh Sep 4 at 11:18
• Okay, you've got something wrong, $\nabla\times\mathbf A = \mathbf B \ne \nabla\cdot\mathbf A$. – acarturk Sep 4 at 12:26
• @Timetraveler11 The vector potential $\mathbf{A}$ is not a "potential" in the sense that you're defining it. It's certainly like a potential, but it doesn't have all that much to do with energy and work. – probably_someone Sep 4 at 12:41

I am using the logic of electric fields for this reasoning

The magnetic field does not have a potential in the logic of electric fields. In fact, by the logic of electric fields the magnetic field is not even a force field since the magnetic field is not proportional to the force.

The “potential” that a magnetic field has is a vector potential, which is a fundamentally different concept than the potential of the electrostatic field. The scalar potential is directly proportional to the electrostatic potential energy, which is a scalar. The vector potential cannot be proportional to the potential energy since it is not a scalar. So “using the logic of the electric field” simply doesn’t work for the magnetic field.

However, per Poynting’s theorem the magnetic field does have an associated energy density. That field is involved in storing electromagnetic energy at a particular location and also in moving electromagnetic energy from one location to another, but it is not clearly involved in doing work on matter. The term that governs work on matter is $$\mathbf E\cdot \mathbf J$$. But, it is important to realize that these quantities are all closely interrelated. It is possible to write $$\mathbf E\cdot \mathbf J$$ in a way that involves the magnetic field as well.

My favorite derivation of Poynting’s theorem is found at section 11.2 here: http://web.mit.edu/6.013_book/www/book.html In the absence of magnetizable or polarizable material the equation is:

$$-\nabla \cdot (\mathbf E \times \mathbf H) = \frac{\partial}{\partial t} \left(\frac{1}{2}\epsilon_0 \mathbf E \cdot \mathbf E\right) + \frac{\partial}{\partial t}\left(\frac{1}{2} \mu_0 \mathbf H \cdot \mathbf H\right) + \mathbf E \cdot \mathbf J$$

Where the term on the left is the flow of energy from one region of the field to another and the first two terms on the right are the change in the energy density of the electric and magnetic fields respectively. Those are purely field terms; the only term involving an interaction with matter is the last term $$\mathbf E \cdot \mathbf J$$ where $$\mathbf J$$ is the free current. This means that the only way that work can be done on a conductor is via the $$\mathbf E$$ field.

This holds as long as Maxwell's equations hold. So whenever you have a situation that looks like there is a magnetic field doing work, you can always "dig deeper" and find where it is actually the $$\mathbf E$$ field doing the work.

The term $$\mathbf E \cdot \mathbf J$$ includes not only the mechanical work, but also the non-mechanical work. Usually we want to maximize the mechanical work and we want to minimize the non-mechanical work. So suppose, instead of trying to find where the $$\mathbf E$$ field is, we try to separate out the non-mechanical work from the mechanical work.

To do so, we will transform to the rest-frame of the conductor, since in that frame there is no mechanical work so all of the work in that frame is non-mechanical. Assuming that $$v\ll c$$ the transformation equations are: $$\mathbf E' = \mathbf E + \mathbf v \times \mathbf B$$ $$\mathbf J' = \mathbf J - \rho \mathbf v$$ where the primed quantities are quantities in the rest frame of the conductor. Substituting those into $$\mathbf E \cdot \mathbf J$$ and simplifying, we get: $$\mathbf E \cdot \mathbf J = \mathbf E' \cdot \mathbf J' + \mathbf v \cdot (\rho \mathbf E + \mathbf J \times \mathbf B)$$

So, that means that the $$\mathbf E \cdot \mathbf J$$ term itself contains within it the mechanical work due to the magnetic field. If you pull out the non-mechanical work, then the mechanical work is exactly what you would expect from the macroscopic Lorentz force law, including a term from the magnetic field.

This result may seem a little surprising or confusing, as it appears to contradict the above, but it does not. The thing is that all of the fields in electromagnetism are closely related to each other. You can often express the same thing in multiple ways, or dig out hidden dependencies. So although Poynting's theorem holds and although it clearly states that the total work is always $$\mathbf E \cdot \mathbf J$$, it is not a mere coincidence that formulas describing only the mechanical work correctly include the $$\mathbf B$$ field and show that it does mechanical work.

• This seems like something I should know (and maybe I do know it under a different name, or maybe I am just having a "brain fart"), but what exactly is the distinction between mechanical and non-mechanical work? – Aaron Stevens Sep 4 at 13:17
• Mechanical power is $F\cdot v$. It is the work done in the sense of Newton’s laws. Non-mechanical work is other forms of energy transfer, such as Ohmic heating or charging a battery. On a microscopic scale there is no distinction, so non-mechanical work could also be classified by work done on a microscopic scale. – Dale Sep 4 at 13:24
• Ok. Yes I agree with all of that. Thanks! – Aaron Stevens Sep 4 at 13:24
• Thanks, this makes sense now! – Megha Sep 5 at 6:41
• Would you agree with the statement that in case of "no electric Field" the Energy that the conductor gains through the accelleration in the magnetic field stems entirely from the electrons inside being slowed down? Your formula (whose derivation I really like) suggests that. – Quantumwhisp Sep 5 at 20:23

You need to understand that units checking out is just a necessary condition. One can define infinitely many different vector fields over $$\mathbb R^3$$ that has the units of potential. But, this does not imply any of these must be a potential fields. Check this question in math.SE for a mathematical perspective.

Specifically for question in hand, since $$\nabla\times\mathbf A = \mathbf B \ne 0$$ (assuming it is not null), so it follows that this specific field cannot serve as a potential.

Idea of the potential is tied to conservative fields (path independence). Other vector fields don't admit potentials.

The vector potential describes potential momentum per charge unit. A better name would be "momentum potential". The confusion arises when one tries to interpret this quantity as another energy potential.