There are many posts on this forum asking whether motional emf is actually an instance of Faraday's Law -- because, confusingly, it is often taught as though it is, with no qualification. The best answer I've seen so far is this really excellent one by @Timaeus. But I would like to go one step further conceptually. I want to know why it SHOULD be possible to lump both phenomena into the single relation (quoting from the linked answer):

$$\oint{(\vec E + \vec v \times \vec B)\cdot \text d\vec\ell} = -\frac{\text d}{\text dt}\iint{\vec B \cdot d\vec A}$$

In other words, it seems like a total coincidence that Faraday's Law just so happens to be unifiable with motional EMF into what @Timaeus calls the "universal flux rule". And in the posts I've seen, they make a point of carefully distinguishing between the two phenomena, since after all they're trying to untangle the confusion engendered by the textbook. Nevertheless, there are usually some deeper, more general considerations at play behind such coincidences in theoretical physics.

I'm guessing it has to do with the special relativistic formulation of E&M, since $\vec E$ and $\vec B$ are combined there into the Faraday tensor / Maxwell bivector $\mathbf{F}$, and the Lorentz force law basically just says that a charge's 4-velocity gets rotated in a given $\mu\nu$ spacetime plane(s) at a rate proportional to $F_{\mu\nu}$. Or perhaps it's enough to invoke the symmetry of Maxwell's equations somehow. Or maybe it's related to the fact that you can derive the Lorentz force law from Maxwell's equations + energy conservation (something I've heard can be done, but never actually seen). But I really don't know, and I feel there is something interesting to be comprehended here!

EDIT: Inspired by both of the answers, I read a bit more about the Lorentz force and found -- and I think this is clear from @Hyperon's answer anyway -- that the macroscopic form of Faraday's Law, in which the loop used for flux is allowed to change over time, actually implies the Lorentz force law. So I guess that leaves me wondering about the primacy of the differential vs. integral versions of Maxwell's equations. I always thought they were equivalent, and if anything I would have said the differential ones were more in the spirit of modern physics, but now I learn that the integral versions have this added benefit of logically absorbing the force law, so that electrodynamics is entirely described by Maxwell's equations alone. Is this an accurate statement? If so, why isn't this stated more clearly in typical treatments of E&M?

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    $\begingroup$ This is actually very profound. Einstein asked the same question on the way to creating his Relativity theory. Se the first paragraph at fourmilab.ch/etexts/einstein/specrel/www $\endgroup$
    – John Doty
    Mar 28, 2023 at 13:33
  • $\begingroup$ @JohnDoty agree. Addition: I think feymans lecture describes the formularea bit (also shows that not arbitraty shape changes lead to current). Only after one has asked oneselfs these questions it makes sense to read it. feynmanlectures.caltech.edu/II_17.html $\endgroup$
    – lalala
    Mar 28, 2023 at 15:22
  • $\begingroup$ @JohnDoty Thanks, that puts the whole thing into perspective. And if you're interested, I added another answer below showing how I was finally able to grasp it in purely relativistic terms. $\endgroup$ Mar 29, 2023 at 2:49

3 Answers 3


Consider an arbitrary parameter-dependent vector field $\mathbf{B}(t, \mathbf{x})$ (if you wish, you may interpret the parameter $t$ as "time") and an arbitrary (orientable) two-dimensional (in general parameter-dependent) manifold $F(t)$ with boundary $\partial F(t)$. Under certain differentiability conditions (usually fulfilled in physical applications), the following purely mathematical theorem holds: $$\begin{align} &{}\frac{d}{dt} \int\limits_{F(t)} \! d \mathbf{A} \cdot \mathbf{B}(t, \mathbf{x}) = \\ &{}\! \!\int\limits_{F(t)} \! d \mathbf{A} \cdot \frac{\partial \mathbf{B}(t, \mathbf{x})}{\partial t}+ \!\!\int\limits_{F(t)} \! d \mathbf{A} \cdot \dot{\mathbf{x}}(t) \, \mathbf{\nabla} \!\cdot \mathbf{B}(t, \mathbf{x})- \!\!\! \!\int\limits_{\partial F(t)}\! \! d \mathbf{x} \cdot \left(\dot{\mathbf{x}}(t) \times \mathbf{B}(t, \mathbf{x}) \right). \tag{1} \label{1} \end{align}$$ This formula can be seen as the three-dimensional generalization of the well-known formula for the differentiation of a one-dimensional parameter integral, $$ \frac{d}{dt} \! \int\limits_{a(t)}^{b(t)} \! \! \! dx \, f(t,x) =\int\limits_{a(t)}^{b(t)} \! \! \! dx \, \frac{\partial f(t,x)}{\partial t} + \dot{b}(t) f(t,b(t)) -\dot{a}(t) f \left(t,a(t) \right). \tag{2}$$

In electrodynamics, eq. \ref{1} can be applied to the magnetic field $\mathbf{B}(t, \mathbf{x})$, fulfilling the Maxwell equations $$ \mathbf{\nabla} \cdot \mathbf{B}(t, \mathbf{x})= 0, \qquad \frac{\partial \mathbf{B}(t, \mathbf{x}) }{\partial t} = - \, \mathbf{\nabla} \times \mathbf{E}(t, \mathbf{x}). \tag{3} \label{3}$$ Using the integral theorem of Stokes, the first integral on the right-hand-side of eq. \ref{1} can be written as $$ \int\limits_{F(t)} \! d\mathbf{A} \, \cdot \frac{\partial \mathbf{B}(t, \mathbf{x})}{\partial t}= - \int\limits_{F(t)} \! d \mathbf{A} \, \cdot \left(\mathbf{\nabla} \times \mathbf{E}(t, \mathbf{x}) \right) = - \! \!\!\int\limits_{\partial{F}(t)} \! d \mathbf{x} \cdot \mathbf{E}(t, \mathbf{x}), \tag{4} $$ yielding the final result $$ \frac{d}{dt} \! \!\int\limits_{F(t)} \! d \mathbf{A} \cdot \mathbf{B}(t, \mathbf{x}) = - \int\limits_{\partial F(t)} d \mathbf{x} \cdot \left( \mathbf{E}(t, \mathbf{x}) + \dot{\mathbf{x}} \times \mathbf{B}(t, \mathbf{x})\right), \tag{5} \label{5}$$ being the simple reason for the definition of the "electromotive force". No big mystery, no wires or other unnecessary confusion. If you wish, you may, of course, rewrite the integral form of Maxwell's equations also in manifestly covariant form (using the field strength tensor $F_{\mu \nu}$), but you will not gain much additional insight concerning the concept of the electromotive force.

Additional answer to your EDIT:

This is a misinterpretation. The term $\mathbf{E}+ \dot{\mathbf{x}} \times \mathbf{B}$ in eq. \ref{5} is NOT the Lorentz force. $\dot{\mathbf{x}}$ in eq. \ref{5} is NOT the velocity of a charged particle but the velocity of a point on the time-dependent (moving) curve $\partial F(t)$. Parametrizing $\partial F(t)$ by $\lambda \to \mathbf{x}(t, \lambda)$, you have to insert $\dot{\mathbf{x}}= \partial \mathbf{x}(t,\lambda)/\partial t$ and $d \mathbf{x} = d \lambda \, \partial \mathbf{x}/\partial \lambda$ in the path integral. Note that eq. \ref{5} holds also in the absence of any charges, being a consequence of the two homogeneous Maxwell equations \ref{3} only.

The Lorentz force has a different origin. In the Heaviside system with $c=1$, the (Lorentz- and gauge-invariant) action integral of the electromagnetic field and (for simplicity) a single particle with mass $m$ and charge $q$ is given by $$ S = - \frac{1}{4} \int \! d^4 x \, F_{\mu \nu} F^{\mu \nu} -m \int \! ds -q \int \! dx_\mu \, A^\mu, \tag{6} \label{6}$$ where $F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$ (being equivalent to the homogeneous Maxwell equation $\partial^\mu \varepsilon_{\mu \nu \rho \sigma}F^{\rho \sigma}=0$). The equation of motion of the charged particle, $$ \frac{d p^\mu}{ds}= q F^{\mu \nu} u_\nu, \qquad u_\nu = \frac{d x_\nu}{ds} \quad \Rightarrow \quad \dot{\mathbf{p}}= q (\mathbf{E} + \mathbf{v} \times \mathbf{B}), \tag{7} \label{7} $$ is obtained by varying the action with respect to the worldline of the charge particle, where $\mathbf{v}$ in \ref{7} is now the velocity of the particle. Finally, the inhomogeneous field equations $\partial_\mu F^{\mu \nu} = j^\nu$ ( where $j^\nu$ is the 4-vector current of the charge particle) are obtained by varying the 4-vector potential $A^\mu$.

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    $\begingroup$ This is very helpful, I wasn't even aware of equation (1). And it allows me to make the question even more precise, because there's still something I'm missing: WHY should electrodynamics (Maxwell's equations + Lorentz force law) be of just the right form that your 1st and 3rd terms in equation (1) just so happen to be the electro-motive and magneto-motive forces around a loop, respectively? For example, why couldn't the magnetic force law be of some other form than $\vec v \times \vec B$, in which case this wouldn't hold? Doesn't it seem like too much of a coincidence??? $\endgroup$ Mar 27, 2023 at 18:59
  • $\begingroup$ @Adam Herbst I have added some futher explanations in my answer. $\endgroup$
    – Hyperon
    Mar 28, 2023 at 6:00
  • $\begingroup$ 5) is not a reason for definition of "electromotive force", it is just a mathematical theorem following from the Maxwell equations, relating known kind of electromotive force due to total EM field (and motion of conductor in it) to change of magnetic flux. Electromotive force per unit length in conductor due to EM field is known to be $\mathbf E + \mathbf v \times \mathbf B$ as generalization of experiments - it gives the correct current in the Kirchhoff second circuital law. $\endgroup$ Mar 28, 2023 at 9:17
  • $\begingroup$ @AdamHerbst there is no "magnetomotive" or magnetic force law in 1) or 5). All the forces per unit charge there are electromotive forces, including the motional EMF, which has value $\mathbf v\times \mathbf B$ per unit length. This is electromotive force because it acts on and can change electric current. $\endgroup$ Mar 28, 2023 at 9:25
  • $\begingroup$ In 6) the $A$ term is discontinuous on the point particle trajectory, and the $FF$ term is divergent. The whole expression is unfortunately meaningless. There is a valid Lagrangian for point particle theory, using adjunct fields, which has all terms integrable. This was studied by Tetrode, independently Frenkel and then later by Feynman & Wheeler. $\endgroup$ Mar 28, 2023 at 9:40

Let's say you have a small loop sitting right at the end of a solenoid enter image description here

The following two situations should induce the same current in the loop:

  1. The solenoid's current is turned down
  2. The solenoid is moved left, further away from the loop, also decreasing the $B$ flux through the loop.

Because in both cases, faraday's law tells you what EMF is generated in the loop. All the loop sees is that the flux through it has decreased, so a curling $E$ field is generated that starts a current. But then the following two situations are also equivalent, and probably more obviously:

  1. The solenoid is moved left, further away from the loop
  2. The loop is moved right, further away from the solenoid

After all, these two things are the same event, but described in different reference frames.

And as you hinted, using the special relativity transformations of $E$ and $B$ fields helps us see another way this operates. When the magnet is turned off or moved away, a curling $E$ field is produced. When the loop is moved away, the motion of charges in the loop relative to the $B$ field makes an $v \times B$ force. But if we take that situation and go to the loop's reference frame, we get a curling E field of magnitude $v \times B$ (assuming $v\ll c$ so $\gamma\approx 1$)

I recently relearned this because I had a similar question - when you have an electron moving in a magnetic field it moves in circles perpendicular to the magnetic field (cyclotron motion). If the magnetic field is decreased slowly, the "magnetic moment" (frequency times area of orbit) is conserved. This is true whether the electron itself moves to a higher magnetic field, or the magnetic field is turned up or down. Like you, I saw this as a complete coincidence - in one case an electric field changes the area of the orbit, and in another case it's an inward pointing magnetic field that allows the field strength to change while maintaining $\nabla\cdot B=0$.

In fact, it doesn't matter whether it's a loop, or an electron, or a bomb that's set to explode when it enters a magnetic field of exactly 1T. In any case, it fundamentally cannot matter whether it is moved from a lower magnetic field to a higher one, the magnetic field is adjusted by changing a current, or if a magnet is moved toward it. The arguments that establish the equivalence of these three things are absolute. All the object in question sees - in its own reference frame - is a magnetic field going through it that has changed, and Faraday's law says an EMF must have also been generated.

  • $\begingroup$ This is definitely helping -- but what about the case of a conducting rod that is oriented along the x-axis, and moving along the y-axis through a uniform, unchanging magnetic field along the z-axis? We still get a motional emf due to the Lorentz force in this frame, but when you transform to the rod's frame, there is no changing magnetic field. Rather, I suppose $\vec B$ transforms directly to $\vec E$ via special relativity. But again, I'm trying to grasp this from a unified perspective, so using Faraday's Law for one example and SR transforms for another doesn't quite satisfy... $\endgroup$ Mar 27, 2023 at 19:40

I think the provided answers pretty much work, but I also want to share the conclusion I've reached after thinking more about this issue in relativistic terms. In relativity, Faraday's Law and Gauss' Law for magnetism are combined into the equation

$$\text{d}\mathbf{F} = 0$$

with $\text{d}$ being the exterior derivative. Since $\mathbf{F}$ is an antisymmetric tensor / bivector, it has one component for any given plane. And the vanishing of $\text{d}\mathbf{F}$ means that, given a bounded 3D hypersurface $V$ within spacetime, bounded by $\partial{V}$, there can be no tendency of $\mathbf{F}$ to "wrap around" $V$, that is, when you surface-integrate its component parallel to $\partial{V}$, you get zero.

Let $V$ include time, e.g. $V$ may be in the $xyt$ subspace, for $z$ a constant. Also let $V$ have "end-caps" along the $t$-direction, that is, its temporal cross-section does not taper to zero, but evolves from one finite area to another. These end caps are then the initial and final shapes of the loop, and the rest of $\partial{V}$ corresponds to the time-evolution of the boundary of the loop $\partial{F}(t)$ mentioned by @Hyperon.

So consider what the surface integral of $\mathbf{F}$ over $\partial{V}$ is. Over the end-caps, it is simply the magnetic flux through the loop at the start and end times, ie. $\Delta\Phi_{B}$. And over the loop trajectory, we are integrating the component of $\mathbf{F}$ that is tangent to the loop trajectory.

Then again, as I said earlier, the Lorentz force law says that the 4-momentum of a charge is rotated in any spacetime plane at a rate given by the component of $q\mathbf{F}$ in that plane. Meanwhile, we know the trajectory of any charge must lie within the loop trajectory. Hence, the component being integrated over the loop trajectory is indeed the force-per-charge along the loop -- and the particular orientation of the tangent plane automatically accounts for the current velocity of the wire. (A charge may also be accelerated perpendicular to this plane, but that will be a constraint force that keeps it in the wire, whereas our tangential component is the electromotive force that moves it along the wire.)

Thus the ambiguity between the differential and integral forms of Maxwell's equations is removed by the fact that the exterior derivative necessarily treats arbitrary bounded hypersurfaces in spacetime, regardless of how their cross-sections may evolve.

So I guess the differential form of Faraday's law alone reduces the exterior derivative of a 2-form to that of a 1-form, or something like that, losing some generality in the process, as it cannot possibly encode the time-evolution of the 1D loop about which the exterior derivative is calculated.

In any case, I stand corrected: the most faithful rendition of Faraday's law does put the time-evolution of the loop on the same footing as that of the magnetic field, so I will no longer worry about which line of reasoning to apply to such problems.


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