I've searched around for this but haven't come across a totally satisfying explanation yet. I'm trying to build a relatively simple model of how the Lorentz Transformations of the $E$ and $B$ fields can explain Faraday's Law, but I'm stumbling on one particular example -- B1 in the figure below.

enter image description here

In these sketches, I have a conducting loop moving in some magnetic field. In A1 and B1, the field is assumed to be purely magnetic. In B1 and B2, the magnetic field is non-uniform. According to Faraday's Law, an emf and current should be induced in the loop in each case.

In A1, the loop velocity is perpendicular to $B$, so the induced current is due to the magnetic force acting on charges in the loop ($q \vec{v} \times \vec{B}$).

A2 is A1 boosted to the frame of the loop. Here, there is no magnetic force on the charge carriers, but there is an electric force due to the electric field given by Lorentz Transforming the magnetic field of A1. The electric force creates the current. So far, so good.

B1 is what I'd like to have explained.

B2 is B1 boosted to the frame of the loop. I think this is similarly explained by saying that there is an electric force due to the electric field given by Lorentz Transforming the magnetic field of B1. Though I'm not entirely sure about this, because it seems like that would also be true if the magnetic field was uniform (in which case there should not be an induced current).

The reason B1 is confusing to me is because, by assumption, there is no electric field, and there is also no force due to the magnetic field since the velocity of the charge carriers is parallel to B. So what is inducing the current here?

  • $\begingroup$ Are you trying to draw B1 as a magnetic field that is getting stronger leftwards? That field violates Maxwell's equations and cannot exist as is without electric fields. In general, no good textbook will want to cover the topic this way, because the Lorentz transformation of the EM field is a tremendously complicated set of equations with a lot of difficulties. For example, you wrote that in A2 there is no magnetic force, but there must be: in A1 it is a pure B field, so there is no Lorentz transform to a pure E field, and the current is moving in a mixed field, so there must be magnetic force $\endgroup$ Commented Nov 21, 2023 at 9:27
  • $\begingroup$ @naturallyInconsistent I was intending B1 to be as you said, and I realized after posting that it would violate Gauss' Law. However, I think that issue can be circumvented with a field like $\vec{B} \propto \phi r \hat{\phi} - z \hat{z}$ in cylindrical coordinates. If I did my math right (very possible that I didn't), I think this would satisfy $\nabla \cdot B = 0$ and still produce no current since the magnetic force would be purely radial on the loop. Is there some other reason it can't exist? $\endgroup$ Commented Nov 21, 2023 at 10:41
  • $\begingroup$ @naturallyInconsistent Regarding the magnetic force in A2: certainly there is a B field there, but I think the magnetic force would still technically be 0 since we're in the frame where the charges in the loop are stationary. To be clear, I mean "initially", before the current is induced. After the current is induced there will be a magnetic force as you say. But the question is what causes this current initially. I believe it is the electric field in the loop frame, which is $\vec{E}_\text{loop} \approx \vec{v} \times \vec{B}_\text{lab}$ for $v \ll c$. $\endgroup$ Commented Nov 21, 2023 at 10:47
  • $\begingroup$ Related : my answer as 'user82794' (former 'diracpaul) here Why the induced field is ignored in Faraday's law?. This separation of the emf into the two parts, one due to the time rate of change of B and the other to the motion of the closed path, is somewhat arbitrary in that it depends on the relative velocity of the observer and the system (and in any case isn't Lorentz invariant). $\endgroup$
    – Frobenius
    Commented Nov 21, 2023 at 21:43
  • 1
    $\begingroup$ @Frobenius Thanks. It's clear to me that the emf due to $\partial_t B$ in A2 can be derived by Lorentz transforming A1, but I'm still confused on how to explain the corresponding effect for when there is no relative motion, but the magnitude of $B$ is increasing (like the solenoid example in my comment below LPZ's answer). Eg take A1, but instead of moving the loop, increase the strength of $B$. This can produce the same emf, but it doesn't seem like that can be related to A1 by a Lorentz transformation. Seems like 2 different phenomena that happen to obey the same equation. $\endgroup$ Commented Nov 21, 2023 at 22:55

1 Answer 1


Just a small detail, Faraday's law is about inducing electromotive forces (emf's), not currents. To make the link between current and emf, you'll need to go beyond Faraday's law and make some assumptions on the impedance of the loop.

For $B_1$, yes your field: $$ B = \rho\phi e_\phi -ze_z $$ satisfies $\nabla\cdot B = 0$. However, it is only locally defined; it does not define globally a magnetic field, since $\phi$ is defined up to an integer multiple of $2\pi$. For your argument, all you need is a globally defined magnetic field with no radial component. The simplest way is to think in terms of magnetic vector potential $A$ with only a radial component: $$ \begin{align} B_\phi &= \partial_zA_\rho & B_z &= -\frac{1}{\rho}\partial_\phi A_\rho \end{align} $$ You can take for example: $$ A_\rho = \rho z\cos\phi $$ to get: $$ \begin{align} B_\phi &= \rho\cos\phi & B_z &= z\sin\phi \end{align} $$ which is a well defined magnetic field.

Your previous argument still works, if the velocity of the loop is along $z$, the magnetic force is radial so there is no magnetic emf. You assume that there is no electric field, so no electric emf. Thus no emf. Alternatively, you can check for the flux. The magnetic field is stationary so you just need to look at flux variations due to the motion. The easiest way to calculate it is by looking at the flux through the disk, which is: $$ \Phi = \int B_z \rho d\rho d\phi $$ and $\Phi=0$ thanks to the $\sin\phi$ factor, in particular, it is constant. Thus Faraday's law checks out. As you can see, even if the magnetic field is non uniform, $\nabla\cdot B$ forces it to have a non changing magnetic flux, which is why Faraday's law is valid. In general, you can prove that: $$ \dot \Phi = -\oint (v\times B)\cdot dx $$ where the variation of the flux is due only to the motion of the loop in the stationary magnetic field, assuming that $\nabla\cdot B = 0$. Since by construction you want the RHS to be zero, your LHS will also be zero.

Btw, without the assumption $\nabla\cdot B = 0$, the formula becomes gets an extra term: $$ \dot \Phi = -\oint (v\times B)\cdot dx+\int(\nabla\cdot B)v\cdot d^2x $$ You can check that this is the case in your naive example $B = ze_z$.

Hope this helps.

  • $\begingroup$ This is great, thanks! It addresses the question I initially posed, so I've accepted it. But I guess the root of my confusion is: Is the statement that "changing magnetic fields induce electric fields" something that you can derive from Lorentz transformations? Specifically in the case where it's just the strength of the magnetic field that's changing. Say the field inside a solenoid whose current increases over time. It seems that Faraday's Law follows from SR when there's relative motion, but I can't see how it follows here, which is bizarre given that it's the same equation either way. $\endgroup$ Commented Nov 21, 2023 at 19:14
  • 1
    $\begingroup$ Glad it helped. For your underlying question, the answer is no. You cannot boost your way so that the magnetic field is static and the emf is solely explainable using the magnetic force. It’s rather the opposite, you’ll always typically have an electric origin and magnetic origin, but thankfully this decomposition is compatible with boosts. $\endgroup$
    – LPZ
    Commented Nov 22, 2023 at 11:59
  • $\begingroup$ Interesting. Compatibility is one thing but it seems remarkable that this one aspect of Faraday's Law cannot be explained in terms of the Lorentz force and boosts while all others can. Ie, in Faraday's 3 experiments as described by Griffith's Sec 7.2.1 (link), experiments 1 and 2 are explained entirely by the Lorentz force and relativity, whereas experiment 3 requires something "new" -- that a changing B field induces an E field. And yet, all 3 obey the same equation. $\endgroup$ Commented Nov 22, 2023 at 22:36
  • $\begingroup$ You should rather view it more as a coincidence. Mathematically, you can check that what I described is the generic case. Look no further than EM waves for example. $\endgroup$
    – LPZ
    Commented Nov 22, 2023 at 23:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.