Say we have a non-uniform magnetic field that is static in time. Specifically, let's make it:

$\overrightarrow{B}(x,y,z) = x \hat{z}$

Now say we have a metal loop in the xy plane which has a non-zero velocity in the +x direction. Say we pick a random point in time, and then at that point in time measure the line integral of the E-field around the metal loop. The derivative of the magnetic flux through the metal loop is non-zero, so by Faraday's law the line integral of the E-field around the metal loop will be non-zero.

Now imagine a different scenario where there is no metal loop, but we decide to measure the line integral anyway along the exact same closed path that we did in the previous scenario. The B-field is constant, so by Faraday's Law we will get zero.

I was wondering if anyone knew the answer to this apparent contradiction. Does the presence of a moving metal loop make the E-field different from what it would be with no moving metal loop? The metal loop is neutral, so you wouldn't think that would be possible.

• Comment to the question (v1): Note that $\nabla\times B\neq 0$. How do you plan to satisfy Maxwell-Ampere's law? – Qmechanic Oct 6 '14 at 18:42
• @Qmechanic: in principle, we could be inside some region of nonzero current density, I guess. – Jerry Schirmer Oct 6 '14 at 21:18
• Am I missing something? In the first situation the loop is moving, in the second it is stationary. Forget about the metal/wire loop, I think it's immaterial. Ask instead about the EMF around an imaginary loop. All of the classic "moving loop" problems would manifest a contradiction if you stopped moving the loop and identified it as "a point in time of a moving loop". – garyp Oct 7 '14 at 3:29

I'm pretty sure that the answer to this is that we're cheating by saying that there's an E-field in the loop in the first case.

Well, we're "cheating" in a very narrow sense of the word. What we're doing is implicitly Lorentz transforming to a reference frame where the loop is stationary. If we do this, then the magnetic field at x = 0 becomes time-variant, and we get a manifest electric field, that can drive a motional EMF.

Why do we do this? Because this description is simpler than the one we would have to make by doing an analysis based on pure magnetic fields pushing charge carriers around in the loop. We can simply say, "hey, we've got an e-field, this pushes electrons natively," and be done.

• I think you're right, ultimately this might boil down to incorrect intuition on my part regarding the Lorentz transforms for electric and magnetic fields. Intuitively I would think that the measured line integral of the E-field in two different frames will be very nearly identical as long as the relative velocity of the two frames isn't close to c. However my 'contradiction' seems to demonstrate that this in fact is not true: In this specific scenario the measured line integral around the metal loop must scale linearly with the velocity of the frame you are in. – Eric L. Oct 7 '14 at 1:20
• Because the higher your velocity in the +x-direction, the larger the time derivative of the B-field will appear in that frame. – Eric L. Oct 7 '14 at 1:26

(This is written in a bit of a rush so I will amplify and or correct later).

The derivative of the magnetic flux through the metal loop is non-zero, so by Faraday's law the line integral of the E-field around the metal loop will be non-zero.

I don't believe this is is a valid conclusion.

Assume, for simplicity, that the metal is an ideal conductor. If we integrate the E-field around the metal loop, the result will be zero.

This must be the case since there can be no E-field within an ideal conductor.

But, that implies that the rate of change of magnetic flux through the surface bounded by the metal loop is zero.

Thus, we conclude, the must be a changing current through the metal loop that induces a opposite changing magnetic flux through the surface such that the net rate of change of magnetic flux through the surface is zero.

is it true that the Lorentz transforms for the E & B fields can differ greatly from galilean transforms even if the relative frame velocity is much lower than c?

In your setup, there is only a non-uniform magnetic field in the unprimed (lab) frame of reference:

$$\vec B = x\hat z$$

But, in the primed frame of reference with uniform relative speed $v$ in the $\hat x$ direction, there is both an electric and magnetic field:

$$\vec B' = \gamma_vx\hat z' = \gamma^2_v(x' + vt')\hat z'$$

$$\vec E' = -\gamma_v vx\hat y'= -\gamma^2_vv(x' + vt')\hat y'$$

The primed electric field is non-conservative:

$$\nabla \times \vec E' = -\gamma^2_v v \hat z' = -\frac{\partial \vec B'}{\partial t'}$$

Thus, the line integral of $\vec E'$ around a closed loop in the $x'y'$ plane yields an emf of:

$$\mathscr E = -\gamma^2_vvA'$$

where $A'$ is the area, as measured in the primed frame, enclosed by the loop.

When $v \ll c$, we have

$$\mathscr E = -vA' = -vA$$

To check this, let's look at the electromagnetic invariant:

$$||\vec B||^2 - \frac{1}{c^2} ||\vec E||^2 = ||\vec B'||^2 - \frac{1}{c^2} ||\vec E'||^2$$

which evalutes to

$$x^2 = \gamma^2_v(x' + vt')^2$$

which is indeed true.

• Interesting point about the ideal conductor! Maybe we should call it a wire loop instead of a metal loop, because we know that all wires have some small resistance, and also can have internal electric fields. I distinctly remember learning in my intro E&M course that a wire loop moving through a non-uniform B-field would have an emf, or non-zero line integral. Continuing from my comment to Jerry, is it true that the Lorentz transforms for the E & B fields can differ greatly from galilean transforms even if the relative frame velocity is much lower than c? – Eric L. Oct 7 '14 at 1:38
• @EricL., I've added to my answer to partially address your comment. – Alfred Centauri Oct 7 '14 at 23:00
• thanks, that was very helpful. I found the E&B transforms on wikipedia en.wikipedia.org/wiki/… and they agree with what you put, so I'm sure that you are right. However, there's one thing still bothering me: In the primed frame the B-field is static time and also the loop is stationary, so wouldn't this again contradict Faraday's law because we have an emf without a magnetic flux derivative? – Eric L. Oct 8 '14 at 16:04
• @EricL., the magnetic field is time dependent in the primed frame since the curl of $\vec E$ is non-zero. Remember, the origin of the primed system is located at $vt$ in the lab frame thus, there is a steadily increasing magnetic field in the primed frame. – Alfred Centauri Oct 8 '14 at 23:46
• @EricL., I've updated my answer. – Alfred Centauri Oct 9 '14 at 1:10

The case here is that, for this situation, we have an emf in the loop that is not created by an electric field. The definition of the emf along a closed circuit $C$ is $$\mathcal{E}=\oint_C\mathbf{f}\cdot d\mathbf{l},$$ where $\mathbf{f}$ is the force per unit charge in the loop. It may be the case that this force is done by a non-conservative electric field, but that's not what happens in the situation you are describing. In your situation, the emf is motional and the work done on the charges is done by the force that is pulling the loop, since the magnetic force done in the loop would slow it down if there wasn't any other force to keep the velocity constant. You may want to check sections 7.1.2, 7.1.3 and 7.2.1 from Griffith's Introduction to Electrodynamics. The discussion there is realy elucidating and clear about these subtleties.

From the comments of @Alfred Centauri and @Mateus, I think I've figured it out: In the stationary frame the B-field is static so there is no E-field, but we still measure an emf because the B-field is acting on the electrons in the wire by the Lorentz force law. In the primed frame the loop appears to be stationary, so the emf comes instead from the non-conservative E field that appears from the Lorentz transformations of the E & B fields. Since v << c, the emf measured in the primed frame will be about the same as the emf measured in the stationary frame, so to get the emf in the stationary frame we can just calculate the emf in the primed frame and use that as the answer.