Apparent Contradiction with Faraday's Law

Question

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.

Answer 1

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.

Answer 2

(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.

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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.

Answer 3

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.

Answer 4

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.