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I am having much distress over Maxwell's 3rd Equation (Faraday's Law of Induction) and a thought experiment I had. Given that Maxwell-Faraday's equation is $$\oint E \cdot ds = -\frac{d\phi}{dt}$$And from the definition by HyperPhysics (emphasis mine),

The line integral of the electric field around a closed loop is equal to the negative of the rate of change of the magnetic flux through the area enclosed by the loop.

If this is the case, please consider the following scenario.

I insert a dense magnetic field into ONLY THE CENTER of a loop of wire (the magnetic field does not touch the actual loop). I was taught that Faraday's Law of Induction could be derived from the Lorentz Force on moving charges exposed to magnetic fields. However, as no magnetic field interacts with the charges in the wire (the field doesn't extend to the coil) there should be no EMF induced. But Maxwell's equations says there should be because there is a change in flux in the area of the loop.

I'm pretty sure Maxwell's equations aren't wrong, so could someone please explain what's wrong here? Does Maxwell's equation assume that the flux change is uniform through the entire area? That doesn't sound like an assumption that he would make, given the universality of his 4 equations.

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  • $\begingroup$ Haha. Coincidentally, this was the topic of debate between Electroboom and Walter Lewin on youtube. Go check out their videos, and I think Electroboom managed to summarise it up quite well, here: youtube.com/watch?v=Q9LuVBfwvzA $\endgroup$ – QuIcKmAtHs Dec 4 '18 at 13:11
  • $\begingroup$ @QuIcKmAtHs I've seen those videos and they're pretty amusing, but I think my question is different, because they both agree that there would be something induced, whereas I'm not sure about that part $\endgroup$ – John Hon Dec 5 '18 at 5:23
  • $\begingroup$ The crux of the matter is whether or not such a magnetic field can exist. If you have an infinite solenoid that's been on for an infinite amount of time, maybe, but such an object does not actually exist. $\endgroup$ – probably_someone Dec 7 '18 at 14:16
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Your assumption that there has to be a magnetic field interacting with the wire is wrong. It is not only the magnetic field what moves the charges, it is also an electric field. Inside the area in which the magnetic field is changing the rotor of the electric field is not zero. This creates a contour condition for the electric field, which results in a non zero value for it outside the region, even if the magnetic field is zero there. The magnetic part of the Lorentz Force on moving charges is only one of the components of the EMF, see https://en.wikipedia.org/wiki/Faraday%27s_law_of_induction#Proof

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Answering your question directly, no Maxwell's equations does not take the variation of the flux as uniform. Actually, you could choose any surface that has the loop as it's boundaries to think about this problem, and it could be a much more complicated surface then just a disk (the simplest, and the one we usually think first). And, in this case, we would not be able to justify the flux as varying uniformly. What Maxwell's equations equations tell us, is that whenever we have variation in time of the magnetic flux (doesn't matter it's origin or if it is uniform through the area enclosed by the loop or not), we will see the action of an electric field in the charges of the loop. So, in the end, as said above, it's an electric field that will move the charges, not the magnetic field directly.

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I insert a dense magnetic field into ONLY THE CENTER of a loop of wire (the magnetic field does not touch the actual loop). ... as no magnetic field interacts with the charges in the wire (the field doesn't extend to the coil) there should be no EMF induced. But Maxwell's equations says there should be because there is a change in flux in the area of the loop.

You are correct, there should be no EMF induced. However, you are making a mistake in your calculation of the flux. Maxwell’s equations hold perfectly well in this scenario, and agree with the EMF analysis. There is no change in flux.

To see how this works, it is easiest to think in terms of magnetic field lines. Recall that the field lines form continuous loops and that their density is proportional to the field strength. This means that your dense field consists of very small tight loops.

As you bring the magnet close to the loop the “forward” pointing lines begin to cross the area. However, because the field is so dense every forward pointing line is paired with a backward pointing line. So the net flux is zero.

In order to have a net flux you need to have some forward lines on the inside and backwards lines on the outside. That only happens if some of the field reaches the wire loop.

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  • $\begingroup$ not necessarily, think of a toroidal solenoid in which the magnetic lines stay inside the torus and do not necessarily "touch" the wire $\endgroup$ – Wolphram jonny Jan 5 at 4:45
  • $\begingroup$ What do you mean “not necessarily”. With respect to my answer a toroidal solenoid is an example not a counterexample. If you move a toroidal solenoid through the area you always have each forward line canceled by a backward line. $\endgroup$ – Dale Jan 5 at 12:46
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    $\begingroup$ I interpreted your answer as meaning that as the B field loop grows it crosses ("touches") the wire, and thus there must be a B over the wire. I believe now I missinterpreted you. In the toroidal solenoid the field never reaches the wire loop. $\endgroup$ – Wolphram jonny Jan 5 at 14:46
  • $\begingroup$ No problem, misunderstandings are easy! $\endgroup$ – Dale Jan 5 at 17:40
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Your distress is understandable, everybody who has ever studied the subject including Faraday was puzzled about this. Indeed, not the Lorentz force but the flux change rate what matters in induction. The Lorentz force law does explain induction where magnetic field intensity is not zero but it does not explain induction where the field is zero. You may have already learned that just as an electrostatic field that satisfies $\textrm{curl}\textbf{E}=0$ can be represented by a non-unique scalar potential $\textbf{E}=-\textrm{grad}{\phi}$ a magnetic (static or non-static) field for which one has always $\textrm{div}\textbf{B}=0$ can be represented by a (non-unique) vector potential $\textbf{B}=\textrm{curl}\textbf{A}$. The interesting consequence of this representation is that Faraday's law can be written as $$\oint \textbf{E}\cdot d\textbf{l} = -\frac{d}{dt}\oint \textbf{A}\cdot d\textbf{l}$$ which means that the induced EMF around any loop is just the time derivative of the loop of integral around the same contour of the vector potential. Now the A field is nowhere zero and can always be written directly as the following volume integral (radiation/propagation effects ignored):$$\textbf{A}(\textbf{r})=\frac{\mu_0}{4\pi}\int\frac{\textbf{J}(\textbf{r}')}{|\textbf{r}-\textbf{r}'|}dV(\textbf{r}')$$

Later you will also learn that the vector potential is of primary importance and it and not the B-field shows up directly in quantum mechanics.

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  • $\begingroup$ Can you physically generate a B that is changing perpendicularly inside a loop but is zero at the loop? $\endgroup$ – Wolphram jonny Dec 4 '18 at 16:23
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    $\begingroup$ for a tightly wound coil the B field along with its time rate outside are essentially zero, but neither the flux (and its rate) nor the vector potential and its rate would be zero. Check out the Aharonov-Bohm experiment where this is realized and the line integral of the A-field is measured as it influences the phase of the quantum mechanical wave function and hence the interference of the matter waves. $\endgroup$ – hyportnex Dec 4 '18 at 16:32
  • $\begingroup$ I understand that, in QM, A takes relevance, but from a purely classical viewpoint, will you have an EMF if you put a loop outside the coil? $\endgroup$ – Wolphram jonny Dec 4 '18 at 17:01
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    $\begingroup$ @Wolphramjonny That's a classic paradox. The answer is no, see here. $\endgroup$ – knzhou Dec 4 '18 at 18:24
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    $\begingroup$ @WOlframjonny I mentioned that the formula for A ignores propagation effects. Here is the full formula $\textbf{A}(\textbf{r},t)=\frac{\mu_0}{4\pi}\int\frac{\textbf{J}(\textbf{r}',t-r/c)}{|\textbf{r}-\textbf{r}'|}dV(\textbf{r}')$, where $r=|\textbf{r}-\textbf{r}'|$ and $t-r/c$ now takes into account the finite propagation time from $\textbf{r'}$ to $\textbf{r}$. $\endgroup$ – hyportnex Dec 4 '18 at 18:25

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