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This Walter Lewin lecture hinges on the solenoid producing a non-conservative field.

I read in the Feynman Lectures that "all the fundamental forces in nature appear to be conservative" (Vol 1, 14.5) which comes decades after Faraday's Law of Induction so surely he would have known about it.

1) Who is right? Or, what did Feynman really mean?

2) In Lewin's final diagram there are two voltmeters which, apart from a section of wire, attach in the same place. What happens as you bring the two attachment points together to meet? I can't see, in a physical on-the-table experiment, how one voltmeter is "measuring in the other direction".

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You should not link videos without giving a point to forward to. In this case it's about 7 minutes in, and the example is a solenoid with a varying current acting as a transformer. This is not contradicting Feynman, obviously, who discusses induction and the nonconservative EMF too. –  Ron Maimon Jun 20 '12 at 17:33
Also of interest here: physics.stackexchange.com/questions/75349/… –  dmckee Nov 9 '13 at 16:47

3 Answers 3

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Feynman meant that conservation of energy always holds, so that if you have a static situation, the force field on a particle is conservative. For magnetic forces, you have moving (and changing) currents in the solenoid, so its not static, and if you extract energy from the field, you just weaken the current and extract energy from the system producing the field, doing work on it.

The fact that magnetically induced EMF is non-conservative is the basis of countless claims of perpetual motion machines, so it is good to say early that you can't do this.

Magnetic fields that are changing give rise to non-conservative forces, the integral around a loop is the change in flux inside, but the process of extracting energy from the EMF reduces the magnetic field, and the amount of energy stored in it.

Feynman discusses transformers and the EMF around a loop. He also discusses something else even more counterintuitive and not at all discussed by other people. He shows two moving charges, A and B, so that A is moving perpendicular to the line joining A and B and B is moving along the line joining A and B.

In this case, the force from A on B is not equal and opposite to the force from B on A! This shows you that the (nonradiative) field is carrying momentum, and is transferring momentum to the two charges as the E and B fields rearrange. The recognition that you need to include fields in the conservation laws was long in coming, and this example is just as useful as the transformer for explaining this. Feynman also discusses a case where the field is carrying angular momentum, a collection of charged balls with a current, and when you switch off the current, the balls start to rotated around.

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The phrase "conservation of energy always holds... in a static situation" seems a bit tautological. Can you point me to the sections in which Feynman which you are referencing? Thanks. –  spraff Jun 25 '12 at 12:08
@spraff: It is not tautological! I am saying that the forces (ignoring velocity dependent forces) are conservative, and you can ignore velocity dependent forces for nonrelativistic motions. I am not citing Feynman, I don't cite. I am explaining Feynman. The sections are scattered throughout vol II, and it's not that long and I don't own it. –  Ron Maimon Jun 27 '12 at 17:32
In a static situation nothing changes, so everything is conserved by definition, that's the tautology I meant. –  spraff Jun 28 '12 at 8:38
@spraff: I didn't mean totally static--- I meant ignoring velocity dependent forces. I know what you intended to say, but I didn't mean the tautology, but that slow adiabatic changes are allowed. This is the limit of nonrelativistic particle mechanics with no magnetic field. –  Ron Maimon Jun 28 '12 at 8:46

Regarding part 2) of your question:

In elementary circuit theory, you learn that when the leads of a voltmeter are connected together, the (ideal) voltmeter will read zero volts. This is true as long as there isn't a changing magnetic flux through the surface bounded by the leads.

So, to simplify the example in the video, imagine connecting the leads of the voltmeter on the right together with the leads enclosing the solenoid. In this case, the voltmeter will read the emf of 1V.

Now, connect the leads of the voltmeter on the left together with the leads enclosing the solenoid. You'll find that this voltmeter reads -1V.

Why? Because the direction around the loop is opposite that of the voltmeter on the right. To find the emf, we integrate the E field around a closed loop. Assuming the direction is into the positive terminal of the voltmeter, we integrate clockwise for the voltmeter on the right and counterclockwise for the voltmeter on the left. This is the reason the two voltmeters read equal but opposite values.

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So, if I pick up either voltmeter and swing it over to the other side, without moving the attachment points, the value will change? –  spraff Jun 25 '12 at 12:06
That's correct and it must be that way. Consider what happens as you "swing it to the other side". There is a point where the surface bounded by the loop formed by the voltmeter leads is parallel to the magnetic flux lines, i.e., there is no magnetic flux through the surface. In that case, the induced voltage must be zero. So, as you rotate the plane of the loop, the voltage falls from 1V through zero and finally to -1V on the other side. –  Alfred Centauri Jun 25 '12 at 19:50

The title of your question, it's a "“Complete Breakdown of Intuition”, refers to the reaction of EE and physics professors in Prof Lewin's department at MIT when he performed this experiment in front of them. Four minutes into part2 http://www.youtube.com/watch?v=1bUWcy8HwpM: "and some did not believe what they saw"... "some accused me of cheating on the demonstration".

The idea of a voltage between two points in a circuit only makes sense when magnetic fields are static because it's independent of the path taken. We can also say the voltage across the terminals of the voltmeter is the same as that across the ends of its leads from Kirchoffs law.

In non conservative circuits where there's changing magnetic fields from changing currents, the voltage between two points is path dependent and therefore meaningless. Neither can we say the voltage across the voltmeter terminals is the same as that across the end of its leads. We can only say the changing flux around the measuring loop induces an emf from Faraday's law, and therefore creates a current which creates a voltage across the resistance of the measuring terminals of the voltmeter, which is displayed by the voltmeter.

What's the voltage across the ends of an inductance L carrying a changing current i?

Most EEs and physicists would answer $Ldi/dt$. The true answer is close to zero volts because the inductance is constructed from a low resistance metal. The voltmeter across it displays the voltage $Ldi/dt$ dropped across its internal resistance from the induced current around the measuring loop.

All the above is explained more fully in:

What do ``voltmeters'' measure?: Faraday's law in a multiply connected region Romer, Robert H. American Journal of Physics, Volume 50, Issue 12, pp. 1089-1093 (1982).

A long solenoid carrying a varying current produces a time-dependent magnetic field and induces electric fields, even in the region exterior to the solenoid where ∂B/∂t and therefore curl E vanish. By paying attention to (a) what it is that a ``voltmeter'' measures and (b) the simplest properties of line integrals (e.g., under what circumstances the line integral of E is path independent), it is easy to use Faraday's law to predict the readings of voltmeters connected to various points in a circuit external to the solenoid. These predicted meter readings at first seem puzzling and paradoxical; in particular, two identical voltmeters, both connected to the same two points in the circuit, will not show identical readings. These theoretical predictions are confirmed by simple experiments.

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"The true answer is close to zero." It's true that the electric field in an inductor's wire is 0 (or very low), so a line integral of E along the wire gives 0, but you shouldn't conclude that the voltage across the inductor is 0; that would be neglecting Faraday's law! In a well-designed inductor, the magnetic field is confined within the device, and any measurement at the terminals with a voltmeter (or scope) will give the L di/dt result. –  Art Brown Nov 9 '13 at 18:12

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