If you place the multimeter as you show in the image, what you get is the **voltage** drop in the resistance of the wire which is very small.

That's right. That's what we are going to measure. We are after

**voltages** here, because we need to ascertain if Kirchhoff's

**VOLTAGE** law holds for Lewin's circuit.

The induced EMF in the ring wire is cancelled by the induced EMF in the voltmeter probes.

OK. Let's suppose for a moment that the only thing going on in the wires (and probes, which are also wires) is the induced EMF. If instead of the meter, I hook up an arbitrary resistive load, for instance an LED, we won't have anything done there, because all the voltage that would be supposedly available by the EMF in the wire, will be cancelled by the leads of the LED, which will take the place of the probes. So for all intents and purposes, the voltage across the wire will be ideally zero, or, in practice, just the resistive voltage drop.

No wonder no one connects a load across a wire on a transformer. EDIT: the way the meter is connected in my previous message, of course.

Since the EMF is cancelled, as you say, for things connected in parallel with wires, let's see how it behaves for the resistors in series with them.

When we have a battery connected to a load, the voltages, and their respective electric fields, across the battery and the resistors are in opposite directions around the circuit. So, when we add them up, they will cancel each other out as dictated by Kirchhoff's VOLTAGE law. Or, in Kirchhoff's own words, the EMF (the voltage across the battery) will have to be equal to the current through the resistor times its resistance.

However, when we replace the battery with a wire and subject the entire circuit to a varying magnetic field, the electric field inside the resistor and inside the wire, and their associated EMFs, will be in the same direction around the circuit. So they do not cancel each other out in the equations. Quite the opposite, they'll accumulate. If I suppose that the induced EMF will have to be equal to the current through the resistor times its resistance, when I go around the loop I will be effectively counting them twice, because they have the same direction.

So it is pretty obvious that something else is going on. We need to have in the wire an electric field that has the opposite direction of the electric field in the resistor along the circuit. Just like we had, when there was only a battery. And we do. It is the electrostatic, or coulombic electric field that is generated by the displacement of charges produced by the induced electric field in the wire.

So now the puzzle is solved. What the wire is doing there is to produce the necessary coulombic field, which is conservative, by the way, to cancel the field in the resistor along the circuit in the equations, and as an aside, provide a return path for the current.

However we have a caveat. Since the coulombic field has the same magnitude as the induced field in the wire, and they are in opposite directions, they will cancel each other out, and the field inside the wire will be zero, rendering part of the induced EMF there as effectively zero. The "rest" of the EMF plus the coulombic field will be in the same direction inside the resistor and they will add up to be equal to the total induced EMF. That's a pretty cool mechanism that works very conveniently.

So yeah, the meter is telling the truth. The total EMF in the wire is just the residual voltage necessary to satisfy Ohm's law. KVL unfortunately for this case (or for this path, if you prefer) kills itself so as to make sense from the point of view of the fields. It survives, however, outside the magnetic field. But inside, for-fluxing-get it. The energy is obviously coming from whatever is generating the varying magnetic field, just like in a transformer or loop antenna, or any arbitrary loop (conductive or not) subjected to an "interference" ( which etymologically means bringing [something] into ).

EDIT: last image replaced, grammar and typos corrected.