I'm not sure I understood all your points. I suggest you to read this beautifull paper

- Romer, R. H. (1982).
*What do “voltmeters” measure?: Faraday’s law in a multiply connected region*, American Journal of Physics, 50(12), 1089. http://dx.doi.org/10.1119/1.12923

if you can find it. It's seems it's exactly what Prof. Levin is doing in his lecture. The proof is clear. If you have some difficulties to obtain this paper, I have some notes about this paper that I can share on SE too.

**Edit:** If you can find this reference, you will see that the only thing which matters is the position of the circuitry relative to the solenoid. More explicitly, this is the *topology* of the circuit which matters. It will then be clear for you that

- you would record nothing if you don't enclose the solenoid, subsequently you would measure nothing until you close (
*i.e.* make a turn) the circuit,
- putting your wires far away from the solenoid does not change the measured voltage, nor putting them close to he solenoid, the only thing which matters is the previous point,
- doubling the number of turn around the solenoid doubles the corresponding voltage,
- the wires participate for nothing in the measurement except for encircling the solenoid (as stated in the previous points). This last point is true only for idealised conditions of course (infinite solenoid, no resistive effect in the wire, ...).

Now, some of your questions:

There is a famous law which says that a potential difference is produced across a conductor when it is exposed to a varying MF. But, how do you measure it to prove? It is quite practical.

You have two ways to see induction, as clearly discuss in the Feynman lecture. To be honest, I do not know a better book to start with.

you close a loop with a moving bar, whereas a coil was previously inside the loop. Then you record a voltage drop through Faraday's law because the circuit is moving (if you prefer, the path you calculate your integral with).

you close a loop and encapsulate a time dependent magnetic field via time dependent current passing through the coil. Then you record a voltage drop because of the time variation of the magnetic field itself.

**In both case you have a time varying flux.** As Prof. Levin says: "Misterrr Farrraday is happy with that !"

Look at dr. Levin's voltmeter. What does it measure? I ask this question partially because he tells about non-conservative measurements, which depends on the path, but does not explain how to set up the paths to measure -0.1 v in one case and +0.9 v in the other, between the same points.

Yes he does ! It is always from top to bottom (A to D point if I remember correctly), first passing on the left, second time passing on the right. **Edit:** The previous statement was confusing. When discussing magnetic flux, you need to define convention for following the circuit path, *i.e.* rotation direction. The path for -0.1 V is the counterclockwise path from A to D, the path giving +0.9 V is the clockwise path from A to D. That explain the signs also.

How the typical voltmeter works? There should be some known high resistance and small current through it shows how large the voltage is. But here, in addition to the D-A induced voltage, the EMF may be added because the is also induced current flowing through the voltmeter. How much is the effect?

I suggest you to read the wikipedia page http://en.wikipedia.org/wiki/Voltmeter. A perfect voltmeter has no EMF. (**Edit:** see also http://en.wikipedia.org/wiki/Electromotive_force for the definition(s) of an EMF and its different interpretations.)

For me your last paragraph is incomprehensible. Faraday's law relates voltage drop to time varying magnetic flux, so it doesn't apply to open circuit.

Final edit: A way to avoid the flux drop: From http://dx.doi.org/10.1119/1.12923, see http://i.stack.imgur.com/X5qPb.png.

Here is one way to think about it:

When a charged particle travels in a magnetic field, it experiences a force. If the particle is stationary but the field is moving, then in the frame of reference of the field the particle should see the same force.

Now let's take a conductor wound into a coil. In order to increase the magnetic field inside, I could take a dipole magnet and move it close to the coil. As I do so, magnetic field lines cross the conductor, and generate a force on the charge carriers.

It is a convenient trick for figuring out "what goes where" to know that the induced current will flow so as to oppose the magnetic field change that generated it. In the perfect case of a superconductor, this "opposing" is perfect - this is the basis of magnetic levitation. For resistive conductors, the induced current is not quite sufficient to oppose the magnetic field, so some magnetic field is left.

The point is that the flowing of the current is instantaneous - it happens as the magnetic field tries to establish in the coil. So it's not "Apply field in coil. Coil notices, and generates an opposing field. " - instead, it is "Start to apply field in coil. Coil notices and prevents field getting to expected strength".

Not sure if this makes things any clearer...

## Best Answer

In $$dV=-\vec E\cdot d\vec r$$ $E$ is

electrostatic electric fieldand the negative sign implies that the electric potential $V$ decreases with increasing electric field $E$. This is the significance of this negative sign.In $$\epsilon=\oint \vec E\cdot d\vec l$$

E is not electrostatic, but

induced electric field. This field is induced due to time-varying magnetic fields or motional electromotive force causing a change in magnetic flux.This electric field is not a conservative field like the electrostatic field. Work done by this induced electric field around a closed loop is not zero.Also this induced electric field does not necessarily decrease in the direction of increasing induced emf. Rather it increases. Hence the question of having a negative sign in this second equation is meaningless.

To explain the last paragraph, let us consider a conducting loop of radius R placed in a time-varying magnetic field $\vec B$ directed perpendicular to the plane of the loop. Now there will be a change in magnetic flux causing an induced emf $\epsilon=-\frac{d\phi}{dt}$.

Think of an electron in this loop.The induced electric field E is directed tangential to each point on the loop since all points on the loop are equivalent.

On one hand, from point of view of induced emf $\epsilon$, work done to move a test charge $q$ once around the loop is $q\epsilon$. Again from point of view of electric field $E$, same work done as above is $qE(2πR)$.These two work must be same.

So,$$q\epsilon=qE(2πR)$$

and combining above results, we have $$E=-\frac{1}{2\pi R}\cdot\frac{d\phi}{dt}$$

Therefore, E decreases with increase in rate of change of magnetic flux , that is, E decreases with decrease in induced emf. So,induced electric field increases in the direction of increasing induced emf.