You're splitting hairs here Ex. Reference the Lewin experiment, a DC source is used to pulse the coil. Consequently, there is a corresponding unipolar response on the loop (polarity dependent on meter connection). i.e. the current does not change direction, it rises, then falls back to zero.

I'm not splitting hairs, I justified what I said. I say that "+" or "-" are not relevant for signals with a null dc component and I explain why it is the case here.

**An induced current never rises and fall back to zero**. It rises and fall back below zero at some time, otherwise there would be a dc component. If transformers could pass DC, it would be known for a long time!

For some odd reason, many here seem obsessed with the solenoid "flux". The flux outside the solenoid is zero, so according to you Ex, the measurement leads will never be influenced by the solenoid "flux" (and you would be correct).

This idea is not according to me, but to your interpretation of what others say, including me, which is different.

The Faraday's law doesn't imply that the circuit be in the flux. The Faraday's law stipulates only that the surface of the circuit must be crossed somewhere by a varying flux. The flux can cross the surface even through only 1% of the surface and anywhere. If your loop is at 1km from the solenoid, still around, the same emf will be induced.

The "correction factor" you propose is nonsense. The scope indicates what is there across the two points on the loop. We never ask or ponder whether the measurement is relative to the R100 or R900 resistor , the measurement simply IS what it IS. The scope certainly doesn't care.

"The measurement simply IS what it IS": yes, I agree. The measurement that you imagine by connecting CH1 to R100 while the wires are enclosing the disk surface is the measure of the potential difference at the resistance terminals + the emf induced in your measurement circuit, because the total flux of the solenoid is crossing your measurement circuit.

It is what it is: 0.9v, which is a real measurement that is what it is, but naturally not the potentiel difference felt by the R100 resistance.

Of course you will not get the same measurement by connecting CH2 to R100 by the other side, which gives the real potential difference at R100.

The nonsense is only to imagine that both measurements should give the same result.

My "correction factor" was only to obtain the potential difference at R100 terminal from the measurement of CH1 (in the same way that when you measure the potential difference at the terminals of two identical resistances in series, you can use a "correction factor" to obtain the potential difference at only one: by dividing by two).

I've stressed this a hundred times it seems, but it does not sink in; **forget about the flux**, it is simple a means to an end, and that END is the resulting induced electric field. The E-field is what counts. No E-field, no induced emf. The E-field is all around the solenoid, inside and outside, the flux is internal only to the solenoid.

You can get the same results by using the E-field, as by using the flux. But you can't oppose one method to the other, because the two methods are strictly equivalent due to the Stokes theorem. If one failed, the other too. I said it somewhere else but you missed the point:

"According to Stockes theorem,

**the surface integral of the curl of a vector field over a surface is linked to the line integral of the vector field over its boundary**.".

Demonstration that they are the same:

**Emf = -dΦ/dt** (Faraday's law = flux crossing method)

**Φ = ∫∫B.n.dS** (definition of the flux, B and n are vectors, n is the unit vector, normal to the circuit surface S).

**(1) Emf = -dΦ/dt = -∂(∫∫B.n.dS)/∂t = -∫∫∂B/∂t .n.dS****Emf = ∮E.dl ** (integral of the electric field over the circuit = E-field method, l is refering to the circuit path).

**∮E.dl =∫∫(∇xE).n.dS** (Stokes theorem applied to the E-field, S being the surface enclosed by l)

and

**∇xE = -∂B/∂t** (Maxwell)

so

**∮E.dl = -∫∫(∂B/∂t).n.dS = equation (1) = -dΦ/dt**QED

The only interest of the E-field, is that it gives the local field acting on the electrons, and therefore the force F=q.E which is the cause of their motion. On the contrary, if we supposed that the flux would be the cause of the current, the phenomenon would be a non local (therefore outside classical physics) because there is no flux at the position of the wire enclosing the solenoid. Nevertheless in our approximation of the quasi-stationary states (we suppose that there is no EM waves), the both methods are strictly equivalent to calculate the emf but the method by the flux is by far the simpler.

The flux has no direct influence on the loop or the measurement leads, ever.

I agree, but the flux variation through the circuit surface is an event concomitant with the local E-field along the circuit, and so, there is no discrepancy in using one method or the other, as proved by the equations.