Start with the long U-shaped conductor. Is there potential across the ends or not?
You are asking the wrong question. It's probably my fault because in a previous post I used the term 'equipotential' in quotes to mean 'the voltage difference between two points is zero' (inside a certain cylinder), so let me explain again.
You can no longer talk about potential function for the total electric field. It has no longer sense because the path integral of the total electric field depends on the path joining two points.
Potential no mas.
So your question has no sense. You should ask, instead: is there a voltage across the ends?
And the answer is:
IT DEPENDS ON THE PATHWhat path? The path along which you want voltage, which is a function of the endpoints
AND THE PATH, to be evaluated.
For all paths that together with the conductor (let me rephrase: together with any path inside the conductor that joins point A and B - it does not matter which because the electric field inside is exactly zero if there is a gap)--- let me repeat: for all paths that together with the conductor form a closed path that DOES NOT go around the dB/dt region, the voltage difference is ZERO.
For all paths that, together with the conductor, DO GO around the dB/dt region, the voltage difference is 1 EMF (or -1 depending on how we go around, or a multiple of the EMF if we go around several times).
Tomorrow I will add a couple of pictures that hopefully will clarify what I wrote.
As I was responding to Sredni and thinkfat it occured to me that there's another issue with the equipotentiality argument. In the static, conservative, irrotational field, there isn't an E-field inside the conductor because the charges instantly rearrange themselves to oppose it. Thus there is no net force on any charge within the conductor. In the case of a charge moving in a magnetic field, you accept that there is an non-conservative field inside the conductor acting on the charges, right? But then we get to the MQS system, and even though there is clearly a local force acting on charges inside the conductor--which is the EMF--
Stop. This is wrong.
It is NOT the EMF. We always end up here: forgetting the role of the displaced charge. The EMF - or better, the induced electric field Eind, is compensated in the conductor by the coloumbian field Ecoul. If there is a gap and no current is flowing, this is always exactly true.
If there is a current flowing (because we closed the circuit) then it is zero only if the conductor is perfect (zero resistivity, infinite conductivity). Charges move without a field. In practice, with real conductors, the compensation is not perfect and there is a residual field E in the conductor, directed along the conductor and whose value complies with the local form of Ohm's law: E = j / sigma_copper.
How does it follow the conductor? Answer: surface charge. (Read here, if you are interested
https://electronics.stackexchange.com/questions/532541/is-the-electric-field-in-a-wire-constant/532550#532550 )
So, it is not Eind that is acting on the carriers. It is the combined effect of the EMF (represented by Eind) and the displaced charge (represented by Ecoul).
and those forces continually push charges through the conductor, that the conductor is nonetheless equipotential in the same way as in the static case, and for the same reason--there's no electric field in a conductor.
and this is again (EDIT: in general) wrong for the reason above. But, if you consider a region of space that contains all of your piece of conductor and none of the magnetic flux, like a cylinder around the rod in the middle of the torus, in that region of space all paths you can imagine will never enclose the magnetic flux region. in this sense you can consider it as equipotential. Shrink the surface to include only the conductor if you want. But the moment you consider a path the forms, with the conductor, a closed path that cuts the flux - NO MORE.
(I have a picture to visualize this as well - but I drank too much tequila to operate a scanner)