Time-varying magnetic field and classical electrodynamics My question relates to magnetism and classical electrodynamics.
The following is a reference. This question says $\downarrow$ (do not answer to this):

$N$ sources of current with different emf's are connected as shown in
the following figure   The emf of the sources
are proportional to their internal resistances, i.e., $E=\alpha R$,
where $\alpha$ is an assigned constant. The lead wire resistance is
negligible. Find:
(a) the current in the circuit
(b) the potential differences between points A and B dividing the
circuit in $n$ and $N−n$ links.

The answer provided is

(a) $\dfrac{E}{r}$, (b) $0$ which I completely agree with.


My question is about transposing the above problem so that it resembles the problem below which concerns with TVMF(Time Varying Magnetic Field).
Consider a circular wire loop in the presence of a time varying magnetic field parallel to its central axis.
(to simplify the question, keep the configuration of $B$ such that $\dfrac{dB}{dt}$ is a constant.)
(a) Can we find relative potential here? (Considering the newly formed infinitesimal cells $\equiv$ to the above question)
(b) When we use $\dfrac{-d\phi}{dt}$ in this question, what kind of potential do we find and how is it distributed/mapped?
(c)To plot the equipotential lines/surfaces (on the outside of the circular loop), I could come up with radial lines emerging from the center, but I am not sure how I should assign the potentials to the lines.
Post some discussions and a previous answer, I was presented with the fact that potential is not defined. Countering that, if I measure the potential along the so proposed radial lines in the part [c] of my questions, what potentials would I measure? And iff, the potential measured=0 , how do we confirm that there is current flow.
If available, a resource suggestion is also welcome.
 A: With the ring of cells we can at least talk sensibly about potential differences. That's because the emfs in a cell arise at the electrodes and not in the bulk of the electrolyte. So as charge flows there are charge density inequalities so potential rises at the electrode/electrolyte interfaces and there are equal potential drops in the bulk of the electrolyte.
I don't think we can talk sensibly about potentials for the ring and magnet (assuming symmetry). By symmetry there is no redistribution of charge around the ring as we advance the magnet (no formation of regions of surplus and regions of deficit), and without charge concentrations we won't have a conservative electrostatic field, so we can't apply the concept of potential.
A: Original question: The current in the circuit should be the sum of the emf 's (each an αR) divided by the sum of the internal resistances (each an R)  If the R's are different but α is constant you can factor out the α and the two sums cancel, leaving the current, I = α.  Then the voltage drop on each resistor is IR = αR which is equal to the corresponding emf. The terminal voltage of each cell is zero and the voltage drop between any two points (outside of the cells) is zero.
Your question: If you consider a loop of wire with a changing magnetic flux, you can think of each segment of wire as being like a cell with an emf proportional to its length (and resistance).  The result is the same: There is no voltage difference between any two points in the loop. (Unless you break the loop. Then the current stops, and the total emf appears across the gap.)
A: Philip gave the right answer. I will just a give a highbrow answer, mostly filled with the jargons:
Potential only makes sense when curl of electric field is zero i.e., $\vec{\nabla}\times\vec{E}=0$, which holds only for electrostatic case. When we have a time changing magnetic field, the right equation is $$\vec{\nabla}\times \vec{E}=-\frac{\partial\vec{B}}{\partial t}$$. Clearly curl of $\vec{E}$ doesn't vanish here, so potential makes no sense here.
What happens if you take a test charge and make it goes around the loop. The force due to electric field is given by $\vec{F}=q\vec{E}$ so work done is $\int\vec{F}\cdot d\vec{l}$ equivalently $$W=q\oint\vec{E}\cdot d\vec{l}$$
where I have used the symbol $\oint$ to denote the work done in completing the loop. You can take another tour around the loop, and you'll spend $2W$ joules. This work done is strikingly different from conservating force work because the latter is zero for a closed-loop.
Electromotive force is defined as $$\oint \vec{f}_s\cdot d\vec{l}$$
where $\vec{f}_s$ is the force responsible for motion of charge after removing electrostatic force since for the latter $\oint \vec{E}\cdot d\vec{l}=0$. In this case electric field produced by changing magnetic field is the sole contributor to $\vec{f}_s$. So there is no issue of going around the loop any times; it is defined only for one round.
Since equipotential lines will make sense only when we have a potential to work on since they simply are $V(x,y)=c$ for some constant $c$. Though if you take radial lines, the work done on the test charge when moved along them is zero since force is perpendicular to the displacement. But it won't suffice they are equipotential lines since there is not potential here. Mathematically, let's say there exists a potential $V$ such that $E=-\vec{\nabla}V$. Then the curl of $\vec{E}$ should give $-\frac{\partial \vec{B}}{\partial t}$
though by identity $$\vec{\nabla}\times\vec{\nabla}V=0.$$ therefore there doesn't exist any scalar potential. Even though there exist paths on which no work is done.
Take an electromagnet run current through it (charging your phone will be enough) and touch the two pointers of the voltmeter so that a closed-loop is formed for current to run through it. You'll read some value and it will change as you change the orientation of loop or loop's shape or loop's distance from the electromagnet.
