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 Figure of cellsThe 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.

  • $\begingroup$ Request clarifications if necessary $\endgroup$
    – Dorothea
    Sep 21, 2020 at 12:47
  • $\begingroup$ Bounty1 (expired): I expect that someone answering the question must provide minute details and respond to cross questions if applicable. By minute details I do no mean that at what point did you apply "Fleming's left hand rule" or maybe "Lenz's law" but I mean the mathematics or the experimental evidences. As a motivation to answer the question, consider the following scenario. In the circular loop (given in the question body), If I randomly select two points, what will be the potential difference between them OR if I join the two points (with a superconductor), will there be current flow? $\endgroup$
    – Dorothea
    Sep 27, 2020 at 15:26
  • 1
    $\begingroup$ "how do we confirm that there is current flow[?]" The ring will get warm. $\endgroup$ Sep 27, 2020 at 20:20

3 Answers 3


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.

  • $\begingroup$ Do you wish to imply that the electric field in the first is conservative?; I did not get this By symmetry there is no redistribution of charge around the ring as we advance the magnet If we progress the magnet, there will surely be a current flow and redistribution of charge $\endgroup$
    – Dorothea
    Aug 16, 2020 at 21:37
  • $\begingroup$ I did wish to imply the first – or at least that there are conservative fields present in the ring of cells. (b) It depends what one means by 'redistribution'. The charge–carriers move, of course, but in the magnetic case their distribution around the ring remains the same as before. There are no regions of surplus or deficit. Sorry about the ambiguity. $\endgroup$ Aug 16, 2020 at 22:07
  • $\begingroup$ I would like to constructively argue on the conservative nature of the first field. My points: The field is net circular and we a considering a limit of $\infty$...so should be non cons. Also, we do find $d\phi / dt$ in the case of a closed loop and it gives us something of the dimensions of potential...so wont it be teeny bit equivalent to the first question? $\endgroup$
    – Dorothea
    Aug 16, 2020 at 22:19
  • $\begingroup$ Also, you said The charge–carriers move, of course, but in the magnetic case their distribution around the ring remains the same as before.. I interpret that in the first question the charge carriers dont move and their are regions of surplus and deficit? I am unclear how localization/delocalisation of charge is used here, sorry $\endgroup$
    – Dorothea
    Aug 16, 2020 at 22:22
  • $\begingroup$ "I interpret that in the first question the charge carriers don't move" By 'the first question', do you mean the ring of cells? The charge carriers do move around the ring; there is a current! (b) In a cell with a resistor across it there is a surplus of electrons on the positive terminal and a deficit on the negative, is there not? I'm arguing that there are surpluses and deficits around the electrodes in the ring of cells, giving rise to a sort of sawtooth graph of potential against distance as we go round the ring, the potential drops being within the electrolytes. $\endgroup$ Aug 16, 2020 at 22:46

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.

  • $\begingroup$ technically this is what I was looking for, thanks for your answer $\endgroup$
    – Dorothea
    Sep 28, 2020 at 3:54

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.)

  • $\begingroup$ I am really pretty convinced by this notion, that's why I asked to confirm...but I have encountered several questions in which people say which point will have a higher potential w.r.t. another which i do not think is defined to ask. So, if you can provide, a paper or an experiment in which it has been tested, It'll be absolutely convincing $\endgroup$
    – Dorothea
    Aug 16, 2020 at 19:26
  • $\begingroup$ Testing this idea becomes problematic because a voltmeter and its connecting wires form a loop which might also be subject to an induced emf. $\endgroup$
    – R.W. Bird
    Aug 17, 2020 at 13:22

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