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Basically in a circuit there is a cell with a positive and negative end, and current flows through it - electrons are pushed by negative electrode and attracted to the positive electrode. But if say a ring rotates about its diameter and say there is an applied magnetic field and as the ring rotates, there is a flux change, so according to lenz law emf will be generated, but the ring is all connected, what part is supposed to be the positive and negative end?

Qmechanic
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Questioningmind
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  • This figure https://postimg.cc/MMjt03b4 shows how the total electric field is rotational in a ring with two different resistors and a changing magnetic field linked by its interior. The total field is the composition of the rotating induced field with the coulombian field (whose field lines start from positive charges and end up in negative charges) of the charges alone. "Potential difference" is defined only for the latter because the total field does not admit a potential function and voltage is path dependent. – Peltio Mar 23 '24 at 22:08
  • @Peltio I don't think your diagram is correct, why would charges accumulate at the ends of the resistor? Will the current stop flowing, as there would be no electric field inside the wire? – Wolphram jonny Mar 24 '24 at 12:36
  • Charges accumulate at the surface and interfaces of resistors. Conduction in a circuit is driven by surface and interface charges. https://physics.stackexchange.com/questions/755990/stationary-charge-near-a-current-carrying-wire-experiment-to-check-it-at-home/756039#756039 .The very small electric field in the wire is that allowed by Ohms law in its local form (E = j / sigma). Since copper - the most used conductor - has a very high conductivity the electric field in the wire will be very small. Ideally, with perfectly conducting wires, the field would be zero (and current would still flow). – Peltio Mar 24 '24 at 15:12
  • @Wolphramjonny to be clear the diagram assumes the ring is stationary and the field is changing, so it's seen from the point of view of an observer stuck to the ring. I apologize for not specifying that - I always fall back to the Lewin ring when discussing non conservative electric fields. An observer that would see the ring rotate would see a field in the copper, but I believe he would still see charge accumulation at the interfaces. – Peltio Mar 24 '24 at 15:34
  • @Peltio So what would make the current flow inside perfectly conducting copper, I guess inertia? – Wolphram jonny Mar 24 '24 at 15:53
  • @Wolphramjonny I see perfectly conducting wires as an abstraction to avoid thinking about minor voltage drops in the connections. From the point of view of circuit theory the question is irrelevant because in lumped circuits all connections are zero-dimensional. To avoid paradoxes in circuits with actual wires I consider the zero conductivity case as the limit for sigma going to zero. The current has a finite limit even if it is the ratio of two quantities that go to zero. – Peltio Mar 24 '24 at 16:16

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The cell's job is to create an electric field that makes the "free" electrons inside a metal move. You don't need a positive and negative end in the case you describe, because the electric field created by the changing flux makes the job of keeping the electrons circulating.

Wolphram jonny
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  • so, regardless of any positive negative end , they will move?! how are the electric fields oriented then? are they closed? and if i use a multimeter will i get to measure any potential difference? – Questioningmind Mar 23 '24 at 14:25
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    The field lines are oriented and form a closed loop. The idea of electric potential is not longer valid, but you will measure one with a voltimeter. the electric field is no longer a conservative field though. See here https://farside.ph.utexas.edu/teaching/316/lectures/node87.html#:~:text=The%20electric%20field%20generated%20by,closed%20loops%20in%20free%20space. – Wolphram jonny Mar 23 '24 at 14:32
  • What you get on the voltmeter depends as much on the exact route of the wires between ring and voltmeter as on where they connect to the ring. – Philip Wood Mar 23 '24 at 15:21
  • @PhilipWood can you elaborate a bit? – Questioningmind Mar 23 '24 at 16:59
  • What part specifically you do not understand? Otherwise I could write an entire chapter. Did you read the link? What level of understanding of physics do you have? – Wolphram jonny Mar 23 '24 at 17:02
  • In the end, there is no positive or negative end, but there is a direction, and you will always measure a voltage difference in the same direction, the total voltage along the entire circuit is not zero, as when you have a battery. – Wolphram jonny Mar 23 '24 at 17:08
  • Elaborating a bit, the wire from one point, A, on the ring to the voltmeter, the voltmeter itself, the wire from the other terminal of the voltmeter to the other point, B, on the ring and the minor arc AB of the ring form one loop in which an emf is induced. Another loop is the same as this one, but with the major arc AB instead of the minor one. Emfs will be induced in both loops. Quite complicated ... – Philip Wood Mar 24 '24 at 17:36