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Say we use a cell to give rise to a current in a circuit and then remove the cell such that the circuit doesn't break. It means that no potential difference exists between any two points in the circuit since a circuit wire with current is neutral. So will the circuit wire have a decreasing current after I remove the battery due to N1L and Galileos definition of inertia that a body's state of rest or motion cannot be changed without an external force acting on it? Also does this mean that a superconductor will have constant current flowing through it even after removing the voltage source since there is no resistance to slow it down?

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    $\begingroup$ A Josephson junction is a device that can switch between being a superconductor and a normal conductor. Superconducting digital logic families often use this property of superconductors to store digital signals as persistent currents in an inductive loop without a voltage source in it. $\endgroup$
    – trent
    Feb 28 at 17:15
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    $\begingroup$ (Note also: the "inertia" that keeps the current flowing in a loop after the source is removed, is called inductance, and the mechanical inertia of the electrons (or Cooper pairs, in a superconductor) is not generally significant, because the energy stored in the electromagnetic field around the wired dominates instead.) $\endgroup$
    – trent
    Feb 28 at 17:18
  • $\begingroup$ How could difference not be needed for flow, including electric current? Could using a cell to give current, then removing the cell such that the circuit didn't break, be useful? How would that mean no potential difference between any two points? If your circuit wire was neutral, how might that help? Does this mean a superconductor should see constant current even after removing the voltage source, with no resistance to slow it down? $\endgroup$ Feb 28 at 23:18

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Persistent currents
Persistent currents is the name given to the currents flowing without energy dissipation and therefore not requiring a voltage bias to be sustained. This is notably the currents in superconductors (already mentioned in the answer by @kruemi), but also currents in some mesoscopic devices, where the energy cannot be dissipated due to the size effects (e. g., lack of efficient coupling to phonon bath and/or restrictions on the electron momentum change). Note however that the currents on microscale generally do not follow the same rules as those for conventional circuits - in particular, the lumped-circuit description fails (see also my answer to Are voltages discrete when we zoom in enough?).

Diffusion currents, etc.
As less exotic options one could mention diffusion current, where the current is due to the difference in electron concentrations, rather than bias. This is however difficult to sustain for a long time, and it usually co-exist with conventional (drift) current, as, e.g., in pn junctions.

Thermoelectric and other phenomena
Onsager relations predict a number of phenomena where the current can arise from spatial variation in other thermodynamic parameters. The diffusion current above can be viewed as the result of a gradient in chemical potential. Another well-known case is thermoelectric phenomena. These are often expressed in terms of voltage at the terminals of a device, due to the accumulated charge, but in a bulk they are actually currents.

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  • $\begingroup$ On a simpler note, induction is current without a voltage. But, any real impedance in the secondary will be measurably inseparable to a voltage difference. It'll look the.same. It gets complicated quickly enough though. Ohms law is an approximation. $\endgroup$ Feb 28 at 20:34
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    $\begingroup$ @StianYttervik I didn't include induction, because it is generating electromitive force - which is a kind of voltage - rather than directly current. This is also seen from the Maxwell equations. ALthough I conceded that the examples with diffusion and thermoelectricity are also not simple at the microscopic level. $\endgroup$ Feb 28 at 20:40
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. . . . we use a cell to give rise to a current in a circuit and then remove the cell such that the circuit doesn't break.

The circuit whether there is a break or not can now being considered as having an inductance, resistance and capacitance with the capacitance easier to visualise if there is a break in the circuit.
However even without a break there is capacitance as explained in the Wikipedia article parasitic capacitance.

What happens next depends on the relative values of capacitance, inductance and resistance but basically it will be a damped LCR system which can be over-damped (current falls to zero exponential), critically-damped (current falls to zero in the shortest time) or over-damped (current executes damped simple harmonic motion).

. . . . does this mean that a superconductor will have constant current flowing through it even after removing the voltage source since there is no resistance to slow it down?

A way of exciting a superconducting magnet is explained below.

enter image description here

The switch heater, the superconducting switch and the superconducting magnet are immersed in a cryogenic liquid below the temperature at which the superconducting switch and the superconducting magnet become superconducting.

The heater is switched on and heats the superconducting switch to a temperature at which it is no longer a superconductor.

The magnet power supply is switched on and the current, passing though the superconducting magnet and not the superconducting switch, is slowly increased to the required value.

The heater is switched off and so the temperature of the superconducting switch drops until it becomes a superconductor.

The magnet power supply is switched off and a constant current passes through the circuit consisting of the superconducting switch and the superconducting magnet.

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  • $\begingroup$ Why does the superconducting switch started as non-superconducting for the constant current to work? Is it to lock the direction of the current? In that case is is possible to have a superconducting diode (if that exists) instead? $\endgroup$
    – justhalf
    Mar 1 at 15:30
  • $\begingroup$ So you get the current into the superconducting coil via conventional conductors. Then complete the superconducting circuit and the current in the superconducting coil has to go somewhere and so it goes through the superconducting switch. I am sorry but I do not know anything about superconducting diodes. $\endgroup$
    – Farcher
    Mar 1 at 16:21
  • $\begingroup$ Ah, ok. I just realized that if superconducting switch started as superconducting, then all the current will go through the switch instead of the magnet. Is this correct? $\endgroup$
    – justhalf
    Mar 2 at 14:10
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For one: removing a voltage source without breaking the circuit is very hard. But in theory, that would be how it works. If you want to give it a try, use an inductor to pass current trough and then just short circuit your battery. The flow trough the inductor will continue for some time until the energy has dissipated because of internal resistances.

For some superconducting systems, that's exactly what they do (for example at CERN for magnets where a static field is needed (for example in some detectors). It's basically a "ring" of superconducting metal. Superconductors need to be below a certain temperature to be superconducting. So most of the ring is cooled down below that tripping point but one part is left "warm" (just above the required temperature). Then the current is introduced trough connections just before and after that "warm" piece. If the current has reached the desired value, it's held by the external source and the "warm" section is cooled down until it is superconducting too, thus closing the circuit. Over time the current is reduced (because changing the path of charged particles takes energy, so the energy from the magnet is slowly depleted) so the magnet has to be "restarted" again... The energies in those coils are enormous (We're talking GJ). So they have constructions to get rid of the energy within a second for example if for some reason part of the magnet is getting too warm. This means, for a short time, they send the current trough iron bars above ground and convert about 1GW of electrical energy into heat (that's equivalent to a small nuclear reactor). In other magnets, they have heaters to heat the whole magnet so the energy is dissipated over the whole length of the magnet in case of a quench event.

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  • $\begingroup$ That is also induced emf that inductors produce $\endgroup$ Feb 28 at 12:49
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Not sure about Galileo here and superconductors, I will answer the otherwise realistic and simple question. As others pointed out there is always a capacity in a real circuit, even parasitic, so the action of disconnecting the cell source (however fast and without leaking currents you might be able to do that in reality) means that the capacity will act as a source and there will be a transitory current AND difference of potential.

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To add to the other answers, bound (or magnetization) currents in permanent magnets are an example of currents that do not dissipate, or at least dissipate very slowly. This can be seen as an example of persistent currents Roger Vadim mentions in his answer.

You can make an electromagnet by winding a wire around a piece of steel. Suppose the steel is initially demagnetized. After you power the electromagnet with a battery and later disconnect the battery, the steel core retains some residual magnetization (assuming the magnetizing current was sufficiently high), producing a magnetic field. Ampere's law reads $$\vec \nabla\times\vec B = \mu_0 \vec J$$ and it is the current density $\vec J$ in the steel core (due to microscopic currents) that is the source of the magnetic field $\vec B$. Although these currents do not flow within the wire, they are still very much real.

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