Speed of electrons in a perfectly conducting wire

Question

So we know that EMF of the battery provides an electric field in the circuit and thus accelerating the electrons inside them, so my question is that lets say if there is a perfectly conducting wire, so there the electric field will continuously accelerate the electrons(as no resistance if there for them to collide and make them lose energy) and we know that nothing can go beyond speed of light, so will there be a current in the circuit or will the current be just too high?

I looked for the explanation at many places, but everywhere people just say that such a wire is not possible, but what if it is?

What do you mean with "perfectly conducting wire"? All wires I know consist of atoms; move the electrons and an electric field will build up. To avoid this and come close to the aim of your question (infinitely accelerating the electrons) you could put your lectrons in a vacuum and then accelerate them with a voltage. This is called particle accelerator then. Which of the two cases is more interesting for you? — sir_khorneflakes, May 05 '23 at 14:58
And are you thinking about applying a fixet voltage, like from a battery, or accelerating them infinitely long (with infinite voltage) over a fixed electric field? Remember that electrons have mass and an electron is accelerated by one Volt to a fixed kinetic energy (1eV), as Volt is a potential difference, not the slope of the potential. — sir_khorneflakes, May 05 '23 at 16:21
@KutubkhanBhatiya A betatron might be what you're seeking: https://en.wikipedia.org/wiki/Betatron — John Doty, May 05 '23 at 20:28
Does this answer your question? What is the drift velocity of an electron in a superconductor? — Agnius Vasiliauskas, May 05 '23 at 20:35

Dale · Accepted Answer · 2023-05-05T20:18:44.153

lets say if there is a perfectly conducting wire, so there the electric field will continuously accelerate the electrons

Your "so" does not follow from your "if" here.

A perfectly conducting wire would be a material with resistivity $\rho=0$. This would mean that $\vec E = \rho \vec J$, so in a perfectly conducting wire there cannot be an electric field! Regardless of $\vec J$ we always have $\vec E=0$ for $\rho=0$. So in a superconductor the current density $\vec J$ is not determined by Ohm's law but rather by the continuity equation (NB $\rho$ above is resistivity while $\rho$ below is charge density): $\frac{\partial}{\partial t}\rho + \nabla \cdot \vec J = 0$.

This is not nitpicking, but it is actually a real issue that people have to address in designing superconducting switches and ramping up superconducting magnets. Usually, in order to get a current into a superconductor it is necessary to make a small part resistive so that you can put an E field across it. This paper by JM Oleski (Persistent Switch Design for MRI MgB2 Superconducting Magnet) is an example of some of the design issues faced in making these kinds of persistent superconducting current switches.

we know that nothing can go beyond speed of light, so will there be a current in the circuit or will the current be just too high?

All superconductors have a finite current density, called the critical current density, at which point the superconductor abruptly becomes non-superconducting. So you would never reach a state where the speed of the Cooper pair approaches $c$

Got most of the part. I agree that E will be zero, but can we comment something about what would be J, or can it be anything. Shouldn't it have some value. I don't get that part. — Kutubkhan Bhatiya, May 05 '23 at 15:57
+1 and a request for reference: can you link or reference some resource where the addition of a resistive path is described with some detail (I am interested in the cross section distribution of current density before and after the switch)? I have heard about it from Walter Lewin and I found this idea of having a resistor "pushing the electrons" freaking awesome. — Peltio, May 05 '23 at 16:07
@KutubkhanBhatiya I have added a sentence describing the role of the continuity equation in that regards — Dale, May 05 '23 at 18:53
@Peltio I have added a reference with some detail on the construction of such devices, but it may not have the level of detail you are looking for regarding cross sectional distribution of current density — Dale, May 05 '23 at 18:54
@Peltio, there's also an answer here that touches on the concepts. https://physics.stackexchange.com/a/179386/55662 — BowlOfRed, May 05 '23 at 19:53

score -1 · Answer 2 · answered May 05 '23 at 14:55

-1

Perfectly conducting wire is not a real thing and, more importantly, not a model of a real thing - rather it is a concept in a lumped-element model - where the essential electromagnetic features are supposed to be concentrated in capacitance, inductance and resistance, whereas wires only indicate how these elements are connected to each other. This is intended to describe a real circuit, where all these are distributed (although some parts might be made explicitly to have higher capacitance, resistance, inductance.) See also Telegrapher's equations.

answered May 05 '23 at 14:55

Roger V.

58,522

Note that the Telegrapher's equations don't involve electrons at all: they describe the propagation of an EM field in a conductive cable. In practical cases, the electrons barely move (electron drift: https://en.wikipedia.org/wiki/Drift_velocity). You may think of them as providing the current that supports the energy transfer via the field. But for problems like this, forget electrons: they are merely a distraction. Only the current matters. – John Doty May 05 '23 at 17:13
@JohnDoty current is electrons. I cited telegraph equations as an example of a distributed circuit. One can see them as describing charge oscillations in a circuit or an EM wave in a waveguide - these are two sides of the same coin. – Roger V. May 05 '23 at 18:53
Superconductors exist. – naturallyInconsistent May 05 '23 at 18:55
@naturallyInconsistent superconductors are very different from ideal wires. – Roger V. May 05 '23 at 19:11
Current is often electrons. Sometimes it's ions, holes, ... But even when it really is electrons, thinking about electrons is a distraction when it's current that matters. The Telegraph equations work the same regardless of what carries the current. – John Doty May 05 '23 at 19:43
@JohnDoty I don't think electrons are a distraction - this is perhaps so, if your background is in classical electrodynamics... but my experience is largely in quantum transport. – Roger V. May 05 '23 at 20:26
If I'm working with batteries and wires, I use the appropriate model: classical electrodynamics. – John Doty May 05 '23 at 20:31
@JohnDoty there's more than one way of thinking and doing things. Just because certain picture is clearer to you doesn't mean that everyone should be doing it your way. – Roger V. May 06 '23 at 05:18
@RogerVadim But it leads to the false idea that electric current flow is always electrons. It leads to confusion about the sign convention for current. It leads to confusion about how much the electrons are moving. And it's simply a distraction: the mathematical formulation of these problems doesn't have quantization of charge in it. – John Doty May 06 '23 at 12:05
@JohnDoty the problem of the OP is that they do not understand what is accounted and what is not accounted by the lumped-element model. The rest of your comments is just nitpicking. – Roger V. May 06 '23 at 20:14
I see nothing in the question that refers to the lumped element model. What I see is the misconception that the physics of the problem is captured by the idea that the battery accelerates electrons. – John Doty May 06 '23 at 20:20
@JohnDoty then write an answer. Your comments here are clearly nit made in good faith. – Roger V. May 06 '23 at 21:15

Speed of electrons in a perfectly conducting wire

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