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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?

Roger V.
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Kutubkhan Bhatiya
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2 Answers2

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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 $c$

Dale
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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.

Roger V.
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