# Max value of I in a circuit

I am a EE student.I have been taught Ohm's law which says $$I = \frac{V}{R}$$.In the limit of$$R \rightarrow 0,I \rightarrow \infty$$.But I have also taken the course "Electromagnetism II" where we learn of the invariance of the speed of light which also acts a limit to information transfer.The current $$I$$ inside a conductor is due to charged particles which move according to the electromotive force of the battery.Now I know that the velocity of these charged particles is also due to temperature but lets assume we cool the conductor at near 0K temperatures so the velocity is mainly due to the electromotive force.But the velocity of the charged particles which create current $$I$$ cannot be greater than c because they would carry information with a velocity greater than c which is universally forbidden.So the current $$I$$ has a limit.Which is it?Has it been calculated?

• Current is charge/second. If I make a bigger wire, I can fit more charges and more current Commented Mar 17, 2023 at 19:25
• Obviously the size of the wire remains constant Commented Mar 17, 2023 at 19:26
• In a band structure, electrons have crystal momentum even at T=0. Commented Mar 17, 2023 at 19:39
• @JonCuster yes and they have momentum anyway due to their zero-point energy....but it is so small it can be ignored... Commented Mar 17, 2023 at 19:44

In the most basic classical picture, the drift velocity $$v_d$$ of electrons in a wire is related to the current $$I$$ via $$v_{d} = I/neA \equiv J/ne$$ where $$n$$ is the density of conduction electrons, $$e$$ is the fundamental charge, $$A$$ is the cross-sectional area of the wire, and $$J\equiv I/A$$ is the current density in the wire. Plugging in $$v_d = c$$ and $$n\approx 8.5 \times 10^{28}$$ m$$^{-3}$$ (the conduction electron density of copper) yields an upper limit of $$J\approx 4\times 10^{18}$$ A/m$$^2$$, which corresponds to a current of $$I\approx 8 \times 10^{13}$$ A in a cylindrical wire with diameter $$5$$ mm.