In a 0 resistance wire, the electric field inside the wire is 0 as there is no potential difference. That means that when charges start to flow, no external force acts on them. But it doesn't mean that the drift velocity of electrons is 0. We know $V_d = (eEt)/m$ where $V_d=$ drift velocity of electrons, $e=$ electronic charge, $E=$ electric field strength, $t=$ average relaxation time. When we put $E=0$, it means $V_d=0$. Why does this happen?

  • 3
    $\begingroup$ In a superconductor, you may have $E=0$ and current flowing, but then $t\rightarrow\infty$, so the right side of your equation becomes undefined. $\endgroup$
    – noah
    Jun 9 '21 at 22:34

All of this falls into place when you consider that the wires have a small resistance. A lot of apparent contradictions in current electricity appear due to taking this approximation of $0$ resistance seriously.

Consider an electrical circuit with a battery of voltage $V$ and a resistance $R$; consider wires to have a resistance $r$.

The current is $V/(r+R)$

The potential drop across the wires is $Vr/(r+R)$

If the wires have a length $l$, the electric field inside is $Vr/l(r+R)$

Since, $$v_d=\dfrac{eE\tau}{m}$$

, you get a non zero drift velocity since field is not exactly $0$, it is just low. Moreover, the drift velocity isn't just non zero, it is finite enough because metals have long mean free paths and thus large relaxation times $\tau$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.