If I apply $1 \text{ V}$ across a $1 \text{ }\Omega$ resistance, I'd get $1 \text{ A}$ flowing. $1 \text{ A}$ is defined as $1 \frac{\text{C}}{\text{s}}$, and $1 \text{ C}$ is equivalent to $6.24150975 × 10^{18}$ electrons.

Therefore, Ohm's Law describes only the number of electrons passing a given branch per second. How do we determine the speed of electrons as they move through a conductor? And what is that speed?


You can't measure speed of electrons from these data alone. If the area of the cross section of a cylindrical conductor is A then the formula would be $v = \frac {I}{QeA}$ where $Q$ is the mobile electrons per cc and $e$ is the charge of electron, $v$ is the speed of electron, $I$ is the current. This is for dc. See this http://amasci.com/miscon/speed.html

So you see, you need to know the diameter of the cylindrical conductor. However the electrons don't move in a conductor in an orderly manner. They are constantly colliding with each other and typically have speed components along the conductor about some millimeters per second.

| cite | improve this answer | |
  • 1
    $\begingroup$ To add to this, one should distinguish between ballistic and diffusive regimes of electron motion. On short enough timescales, electrons move with colliding with a momentum around that of the Fermi momentum (which may be measured most directly with something like ARPES); the mass of the electrons may be deduced from heat capacity and then the (semi-classical) velocity is the ratio. In this regime (sub-mesoscopic) the speed is quite high. In the diffusive regime, one gets the more often quoted "millimeters per second", as discussed in the link. Fun quantum things can happen at low temperatures. $\endgroup$ – genneth Mar 1 '11 at 14:46
  • $\begingroup$ Electromagnetic Fields And Waves page 73 calculates a figure of 300mm/hour, 0.1mm/sec for 1A flowing through 1mm^2 of copper. $\endgroup$ – John McVirgooo Mar 1 '11 at 22:53
  • $\begingroup$ How do you get Q? What does "per cc" mean? $\endgroup$ – Martin Thoma Dec 26 '12 at 10:27

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.