Re: drift speed in superconductors

[Larry Woolf <larry.woolf@GAT.COM>]

To some extent, it is not defined for a superconductor.
Drift velocity is defined by v(drift)=mobility * electric field.
A superconductor cannot support a voltage difference, since R=0.


On the other hand, one can define a local average drift velocity for a
superconductor, following the London formulation, as:
J(superconductor) = n(superconductor)* e * v (superconductor)

A typical superconductor, when operating, may have a current density of
10^6A/cm^2.
In a two-fluid approximation (there are superconducting electrons and normal
electrons), at low temperatures all of the electrons are in the
superconducting state so n~10^22/cm^3.
The charge is the electric charge (really 2e since they are paired into
Cooper pairs), so take e=1.6x10^-19 C
Then we find that v(superconductor)=600cm/s as a rough order of magnitude
answer.

Larry Woolf;General Atomics;San Diego CA
92121;Ph:858-526-8575;FAX:858-526-8568; www.ga.com; www.sci-ed-ga.org

> -----Original Message-----
> From: Larry Smith
> Sent: Wednesday, February 19, 2003 12:54 PM

> Subject: drift speed in superconductors

> What is the drift speed of electrons in superconductors?