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Re: drift speed in superconductors



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.

That seems a little backward. Speed is *defined* as distance
traveled per unit time. The relationship given above is the
definition of mobility, not drift velocity.

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)

That seems like a better approach, but there are no definitions here.
This is a *relationship* between quantities that are well-defined
elsewhere.

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.

Wow! That *is* a pretty amazing result, at least for those of us
used to encountering drift velocities on the order of fractions of
millimeters per second. It is, however, very simply related to the
fact that nonsuperconducting wires can't begin to support those kinds
of current densities.

Just as a point of comparison, if one were to run that kind of
current density through room temperature copper, one would be
depositing j^2 * rho = 1.6 MW/cm^3 in the copper due to Ohm heating
and it's temperature would be rising at a rate of about half a
million K/s.

--
A. John Mallinckrodt http://www.csupomona.edu/~ajm
Professor of Physics mailto:ajm@csupomona.edu
Physics Department voice:909-869-4054
Cal Poly Pomona fax:909-869-5090
Pomona, CA 91768-4031 office:Building 8, Room 223