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Re: Science fiction or a wrong model again.



Ludwik,
Your Ohms law solution reminded me of an article I read years ago
in which the author tried to measure the resistance of a superconductor
with a DVM's ohmmeter (not four terminal) and found he always got a
resistance it didn't go to zero. He was measuring the meters internal and
leads resistance. A four terminal meter gets around this be using two leads
to supply a constant current and two to measure the voltage drop. Getting a
one volt drop in a super conductor is an impossible task without
exceeding the critical current.
In superconducting magnets, there is a shunt across the end of the
coil that can be heated to make it non-superconducting. With the shunt non-
superconducting a current is applied to the magnet's coil. The current is
limited by its power supply. The shunt is then cooled to superconducting
state and the power supply can be turned off. The current will continue to
travel around the coil.
In a super conductive ring, if a magnet is place in its center and
the ring cooled to super conduction. Then removing the magnet will induce a
current in the ring, which can be sensed using a magnetometer.
I hope this helps.

Gary

A conceptual model of electric current developed in an introductory
physics course is too naive to be useful in making predictions about
superconductors. Here is another illustration. Suppose that the emf=1
volt is suddenly inserted into a superconducting loop. Predict the
electric current. Ohm's law predicts that I should keep increasing
when R --> 0. To overcome the "division by zero" dilemma one can
reason in a different way. The current is normally calculated as:

I=n*q*A*v,

where n is the density of free electrons (such as 10^28 per cubic
meter), q is the charge of one electron (1.6*10^-19 C), A is the
wire cross section (for example, 1 mm^2=10^-6 m^2) and v is
the velocity of electrons. To estimate v let me assume that the
length of the wire, L is one meter. After covering this distance
(along the wire) the kinetic energy of each electron will be 1 eV
and the corresponding v will be nearly 6*10^5 m/s. With this v
the above formula yields I~10^9 A, which is enormous and
ridiculous. Why ridiculous? Because 1 eV per electron would
mean a total kinetic energy increase of 10^28 eV =1.600,000 J
(enough to elevate 10,000 kg to a height of one mile)!

A "Modern Physics" textbook I consulted (Serway, Moses and
Moyer, 1989) states that "supercurrents have been observed to
persist in a superconducting loop for several years with no
measurable decay." Does anybody know how such currents are
started and how large they are, typically? The textbook table
shows that in several Type 2 superconductoprs Bc2 exceeds
30 Teslas. What kind of power supply is used in commercial
superconducting electromagnets? Such devices are often used
in research to produce B~2 T, and above.

I hope those familiar with practical supercoductivity will answer my
questions and will criticize my naive speculations in this thread.
Ludwik Kowalski