2
$\begingroup$

The electrical resistance being zero in a superconductor, if a magnetic field is strong enough to generate vortices where the flux lines will pass through the material, and the current flow is perpendicular in the superconductor(type II) will it experience a Lorentz force?

I assume the equation: $$ F_L = IL \times B$$ is not accurate to this case?

$\endgroup$
4
  • $\begingroup$ I would guess that there is small to no Lorentz force in the case that you described because the material within the vortices themselves are normal, not superconducting. Since the Cooper pairs which carry the supercurrent avoid those normal cores, they never experience the B-fields. $\endgroup$
    – user93237
    Dec 28, 2015 at 6:18
  • $\begingroup$ So, in all types of superconductors they cannot experience a Lorentz force? Due to the magnetic field being expelled? $\endgroup$
    – AxtII
    Dec 28, 2015 at 10:12
  • $\begingroup$ @MA - That's my educated guess, but I'm not an expert on superconductivity and perhaps there are some other factors such as the fact that the penetration depth of a superconductor is non-zero which could lead to the moving electrons experiencing some B-field and, hence, Lorentz force. $\endgroup$
    – user93237
    Dec 28, 2015 at 18:23
  • $\begingroup$ Perhaps this helps physics.stackexchange.com/questions/189499/… $\endgroup$ Dec 29, 2015 at 19:22

1 Answer 1

-1
$\begingroup$

As long as the external magnetic field does not destroy the Cooper pair of electrons, this pair does not experience the Lorentz force.

Not only charged particles, when moved through a not parallel to the movement magnetic field, get deflected, but also neutrons get deflected. This happens for all particles and also for atoms and molecules, if their summarized magnetic dipole moment is not zero.

The magnetic dipole moment of each involved proton, electron and neutron is connected to an intrinsic spin. As has been shown by the Einstein-de-Haas experiment (not a thought experiment, but really carried out), the intrinsic spin has to do with spinning around an axis and this spin is pointing in the same direction as the magnetic dipole moment (has the same axis).

In analogy to a gyroscope, the intrinsic spin resists against deflection from a straight line. The result is a precession: When an external force try to change the direction of the axis of a gyroscope, the spin axis does not follow the direction of this force, but get deflected from a right angle to it.

But this holds only if the intrinsic spins of the involved constituents does not cancels out. An example for not canceling out spins is the Stern-Gerlach experiment. Cooper pairs are the perfect example for a spin sum equal to zero. How the repelling electric forces do not destroy the Cooper pair is another question.

$\endgroup$
6
  • $\begingroup$ If someone is interested in more details and to point in the direction we have to make researches, read my elaboration About the distribution of electrons magnetic dipole moments in atoms. $\endgroup$ Dec 28, 2015 at 6:56
  • $\begingroup$ In that case... how can electric motors improve with the use of superconductors or generators? $\endgroup$
    – AxtII
    Dec 28, 2015 at 10:27
  • $\begingroup$ @M.A Good point. I have to think about this. $\endgroup$ Dec 28, 2015 at 10:56
  • $\begingroup$ In a Bose-Einstein-condensate approx. 20% of the particles have the same state and behave like one unit. All other particles are individual particles. The same situation should be in superconducting coils. The particles, which are not Cooper pairs are making the magnetic field. For this you need energy. So a superconductor is or a strong magnet only and he behave like a spring or a superconductor work like a generator/drive and for this he need to be powered. Than together with the Cooper pairs ordinary electrons came into play. They of course get influenced by magnetic field while moving. $\endgroup$ Dec 28, 2015 at 11:12
  • $\begingroup$ @M.A Does a generator/motor with superconducting coils exist at all? $\endgroup$ Dec 28, 2015 at 11:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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