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Super conductors are perfectly diamagnetic in that they perfectly expel a magnetic field. To me that means that if the source of a magnetic field is on one side of a super conductor, the other side should have no effect from it. However, the images that I can find online showing the field lines indicate otherwise:

2 images of magnetic field lines around super conductor

Is it because the air is paramagnetic and it can conduct the field lines around the superconductor? Or is the super conductor drawing them around itself? Or something else?

Given that where it is curving around is cut off from view of the source, I would have expected it to just redirected completely away, like this: What I would expect to happen

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To me that means that if the source of a magnetic field is on one side of a super conductor, the other side should have no effect from it.

The physical mechanism for the expulsion of magnetic fields from the interior of a superconductor is the generation of surface currents which produce a secondary magnetic field, cancelling the field in the interior. There's no particular reason to think that they will cancel any part of the field in the exterior region as well, and in fact they don't.

You may be making the mistake of assigning more physicality to the field lines than they deserve. Remember that the magnetic field is a vector field - an arrow attached to every point in space. The magnetic field lines are a convenient way of visualizing the field, but you must be careful not to assume that they behave like physical objects do. In particular, your illustration indicates that you feel that they should behave like the trajectories of massive objects, which they do not.

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  • $\begingroup$ So is the answer that the vector where I circled caused by the super conductor instead of the original source? Why does B=(mu)H not zero there? Is the H coming from the superconductor? $\endgroup$ – user1934868 Aug 5 '19 at 18:59
  • $\begingroup$ @user1934868 Yes - the field there is equal to the external field plus the contribution due to the surface currents, which can be calculated from Maxwell's equations. In the interior, the latter contribution cancels the former exactly; in the exterior, this is no longer true. $\endgroup$ – J. Murray Aug 5 '19 at 19:50

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