As shown in the figure, in the Meisner effect, the magnetic field of the magnet bypasses the superconductor. My question is, does the magnetic B field belonging to the magnet increase at the arrow indicating position?

Or is there no change in the intensity of the magnetic field that belongs to the magnet? Is it just the superposition of the magnetic field of the magnet and the magnetic field of the superconductor?.

Is it also possible to ask, assuming that the energized conductor is placed at the position of the arrow, regardless of the force between the energized conductor and the superconductor, the ampere force between the energized conductor and the magnet is F1. The ampere force between the energized wire and the magnet after removal of the superconductor is F2,can F1 be greater than F2?

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There is no magnetic field belonging to the magnet or superconductor. There is just the magnetic field, a magnet and a superconductor.

Basically you are solving



where magnetic field ($\mathbf{H}$) vanishes at the infinity, where magnetization $\mathbf{M}\neq\mathbf{0}$ inside the magnet, and where $\left(\mathbf{H}+\mathbf{M}\right)=\mathbf{0}$ inside the superconductor. As you can see there is nothing to tie magnetic field to either the supeconductor or the magnet.

I would expect the magnetic field between the superconductor and the magnet to be stronger than it was before the superconductor was brought closer to the magnet (did not do the actual calculations though).

I don't know what you mean by energized conductor. Charged? I do not understand your final question. I would need several diagrams to understand it (with all the forces and conductors annotated before and after any changes you make to the system).

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  • $\begingroup$ thank you very much,is to place the wire at the position of the arrow, the two ends of the wire connected to the battery into the current, the wire and the magnet between the Ampere force, in the case of superconductors and no superconductors, ampere force is different? $\endgroup$ – dan li Jan 29 '19 at 1:44
  • $\begingroup$ I would expect it to be different. I don't have time to do calculations, but my assertion is based on the fact that force would be different even without a magnet. Basically, if you take a piece of Niobium, or some other superconductor, at room temperature and place it next to current-carrying wire, the wire will experience no significant disruption (in the steady regime). Now if you cool the Niobium down, as soon as it will go superconducting, a force will start pushing the wire away form the superconductor. $\endgroup$ – Cryo Jan 29 '19 at 2:29
  • $\begingroup$ You can estimate this force by integrating the magnetic field squared inside the volume of the superconductor, i.e. find $\frac{\mu_0 H^2}{2}$ inside the volume of superconductor (when it is in normal state), due to current in the wire. This will be the energy. Now find this energy for two positions of the wire ($U_1,\, U_2$). Then the force will be $F=\frac{(U_2-U_1)}{\Delta x}$, where $\Delta x$ is the difference in wire position $\endgroup$ – Cryo Jan 29 '19 at 2:33
  • $\begingroup$ Will the ampere force between the current-carrying wire and the magnetic field act on the superconductor? $\endgroup$ – dan li Feb 6 '19 at 16:01

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