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The Meissner effect expels magnetic field lines in a super conductor, see the picture below. Left is normal conducting, right is the superconducting state.

enter image description here

If I have a superconducting wire of radius $r_0$ now, the Meissner effect leads to $B(r<r_0)=0$. When I drive a current $J(r<r_0)=J_0$ this current leads to zero forces when calculating the force using the Lorentz force: $f=j\times B = 0$.

This would mean that all superconducting coils experience no forces. What am I missing here?

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  • $\begingroup$ For condensed matter, there is also (since only 2 months ago!): materials.stackexchange.com, but I agree with the answer you have already received here :) $\endgroup$ Jul 17 '20 at 0:52
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A great question!

The answer is that there is a still a force of ${\bf I}\times {\bf B}_{\rm external}$ per unit length of the wire. This is because the ${\bf B}$ field does penetrate some distance (naturally this is called the penetration depth) into the supercondcuting wire, and, for reasons similar to the Meissner effect itself, this near-surface penetration depth region is also where the current carried by the wire flows. That the location of the current and strength of the penetrating ${\bf B}$ field conspire to give exactly the same answer for the force as if there were no Mesissner effect is not exactly obvious. It is, however, a magnetic analogue of the statement that if you put a charge $Q$ on a conducting body and immerse the body in a uniform electric field ${\bf E}_{\rm external}$ then the force on the body is still exactly $Q {\bf E}_{\rm external}$ despite the fact that there is no ${\bf E}$ field inside the conducting body.

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