# Meissner Effect and Lorentz Force Paradox?

The Meissner effect expels magnetic field lines in a super conductor, see the picture below. Left is normal conducting, right is the superconducting state. If I have a superconducting wire of radius $$r_0$$ now, the Meissner effect leads to $$B(r. When I drive a current $$J(r 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?

• 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 :) Jul 17 '20 at 0:52

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.