Physics Asked on October 29, 2021
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 $B(r<r_0)=0$. 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: $f=jtimes B = 0$.
This would mean that all superconducting coils experience no forces. What am I missing here?
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.
Answered by mike stone on October 29, 2021
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