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The reason is that those are two different calculations with two different integrals but the hypothesis that link them is made before arriving at the final form. In the Cooper Problem the hypothesis on the potential states that it is non-zeri and attractive when the energies of the electrons are in $[E_F ; E_F+ \hbar \omega_D ]$ and comes from the rigid ...

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A superconductor can indeed carry current without dissipation. For example, in a superconducting loop you could have a current circulate essentially forever. If by "a perpetual electricity stream" you mean you could extract perpetual work out of this current, then the answer is no. By definition, extracting work from a system means adding an ...

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Superconductivity is one of those topics where it's really easy to give an intuitive & wrong explanation. One can go with a simplified picture at first, but when you start understanding it, you are going to come up with more questions, the answers to which will start undoing the (incorrect) premises to the initial, intuitive, answer. So, I will give a ...

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Not a complete answer, but your solution for j$_s$(r) is wrong. You seem to operate in cylindrical coordinates, but tried to solve for r as it it were a Cartesian coordinates. In other words, you tried to solve j''(r) = j(r), but this is not the Laplacian in cylindrical coordinates. If you write your equation in cylindrical coordinates, you should get Bessel ...

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In transition-edge sensors (TES), the superconducting detector is held very close to its critical temperature, so that it's resistance is a very steep function of the temperature. The figure below, from a review paper on TES, demonstrates the resistance vs. temperature in the right panel. In operation, the sensor temperature will be kept somewhere on the ...

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I can see two big differences between a hypothetical perfect conductor and a superconductor: 1. A "perfect" conductor will keep its magnetic flux condition. A superconductor will always expel the magnetic flux (Meissner effect). What does this mean? Let's say that you have a regular conductor in an external magnetic field, which, of course, is also ...

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