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I suppose that the moving wire is a closed circuit and that the magnetic flux enclosed is time-dependent. The Faraday’s law is of course always applicable. The current will not be infinite. Yes R=0 but, what about the self inductance L? It is never zero, in such a manner that the total E field will be null. If you make the calculations the electric field in ...

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The free energy of a circuit looks like, in the thermodynamic limit $$F\sim\dfrac{I\Phi}{2}+\dfrac{VQ}{2}$$ where $\sim$ means that I forget the temperature and pressure terms, which are not important usually in a circuit. $I$ is the current, and $\Phi$ the magnetic flux, whereas $V$ and $Q$ are voltage and charge. For instance, for an inductance, we have ...

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Introduction (BCS Theory): Cooper pairs in BCS theory are explained like this: the energy that can break cooper pairs in material is for example $10^{-4}\text{ }eV$ and thermal energy of that material is $E=kT$ where $k$ is Boltzman's Constant, so if thermal energy is lower than the energy that can break cooper pairs (in this example $kT<10^{-4}\text{ ... 0 The original Cooper instability as treated by Cooper in his seminal paper is indeed an approximation for only two electrons on top of the Fermi sea. Nevertheless, you can make the notion precise for the whole gas using mean-field approximation and renormalisation procedure. The Cooper pairs form only in the vicinity of the Fermi level. Then this region is ... 1 The BiSb type topological insulators are also protected by particle-number/charge conservation symmetry which a superconductor would break. http://arxiv.org/pdf/0901.2686v2.pdf 0 It is known from the tunneling of Cooper pairs through Josephson junctions / Squids that there are excitations in superconductors that (a) have charge exactly 2 electron charges and (b) are in a condensate. Thus, examining the oscillations in an AC Squid establishes that there are Cooper Pairs. It does not tell us about the symmetry or similar but does ... 3 Hints: Define difference$\delta:=\Delta-\Delta_0$. Deduce from$|\delta|\ll |\Delta_0|$that the lhs. of eq. (1) is $$\tag{A}\text{lhs}~\approx~ -\frac{\delta}{\Delta_0}.$$ Substitute$\xi=x\Delta $in the integral on the rhs. of eq. (1). Deduce using$\hbar \omega_D \gg \Delta$that the rhs. is$\$\tag{B} \text{rhs}~\approx~ \int_{\mathbb{R}} \! ...

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