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This is a somewhat more detailed question related to this one. The problem I want to solve is problem 1 here. What I tried:

a) From $V=-L\frac{\mathrm{d} i}{\mathrm{d} t}$, we can integrate and obtain t, right?

b) Here start my doubts: in a normal coil, the energy inside the coil is given by $U=\frac{1}{2}LI^{2}$, right? Is it the same for a superconducting coil? In a superconductor, the magnetic field inside the superconductor is $B=0$, am I right? Does this answer the question?

c) The energy released when the superconductor goes to normal state is $\Delta U=\frac{1}{2}L(I^{2}_{super} - I^{2}_{normal})$, right? And this is the energy that causes the helium to evaporate, right?

Could anybody please tell me if I am on the right direction? Or am I missed something?

Thank you in advance,

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A hint for c) what happens to $I_\text{normal}$? It's just a normal coil above the critical temperature. – Alexander Jun 30 '13 at 16:52
up vote 0 down vote accepted

The statement that $U = \frac{1}{2}L I^2$ does not depend on specifics of the system. It follows directly from the definition $V = L\partial_t I$ and the fact that $P = IV$.

There are three sources of energy which could be accounted for in the process of a superconducting quench.

  1. The energy stored in the magnetic field, i.e. the energy that comes from the inductance.
  2. The thermodynamic energy released/absorbed as the superconductor changes state.
  3. The energy supplied by the voltage source over the time the quench happens.

You can think about the relative importance of these terms.

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@BepopButUnsteady Thank you very much! With respect to the second point, is this energy $g_s - g_n = \frac {\mu_0}{2}H_c^{2}$? And with respect to the third point, the quench can be considered "instantaneous"? – neutrino May 31 '13 at 20:13

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