Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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,

share|improve this question
    
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

1 Answer 1

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.

share|improve this answer
    
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.