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I am a first year A-level student and I am doing a project about the possibility of storing electrical energy in a superconductor. I have researched and I am aware of the critical current density and the critical magnetic field of different superconductors, where the magnetic field created by the wire (Ampere’s law) interacts with the magnetic field of the superconductor (Meissner effect). But, if I had a loop of superconductor and I cool it down enough (with that I mean enough so its resistance becomes completely 0) then could I use a huge voltage and a relatively small current (not big enough to reach its critical current) in order to store electrical energy in the superconductor?

If it is not possible I would like to know why not and if it is possible I would like to know why we are not using it nowadays as a form of storing electrical energy (is it only the practical issues of cooling it down)?

Thank you very much

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Because of the zero resistivity, you can not apply a huge voltage on a superconductor. –  Everett You Apr 17 '13 at 1:19
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As Everett says in his comment, you can't apply a huge voltage, or indeed any voltage, across a superconductor. Because the resistance is zero the potential difference between any two points in the superconductor is also zero.

If you have a superconducting loop/coil, when you put energy into it you are basically storing the energy in the magnetic field generated by the loop. This is just like any inductor, except that in a superconductor the energy isn't dissipated by resistive losses. The energy stored is proportional to the current squared, and there isn't any way around this. See this Wikipedia article for more details.

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Forgive me but that is one of the problems that I don't understand why I cannot put a huge voltage on a superconductor? Making first clear that by voltage in this question I am talking about the e.m.f. and not about the p.d. which would be 0. Because from my point of view if you put a huge e.m.f. on the superconductor, understanding by e.m.f. the potential of a coulomb to do work, this potential (regardless of the amount of e.m.f.) should stay the same since it haven’t got to do work against the resistance of the superconductor. So why couldn’t I put a great e.m.f. through the superconductor? –  Josep Mondaca Apr 17 '13 at 8:44
    
I think you're getting mixed up with the power dissipated in a circuit. For some (resisitive) circuit the power dissipated is given by $W = IV$, so for the same power dissipation you can have large $I$ and small $V$ or small $I$ and large $V$. But we aren't talking about power dissipation: we are talking about power storage. The only circuit elements that store power are capacitors or inductors, that store power as electric and magnetic fields respectively. –  John Rennie Apr 17 '13 at 9:18
    
Thank you very much for your help and time. It was really helpfull for my project and helped me to understand how energy storage works. –  Josep Mondaca Apr 26 '13 at 11:32
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