Persistent current (superconductivity) (source code)

= Persistent current
{disambiguate=superconductivity}
{title2=$I$}

= Superconducting persistent current
{synonym}

A <superconducting persistent current> is an <electric current> circulating in a closed path with <superconductivity> after the external driving <electromotive force> has ceased. Its associated <magnetic field> makes the circulation observable without inserting a measuring instrument into the path. This article concerns macroscopic circulation in <superconductivity>. A finite experiment establishes persistence over its observation window with a stated <measurement error>, rather than directly observing an infinite lifetime.

The ideal circuit connection is <flux conservation in a superconducting loop>. In a fixed loop with <self-inductance> $L$, a single circulating <electric current> $I$, and externally imposed linked <magnetic flux> $\Phi_{\rm ext}$, the lumped circuit approximation gives
$$
L\frac{dI}{dt}+RI=-\frac{d\Phi_{\rm ext}}{dt}.
$$
When the <electrical resistance> $R$ vanishes, integrating gives $LI+\Phi_{\rm ext}=\text{constant}$. Removing the external linked <magnetic flux> therefore changes the circulating <electric current>; holding it fixed afterwards preserves that <electric current>. This is a circuit approximation with a specified <self-inductance>, not a derivation of the microscopic state of <superconductivity>. For thin paths where the carrier contribution matters, <fluxoid and flux quantization in a superconductor> supplies the more careful description.

For constant $L>0$, constant effective <electrical resistance> $R\geq0$, and no changing external linked <magnetic flux>, the <RL circuit> instead predicts
$$
I(t)=I_0e^{-Rt/L},\qquad \tau=\frac{L}{R}\quad(R>0).
$$
This <exponential decay> follows by separating $dI/I=-(R/L)dt$. A long relaxation time tests extremely small <electrical resistance> without trying to resolve an equally tiny steady <electric potential difference>. The stored <magnetic energy> is $U=LI^2/2$, so the same model gives $dU/dt=-RI^2$, exactly the <Joule heating> loss. The finite <energy> stored in the loop supplies the magnetic readout; the circulating <electric current> is not a continuously powered source of useful work.