Superconductors — Question 2

Question 2

A persistent-current experiment gives a very small resistance bound

A closed loop is modeled by $L\,dI/dt+RI=0$, where L=2.00 H is its inductance and R is a possible constant residual resistance. Over $T=10^6$ s, current falls by less than one part in $10^5$.

1. Derive the current decay and the upper bound on R.

2. Compare with the approximation obtained from $-\ln(1-\delta)\simeq\delta$, where $\delta=10^{-5}$.

Question 2 - Solution

A persistent-current experiment gives a very small resistance bound

(a) Method and calculation

$$I(t)=I_0e^{-Rt/L},\quad I(T)/I_0>1-\delta.$$$$R<-\frac LT\ln(1-\delta)=2.00001\times10^{-11}\,\Omega.$$

(b) Interpretation and further calculation

$$R_{\mathrm{approx}}<L\delta/T=2.00\times10^{-11}\,\Omega.$$$$\frac{-\ln(1-\delta)}\delta=1+\delta/2+O(\delta^2).$$The approximate bound differs by about 5 parts per million.

Physical check

Long observation time constrains small resistance even without resolving an instantaneous voltage.