Superconductors — Question 1

Section 26-8 · Physics pages 4292–4293

Question 1 A null voltage reading bounds resistance but does not establish exact zero A four-terminal measurement drives I=2.00 A through a specimen. After offsets are removed, its voltage magnitude is below a hard detection limit Vlim = 20.0 nV. Assume any residual response is ohmic. specimen + − I voltage probes Schematic; labels, not drawing lengths, specify the data. (a) Find the upper bound on resistance and dissipated power. (b) If current increases tenfold without exceeding the critical current, find the new resistance bound for the same voltage limit. Explain the logical limit of either measurement.
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Question 1 - Solution A null voltage reading bounds resistance but does not establish exact zero specimen + − I voltage probes (a) Method and calculation 0 ≤R < Vlim/I = (20.0 × 10−9)/2.00 = 10.0 nΩ. P = I2R < IVlim = 40.0 nW. (b) Interpretation and further calculation I′ = 20.0 A ⇒R < Vlim/I′ = 1.00 nΩ. R < Rbound̸ ⇒R = 0. Finite resolution establishes a bound; superconductivity also has characteristic magnetic behavior. Physical check The current must stay within the superconducting regime; a stronger measuring current can otherwise change the state.

Question 1

A null voltage reading bounds resistance but does not establish exact zero

A four-terminal measurement drives I=2.00 A through a specimen. After offsets are removed, its voltage magnitude is below a hard detection limit Vlim=20.0V_{\mathrm{lim}}=20.0 nV. Assume any residual response is ohmic.

Diagram shown in the original SVG worksheet.

  1. Find the upper bound on resistance and dissipated power.

  2. If current increases tenfold without exceeding the critical current, find the new resistance bound for the same voltage limit. Explain the logical limit of either measurement.

Question 1 - Solution

A null voltage reading bounds resistance but does not establish exact zero

Diagram shown in the original SVG worksheet.

(a) Method and calculation

0≤R<Vlim/I=(20.0×10−9)/2.00=10.0nΩ.0\le R< V_{\mathrm{lim}}/I=(20.0\times10^{-9})/2.00=10.0\,\mathrm{n\Omega}.P=I2R<IVlim=40.0nW.P=I^2R<I V_{\mathrm{lim}}=40.0\,\mathrm{nW}.

(b) Interpretation and further calculation

I′=20.0A⇒R<Vlim/I′=1.00nΩ.I\prime=20.0\,\mathrm A\Rightarrow R<V_{\mathrm{lim}}/I\prime=1.00\,\mathrm{n\Omega}.R<Rbound⇏R=0.R<R_{\mathrm{bound}}\ \not\Rightarrow\ R=0.Finite resolution establishes a bound; superconductivity also has characteristic magnetic behavior.

Physical check

The current must stay within the superconducting regime; a stronger measuring current can otherwise change the state.