Superconductors — Question 4

Section 26-8 · Physics pages 4298–4299

Question 4 Temperature margin limits a superconducting tape current A tape has a supplied engineering critical-current-density law Jc(T) = Jc0[1 −(T/Tc)2] for 0 ≤T < Tc. The engineering area A is fixed. It operates at current I = .60AJc0, and the admissible condition is I < AJc(T). T/Tc Jc Axes and labels specify the model; the sketch is schematic. (a) Find the largest allowed temperature as a fraction of Tc. (b) At T = .50Tc, find the fractional current margin (Ic −I)/Ic and the increase in critical current if temperature is lowered to .30Tc.
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Question 4 - Solution Temperature margin limits a superconducting tape current T/Tc Jc (a) Method and calculation .60 < 1 −(T/Tc)2 ⇒T/Tc < √ .40. Tboundary = .632456Tc (operate below this boundary). (b) Interpretation and further calculation Ic(.50Tc) = .75AJc0, Ic −I Ic = .75 −.60 .75 = .200. Ic(.30Tc)/Ic(.50Tc) = .91/.75 = 1.21333. Physical check This is an area-averaged engineering limit; it does not assume microscopic current density is uniform through the tape.

Question 4

Temperature margin limits a superconducting tape current

A tape has a supplied engineering critical-current-density law Jc(T)=Jc0[1−(T/Tc)2]J_c(T)=J_{c0}[1-(T/T_c)^2] for 0≤T<Tc0\le T<T_c. The engineering area A is fixed. It operates at current I=.60AJc0I=.60AJ_{c0}, and the admissible condition is I<AJc(T)I<AJ_c(T).

Diagram shown in the original SVG worksheet.

  1. Find the largest allowed temperature as a fraction of Tc.

  2. At T=.50TcT=.50T_c, find the fractional current margin (Ic−I)/Ic(I_c-I)/I_c and the increase in critical current if temperature is lowered to .30Tc.30T_c.

Question 4 - Solution

Temperature margin limits a superconducting tape current

Diagram shown in the original SVG worksheet.

(a) Method and calculation

.60<1−(T/Tc)2⇒T/Tc<.40..60<1-(T/T_c)^2\Rightarrow T/T_c<\sqrt{.40}.Tboundary=.632456Tc(operate below this boundary).T_{\mathrm{boundary}}=.632456T_c\quad\text{(operate below this boundary)}.

(b) Interpretation and further calculation

Ic(.50Tc)=.75AJc0,Ic−IIc=.75−.60.75=.200.I_c(.50T_c)=.75AJ_{c0},\quad\frac{I_c-I}{I_c}=\frac{.75-.60}{.75}=.200.Ic(.30Tc)/Ic(.50Tc)=.91/.75=1.21333.I_c(.30T_c)/I_c(.50T_c)=.91/.75=1.21333.

Physical check

This is an area-averaged engineering limit; it does not assume microscopic current density is uniform through the tape.