--- title: "Critical current decreases linearly with field" subject: "Electrical Engineering Material" chapter: 5 key: "semester_3/electrical_engineering_material:critical-current-linear-decrease" source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/critical-current-linear-decrease site: BCT Engineering --- # Critical current decreases linearly with field Electrical Engineering Material · Chapter 5: Superconductivity - Asked in: 2071 Shrawan, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/critical-current-linear-decrease - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-5 ## Question What is critical current? Prove that the critical current decreases linearly with the increase in applied field for a wire. ## Short answer Critical current Ic is the maximum supercurrent a wire can carry. Its own surface field I/(2πr) plus applied field Ha must not exceed the critical field Hc. Hence Ic = 2πr(Hc – Ha), which decreases linearly with Ha from 2πrHc at zero field to zero at Hc. ## Answer The **critical current** $I_c$ is the maximum current a superconductor can carry while remaining superconducting. Beyond it, the magnetic field produced by the current (plus any applied field) exceeds the critical field $H_c$ and the material becomes normal (**Silsbee's rule**). #### Field produced by the current Take a long superconducting wire of radius $r$ carrying current $I$. ```text H_a (applied) ---------------> .-~~~~~~-. / .----. \ H_I = I/(2πr) | ( I ⊙ ) | circles round \ '----' / '-~~~~~~-' ---------------> ``` By Ampere's circuital law, the self-field at the surface is $$ \oint H\,dl = I \;\Rightarrow\; H_I = \frac{I}{2\pi r} $$ Without an applied field, superconductivity is lost when $H_I = H_c$: $$ I_{c0} = 2\pi r H_c $$ #### With an applied field Let a uniform external field $H_a$ act on the wire. At the point of the surface where the self-field and applied field point the same way, the total field is the sum: $$ H_{total} = H_a + \frac{I}{2\pi r} $$ The wire stays superconducting only while $H_{total} \le H_c$. The critical condition is $$ \begin{aligned} H_a + \frac{I_c}{2\pi r} &= H_c \\ I_c &= 2\pi r\,(H_c - H_a) \\ I_c &= I_{c0} - 2\pi r\,H_a \end{aligned} $$ This is a straight line in $H_a$ with negative slope $-2\pi r$. So the critical current **decreases linearly** with the applied field: - at $H_a = 0$: $I_c = 2\pi r H_c$ (maximum), - at $H_a = H_c$: $I_c = 0$. ```text Ic 2πrHc|\ | \ | \ | \ +--------\---- Ha 0 Hc ``` Since $H_c = H_0[1 - (T/T_c)^2]$, the whole line moves down as temperature rises. Critical current density is $J_c = I_c/\pi r^2 = 2(H_c - H_a)/r$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the critical current? A. The maximum current a superconductor carries while remaining superconducting B. The current that first produces resistance in a normal metal C. The current at which the Meissner effect starts D. The minimum current needed for superconductivity Answer: A. Beyond Ic the total field exceeds Hc. (Key point: Critical current meaning) 2. With applied field Ha along the self-field, what is the total surface field? A. Ha - I/(2 pi r) B. Ha + I/(2 pi r) C. Ha x I/(2 pi r) D. I/(2 pi r) only Answer: B. Fields in the same direction add. (Key point: Total field) 3. Setting total = Hc gives which relation? A. Ic = 2 pi r (Hc + Ha) B. Ic = 2 pi r Hc Ha C. Ic = 2 pi r (Hc - Ha) D. Ic = (Hc - Ha)/(2 pi r) Answer: C. Ha + Ic/(2 pi r) = Hc rearranges to Ic = 2 pi r (Hc - Ha). (Key point: Result) 4. Why does Ic decrease linearly with Ha? A. Ic is proportional to Ha squared B. Ic is constant until Ha = Hc C. Ic falls exponentially with Ha D. Ic versus Ha is a straight line of slope -2 pi r, reaching zero at Ha = Hc Answer: D. Ic = Ic0 - 2 pi r Ha is linear with negative slope. 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