--- title: "Strong magnetic field and critical current" subject: "Electrical Engineering Material" chapter: 5 key: "semester_3/electrical_engineering_material:critical-current-strong-field" source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/critical-current-strong-field site: BCT Engineering --- # Strong magnetic field and critical current Electrical Engineering Material · Chapter 5: Superconductivity - Asked in: 2079 Bhadra, 8 marks; 2075 Chaitra, 4 marks; 2072 Chaitra, 8 marks; 2072 Kartik, 8 marks; 2069 Chaitra, 8 marks (asked 5 times) - Video: [Critical Magnetic field in superconductivity](https://www.youtube.com/watch?v=VjpCnVfteqI) — Jeena K - 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-strong-field - Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-5 ## Question How strong magnetic fields effect the superconductor? Derive the relation of critical current in superconductor with necessary diagram. ## Short answer A strong magnetic field destroys superconductivity; the critical field depends on temperature as Hc(T)=H0[1−(T/Tc)²]. For a wire of radius r, Ampere's law gives H=I/(2πr). Setting H=Hc yields critical current Ic=2πrHc and current density Jc=2Hc/r. With an external field Ha, Ic=2πr(Hc−Ha). ## Answer A superconductor stays superconducting only while the magnetic field at every point is below the **critical field** $H_c$. A strong field destroys superconductivity, and the critical field depends on temperature: $$ H_c(T) = H_0\left[1 - \left(\frac{T}{T_c}\right)^2\right] $$ where $H_0$ is the critical field at 0 K. #### Effect of a strong magnetic field - For $H < H_c$: flux is expelled (Meissner effect), resistance is zero. - For $H \ge H_c$ (type-I): flux penetrates completely and the material becomes normal, with normal resistance. - For type-II: flux starts entering at $H_{c1}$ (mixed state), and superconductivity is fully lost at $H_{c2}$. - A stronger field means a lower temperature is needed to stay superconducting, so $T_c$ falls as $H$ rises. ```text H H0 |* | * | Super- * Normal | conducting * | * +-----------------*---- T 0 Tc ``` #### Critical current (Silsbee's rule) A current in the wire itself produces a magnetic field. If this self-field at the surface reaches $H_c$, superconductivity is destroyed. The current at which this happens is the **critical current** $I_c$. Consider a long straight superconducting wire of radius $r$ carrying current $I$: ```text H (circles around wire) .-~~~~~-. / .---. \ | ( I ⊙ ) | radius r \ '---' / '-~~~~~-' ``` By Ampere's circuital law, the field at the surface is $$ \oint H\,dl = I \;\Rightarrow\; H \cdot 2\pi r = I \;\Rightarrow\; H = \frac{I}{2\pi r} $$ Superconductivity is lost when $H = H_c$, so $$ \begin{aligned} I_c &= 2\pi r H_c \\ &= 2\pi r H_0\left[1 - \left(\frac{T}{T_c}\right)^2\right] \end{aligned} $$ The **critical current density** is $$ J_c = \frac{I_c}{\pi r^2} = \frac{2H_c}{r} $$ If an external field $H_a$ is also applied (along the same direction as the self-field at the surface), the total must stay below $H_c$: $$ H_a + \frac{I}{2\pi r} \le H_c \;\Rightarrow\; I_c = 2\pi r\,(H_c - H_a) $$ So $I_c$ falls linearly as the applied field increases, and falls as temperature rises. This sets the maximum current a superconducting cable or magnet winding can carry. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does a field H >= Hc do to a type-I superconductor? A. Flux penetrates completely and it returns to the normal state with normal resistance B. It becomes a perfect conductor with zero resistance C. Only a surface layer becomes normal D. It enters a mixed state of vortices Answer: A. For H >= Hc, a type-I superconductor loses superconductivity entirely. (Key point: Effect of strong field) 2. Which expression gives the critical field at temperature T? A. Hc(T) = H0[1 - (T/Tc)] B. Hc(T) = H0[1 - (T/Tc)^2] C. Hc(T) = H0[(T/Tc)^2 - 1] D. Hc(T) = H0 (T/Tc)^2 Answer: B. Hc is maximum (H0) at 0 K and falls parabolically to zero at Tc. (Key point: Temperature dependence of Hc) 3. For a wire of radius r carrying I, what is the self-field at the surface? A. H = I / (pi r^2) B. H = 2 pi r I C. H = I / (2 pi r) D. H = I r / 2 Answer: C. Ampere's law: H x 2 pi r = I gives H = I/(2 pi r). (Key point: Ampere's law derivation) 4. With an applied field Ha along the self-field at the surface, how does Ic change? A. Ic = 2 pi r (Hc + Ha), increasing with Ha B. Ic stays 2 pi r Hc regardless of Ha C. Ic = 2 pi r Hc Ha D. Ic = 2 pi r (Hc - Ha), falling linearly as Ha increases Answer: D. The total field Ha + I/(2 pi r) must stay below Hc, so Ic decreases linearly with Ha. 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