From b884a9a9dee56a29df5667681d51c238e0a4c0b5 Mon Sep 17 00:00:00 2001
From: Kees van Kempen <k.vankempen@student.science.ru.nl>
Date: Thu, 17 Mar 2022 18:16:45 +0100
Subject: [PATCH] ass3: Task 10a solved in the ugliest manner

---
 superconductivity_assignment3_kvkempen.tex | 25 ++++++++++++++++++++++
 1 file changed, 25 insertions(+)

diff --git a/superconductivity_assignment3_kvkempen.tex b/superconductivity_assignment3_kvkempen.tex
index 49fb293..0dee10f 100755
--- a/superconductivity_assignment3_kvkempen.tex
+++ b/superconductivity_assignment3_kvkempen.tex
@@ -130,7 +130,32 @@ This is very close to what is written in the assignment, but not precisely the s
 \section{Fine type-II superconducting wire}
 
 \section{Critical currents}
+\textbf{(a)}
+Silsbee's rule states that the supercurrents through the wire must not generate magnetic fields in excess of $B_c$ at the surface of the wire.
+We assume that the supercurrent is maximal at the surface with a maximum value of $J_{max}$, and that the supercurrent decays linearly from the surface to zero at a penetration depth $\lambda$ deep.
+We thus find a relation for the supercurrent as function of the cylindrical radius $r$ as
+\[
+	J_s(r) = \frac{J_{max}}{\lambda} \left[ r - R + \lambda \right].
+\]
 
+Now we can use the Maxwell-Amp\`ere law to find this value for $J_{max}$.
+\[
+	\oint \vec{B}\cdot d\vec{\ell} = \mu_r\mu_0\iint\vec{J_s}\cdot d\vec{S}
+\]
+Using $\vec{B} = \vec{B_c}$, $\mu_r = 1$ as we're calculating the field outside the sc, $J_s = J_s(r)$, the area over which $d\vec{S}$ runs to be the small ring from $r = R - \lambda$ to $r = R$, and the path along which $d\vec{\ell}$ runs to be the loop $2\pi R$ along the surface of the wire.
+This gives us
+\[
+	2\pi R B_c = \mu_0 \int_{\phi = 0}^{2\pi}\int_{r=R-\lambda}^R \frac{J_{max}}{\lambda}\left[ r - R + \lambda \right] rdr d\phi.
+\]
+Solving the integral over $\phi$ results in
+\[
+	R B_c = \mu_0 \int_{r=R-\lambda}^R \frac{J_{max}}{\lambda}\left[ r - R + \lambda \right] rdr = \frac{\mu_0J_{max}}{\lambda} \left[ \frac{r^3}{3} - \frac{(R + \lambda)r^2}{2} \right]_{r = R-\lambda}^R.
+\]
+Solving for $J_{max}$, this yields the beautiful expression
+\[
+	J_{max} = \frac{6B_c\lambda R}{\mu \left[ 4\lambda^3 - 9\lambda^2R + 3\lambda R^2- 3\lambda R + 3R^3 -3R^2 \right]}.
+	% https://www.wolframalpha.com/input?i=R*B+%3D+m*x%2Fl*%28%28R%5E3-%28R-l%29%5E3%29%2F3+-+%28R%2Bl%29*%28R+-+%28R-l%29%5E2%29%2F2%29
+\]
 \section{A weak junction}
 See the code in appendix \ref{appendix:program-task-12}.