From 6ef31a964413ef57a524a8068de0dbbada45f7f7 Mon Sep 17 00:00:00 2001
From: Kees van Kempen <k.vankempen@student.science.ru.nl>
Date: Fri, 18 Feb 2022 10:31:15 +0100
Subject: [PATCH] ass1: Finish the assignment ugh deadline. I was 28 minutes
 late.

---
 superconductivity_assignment1_kvkempen.tex | 21 +++++++++++++++++++--
 1 file changed, 19 insertions(+), 2 deletions(-)

diff --git a/superconductivity_assignment1_kvkempen.tex b/superconductivity_assignment1_kvkempen.tex
index c6b62d3..4962825 100755
--- a/superconductivity_assignment1_kvkempen.tex
+++ b/superconductivity_assignment1_kvkempen.tex
@@ -169,7 +169,7 @@ This way we find our result:
 %If we now substitute in the second London equation, assuming that $\Lambda$ is constant over the material,
 \end{em}
 
-\item Assume we have the situation as sketched in figure \ref{fig:slab}. A superconducting slab is placed at $x = 0$ extending to infinity in both $x$ and $z$ directions. (Note that the $y$ dimensions do not matter.) A uniform external magnetic field $\vec{B} = B \hat{x}$ is applied. Use the just derived screening equation to calculate the field $\vec{B}$ inside the superconductor.
+\item Assume we have the situation as sketched in figure \ref{fig:slab}. A superconducting slab is placed at $x = 0$ extending to infinity in both $x$, $\pm y$ and $\pm z$ directions. A uniform external magnetic field $\vec{B} = B \hat{x}$ is applied. Use the just derived screening equation to calculate the field $\vec{B}$ inside the superconductor.
 
 \begin{figure}[H]
 	\label{fig:slab}
@@ -178,7 +178,24 @@ This way we find our result:
 \end{figure}
 
 \begin{em}
-
+Due to symmetry in both $z$ and $y$ directions, we can already tell that the result will depend on $x$ only.
+So for a $\vec{B} = \vec{B}(x)$,
+\[
+	\nabla^2\vec{B} = \frac{\partial^2 \vec{B}}{\partial x^2},
+\]
+Now from the screening equation we find
+\[
+	\nabla^2\vec{B} = \frac{\partial^2 \vec{B}}{\partial x^2} = \frac{\vec{B}}{\lambda_L^2}.
+\]
+The familiar solution to this is the superposition of exponents to the left and right:
+\[
+	\vec{B}(x) = [c_1 e^{x/\lambda_L} + c_2 e^{x/\lambda_L}]\hat{z}.
+\]
+Using that $\vec{B}(0) = \vec{B}$, and that the field cannot diverge for large $x$,
+we find that $c_1 = 0$ and $c_2 = B$, so that
+\[
+	\vec{B}(x) = \vec{B} e^{-x/\lambda_L}.
+\]
 \end{em}
 
 \end{enumerate}