finished solutions 2

[lectures/latex.git] / solid_state_physics / tutorial / 2_02s.tex

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+\pdfoutput=0
+\documentclass[a4paper,11pt]{article}
+\usepackage[activate]{pdfcprot}
+\usepackage{verbatim}
+\usepackage{a4}
+\usepackage{a4wide}
+\usepackage[german]{babel}
+\usepackage[latin1]{inputenc}
+\usepackage[T1]{fontenc}
+\usepackage{amsmath}
+\usepackage{ae}
+\usepackage{aecompl}
+\usepackage[dvips]{graphicx}
+\graphicspath{{./img/}}
+\usepackage{color}
+\usepackage{pstricks}
+\usepackage{pst-node}
+\usepackage{rotating}
+
+\setlength{\headheight}{0mm} \setlength{\headsep}{0mm}
+\setlength{\topskip}{-10mm} \setlength{\textwidth}{17cm}
+\setlength{\oddsidemargin}{-10mm}
+\setlength{\evensidemargin}{-10mm} \setlength{\topmargin}{-1cm}
+\setlength{\textheight}{26cm} \setlength{\headsep}{0cm}
+
+\renewcommand{\labelenumi}{(\alph{enumi})}
+\renewcommand{\labelenumii}{\arabic{enumii})}
+\renewcommand{\labelenumiii}{\roman{enumiii})}
+
+\begin{document}
+
+% header
+\begin{center}
+ {\LARGE {\bf Materials Physics II}\\}
+ \vspace{8pt}
+ Prof. B. Stritzker\\
+ SS 2008\\
+ \vspace{8pt}
+ {\Large\bf Tutorial 2 - proposed solutions}
+\end{center}
+
+\section{Critical current in the surface region of a type 1 superconductor}
+\[
+ j_s(r)=j_s(R)\exp\left(\frac{-(R-r)}{\lambda}\right)
+\]
+$R$: radius of the wire, $r$: distance from the cylinder axis,
+and $\lambda$: London penetration depth.
+
+\begin{enumerate}
+ \item $I_c=\int_0^R dr \int_0^{2\pi} d\phi \, j_c(r) r$\\
+       $\Rightarrow I_c=\int_0^R dr \, 2\pi j_c(r) r
+        = j_c(R)2\pi \int_0^R dr \, r \exp(-(R-r)/\lambda)
+       = j_c(R)2\pi \exp(-R/\lambda) \int_0^R dr \, r \exp(r/\lambda)$
+       $x=\frac{r}{\lambda} \rightarrow \frac{dx}{dr}=\frac{1}{\lambda}
+        \Rightarrow dr=\lambda dx$, $r=\lambda x$
+       $\Rightarrow
+        I_c=j_c(R)2\pi \lambda^2 \exp(-R/\lambda) \int_0^R d(\frac{r}{\lambda})
+           \, \frac{r}{\lambda} \exp(\frac{r}{\lambda})$
+       Integration by parts: $\int uv' = uv - \int vu'$\\
+       $\int xe^x dx = xe^x-\int e^x dx=xe^x-e^x+c=e^x(x-1)+c$\\
+       $\Rightarrow
+        I_c=j_c(R)2\pi \lambda^2\exp(-R/\lambda) \left[
+        \exp(\frac{r}{\lambda})(r\lambda-1) \right]_0^R$\\
+       $\left[ \exp(\frac{r}{\lambda})(r\lambda-1) \right]_0^R=
+        \exp(R/\lambda)(R/\lambda-1)-1(-1)$\\
+       $\Rightarrow I_c=j_c(R)2\pi \lambda^2 \left[
+        \underbrace{\exp(-R/\lambda)\exp(R/\lambda)}_{=1}
+       (R/\lambda-1)+
+       \underbrace{\exp(-R/\lambda)}_{\approx 0 \text{, since } \lambda\ll R} \right]$\\
+       $\Rightarrow
+        I_c\approx j_c(R)2\pi \lambda^2
+       \underbrace{(R/\lambda-1)}_{\approx R/\lambda}\approx
+       j_c(R)2\pi \lambda R$\\
+       $\Rightarrow j_c(R)\approx\frac{I_c}{2\pi R\lambda}$
+ \item $j_c(R,T=0)=\frac{I_c(T=0)}{2\pi R\lambda(T=0)}
+        =7.9\cdot 10^7\frac{A}{cm^2}$
+\end{enumerate}
+
+\section{Penetration of the magnetic field into a type 1 superconductor}
+\[
+ {\bf B}_s=\mu_0 \left({\bf H}_a + {\bf M}_s\right)
+\]
+${\bf H}_a$: strength of the applied magnetic field,
+${\bf M}_s$: magnetization of the superconductor.
+
+\begin{enumerate}
+ \item $\frac{1}{\mu_0} rot {\bf B}_s =
+        rot {\bf H}_a + rot {\bf M}_s
+       \stackrel{rot {\bf H}={\bf j}+\frac{d{\bf D}}{dt}}{=}
+       \underbrace{{\bf j}_a}_{=0}
+       +{\bf j}_s$\\
+       $\Rightarrow
+        rot {\bf B}_s=\mu_0 {\bf j}_s \qquad | rot ...$\\
+       $\Rightarrow
+        \underbrace{rot rot {\bf B}_s}_{grad \underbrace{div {\bf B}_s}_{=0}
+       -\Delta {\bf B}_s}
+       =\mu_0rot {\bf j}_s\stackrel{\text{London II}}{=}
+       -\frac{\mu_0}{\Lambda}{\bf B}_s$\\
+       $\Rightarrow
+        \Delta {\bf B}_s=\frac{\mu_0}{\Lambda}{\bf B}_s$
+ \item ${\bf B}_a=\mu_0 H_a {\bf e}_z$, ${\bf B}_s=B_{s_z}(x) {\bf e}_x$\\
+       Diff. equation: $\frac{d^2}{dx^2}B_{s_z}(x)-
+                        \frac{\mu_0}{\Lambda}B_{s_z}(x)=0$\\
+       Solution: $B_{s_z}(x)=B_{s_z}(0)\exp(-\frac{x}{\lambda})$, with
+                 $\lambda=\sqrt{\frac{\Lambda}{\mu_0}}$\\
+       \includegraphics[width=16cm]{mfsc.ps}
+ \item $\mu_0 {\bf j}_s=rot{\bf B}_s=
+        \frac{-\partial B_{s_z}(x)}{\partial x} {\bf e}_y=
+       \frac{1}{\lambda} B_a \exp(-\frac{x}{\lambda}){\bf e}_y$\\
+       $\Rightarrow$ Direction of screening current is ${\bf e}_y$.\\
+       $\Rightarrow$ Exponential decay inside the SC.\\
+       Interface: ${\bf j}_s(x=0)=\frac{B_a}{\lambda\mu_0} {\bf e}_y$
+\end{enumerate}
+
+\end{document}