solid_state_physics/tutorial/2_02s.tex

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  30 \begin{document}
  31
  32 % header
  33 \begin{center}
  34  {\LARGE {\bf Materials Physics II}\\}
  35  \vspace{8pt}
  36  Prof. B. Stritzker\\
  37  SS 2008\\
  38  \vspace{8pt}
  39  {\Large\bf Tutorial 2 - proposed solutions}
  40 \end{center}
  41
  42 \section{Critical current in the surface region of a type 1 superconductor}
  43 \[
  44  j_s(r)=j_s(R)\exp\left(\frac{-(R-r)}{\lambda}\right)
  45 \]
  46 $R$: radius of the wire, $r$: distance from the cylinder axis,
  47 and $\lambda$: London penetration depth.
  48
  49 \begin{enumerate}
  50  \item $I_c=\int_0^R dr \int_0^{2\pi} d\phi \, j_c(r) r$\\
  51        $\Rightarrow I_c=\int_0^R dr \, 2\pi j_c(r) r
  52         = j_c(R)2\pi \int_0^R dr \, r \exp(-(R-r)/\lambda)
  53         = j_c(R)2\pi \exp(-R/\lambda) \int_0^R dr \, r \exp(r/\lambda)$
  54        $x=\frac{r}{\lambda} \rightarrow \frac{dx}{dr}=\frac{1}{\lambda}
  55         \Rightarrow dr=\lambda dx$, $r=\lambda x$
  56        $\Rightarrow
  57         I_c=j_c(R)2\pi \lambda^2 \exp(-R/\lambda) \int_0^R d(\frac{r}{\lambda})
  58             \, \frac{r}{\lambda} \exp(\frac{r}{\lambda})$
  59        Integration by parts: $\int uv' = uv - \int vu'$\\
  60        $\int xe^x dx = xe^x-\int e^x dx=xe^x-e^x+c=e^x(x-1)+c$\\
  61        $\Rightarrow
  62         I_c=j_c(R)2\pi \lambda^2\exp(-R/\lambda) \left[
  63         \exp(\frac{r}{\lambda})(r\lambda-1) \right]_0^R$\\
  64        $\left[ \exp(\frac{r}{\lambda})(r\lambda-1) \right]_0^R=
  65         \exp(R/\lambda)(R/\lambda-1)-1(-1)$\\
  66        $\Rightarrow I_c=j_c(R)2\pi \lambda^2 \left[
  67         \underbrace{\exp(-R/\lambda)\exp(R/\lambda)}_{=1}
  68         (R/\lambda-1)+
  69         \underbrace{\exp(-R/\lambda)}_{\approx 0 \text{, since } \lambda\ll R} \right]$\\
  70        $\Rightarrow
  71         I_c\approx j_c(R)2\pi \lambda^2
  72         \underbrace{(R/\lambda-1)}_{\approx R/\lambda}\approx
  73         j_c(R)2\pi \lambda R$\\
  74        $\Rightarrow j_c(R)\approx\frac{I_c}{2\pi R\lambda}$
  75  \item $j_c(R,T=0)=\frac{I_c(T=0)}{2\pi R\lambda(T=0)}
  76         =7.9\cdot 10^7\frac{A}{cm^2}$
  77 \end{enumerate}
  78
  79 \section{Penetration of the magnetic field into a type 1 superconductor}
  80 \[
  81  {\bf B}_s=\mu_0 \left({\bf H}_a + {\bf M}_s\right)
  82 \]
  83 ${\bf H}_a$: strength of the applied magnetic field,
  84 ${\bf M}_s$: magnetization of the superconductor.
  85
  86 \begin{enumerate}
  87  \item $\frac{1}{\mu_0} rot {\bf B}_s =
  88         rot {\bf H}_a + rot {\bf M}_s
  89         \stackrel{rot {\bf H}={\bf j}+\frac{d{\bf D}}{dt}}{=}
  90         \underbrace{{\bf j}_a}_{=0}
  91         +{\bf j}_s$\\
  92        $\Rightarrow
  93         rot {\bf B}_s=\mu_0 {\bf j}_s \qquad | rot ...$\\
  94        $\Rightarrow
  95         \underbrace{rot rot {\bf B}_s}_{grad \underbrace{div {\bf B}_s}_{=0}
  96         -\Delta {\bf B}_s}
  97         =\mu_0rot {\bf j}_s\stackrel{\text{London II}}{=}
  98         -\frac{\mu_0}{\Lambda}{\bf B}_s$\\
  99        $\Rightarrow
 100         \Delta {\bf B}_s=\frac{\mu_0}{\Lambda}{\bf B}_s$
 101  \item ${\bf B}_a=\mu_0 H_a {\bf e}_z$, ${\bf B}_s=B_{s_z}(x) {\bf e}_x$\\
 102        Diff. equation: $\frac{d^2}{dx^2}B_{s_z}(x)-
 103                         \frac{\mu_0}{\Lambda}B_{s_z}(x)=0$\\
 104        Solution: $B_{s_z}(x)=B_{s_z}(0)\exp(-\frac{x}{\lambda})$, with
 105                  $\lambda=\sqrt{\frac{\Lambda}{\mu_0}}$\\
 106        \includegraphics[width=16cm]{mfsc.ps}
 107  \item $\mu_0 {\bf j}_s=rot{\bf B}_s=
 108         \frac{-\partial B_{s_z}(x)}{\partial x} {\bf e}_y=
 109         \frac{1}{\lambda} B_a \exp(-\frac{x}{\lambda}){\bf e}_y$\\
 110        $\Rightarrow$ Direction of screening current is ${\bf e}_y$.\\
 111        $\Rightarrow$ Exponential decay inside the SC.\\
 112        Interface: ${\bf j}_s(x=0)=\frac{B_a}{\lambda\mu_0} {\bf e}_y$
 113 \end{enumerate}
 114
 115 \end{document}