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% 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}

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