From: hackbard <hackbard@sage.physik.uni-augsburg.de>
Date: Thu, 26 Jun 2008 09:41:26 +0000 (+0200)
Subject: small fixes
X-Git-Url: https://hackdaworld.org/gitweb/?p=lectures%2Flatex.git;a=commitdiff_plain;h=e53bf68c01b1f2408f15dcc155ca0715fdf7aa20

small fixes
---

diff --git a/solid_state_physics/tutorial/2_01s.tex b/solid_state_physics/tutorial/2_01s.tex
index 22abb78..033c8ae 100644
--- a/solid_state_physics/tutorial/2_01s.tex
+++ b/solid_state_physics/tutorial/2_01s.tex
@@ -62,7 +62,7 @@
         \item $I = (\textrm{charge}) \cdot (\textrm{loops per time})
 	       \stackrel{1/T=\omega_L/2\pi}{=}
 	       (Ze)(\frac{1}{2\pi}\frac{-e}{2m}B)$\\
-	      $\mu=IA=I2\pi<\rho^2>=-\frac{Ze^2B}{4m}<\rho^2>$\\
+	      $\mu=IA=I\pi<\rho^2>=-\frac{Ze^2B}{4m}<\rho^2>$\\
 	      $<x^2>=<y^2>=<z^2> \Rightarrow <r^2>=3<x^2>=3<y^2>$\\
 	      $<\rho^2>=<x^2>+<y^2>=\frac{2}{3}<r^2>$\\
 	      $\mu=-\frac{Ze^2B}{6m}$
diff --git a/solid_state_physics/tutorial/2_02.tex b/solid_state_physics/tutorial/2_02.tex
index ecafd2a..b9b5c5a 100644
--- a/solid_state_physics/tutorial/2_02.tex
+++ b/solid_state_physics/tutorial/2_02.tex
@@ -58,7 +58,8 @@ and $\lambda$ is the London penetration depth.
        {\bf Hint:}
        Use the relation $I_c=\int_0^R dr \int_0^{2\pi} d\phi \, j_c(r) r$
        and integration by parts.
- \item Calculate $j_c(R,T=0K)$ for a wire of Sn with a radius of 1 mm at $T=0K$.
+ \item Calculate $j_c(R,T=0K)$ for a wire of Sn with a diameter of 1 mm
+       at $T=0K$.
        The critical current and penetration depth at $T=0K$  are
        $I_c=75\, A$ and $\lambda =300\cdot 10^{-10}\, m$.
 \end{enumerate}
diff --git a/solid_state_physics/tutorial/2_02s.tex b/solid_state_physics/tutorial/2_02s.tex
index 8c12196..8f3eda0 100644
--- a/solid_state_physics/tutorial/2_02s.tex
+++ b/solid_state_physics/tutorial/2_02s.tex
@@ -55,7 +55,7 @@ and $\lambda$: London penetration depth.
         \Rightarrow dr=\lambda dx$, $r=\lambda x$
        $\Rightarrow
         I_c=j_c(R)2\pi \lambda^2 \exp(-R/\lambda) \int_0^R d(\frac{r}{\lambda})
-	    \, \frac{r}{\lambda} \exp(\frac{r}{\lambda})$
+	    \, \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