small fixes

diff --git a/solid_state_physics/tutorial/2_01s.tex b/solid_state_physics/tutorial/2_01s.tex
index 22abb78af12bee66fb6baec09b2bbaf333f17f3a..033c8ae2702dfe6cab0e232e27b839759de98d33 100644 (file)
--- 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 ecafd2a925bb40885c7b046efb386fa4aee3e9f3..b9b5c5a93af222bb4722c4d5382faceb861b7866 100644 (file)
--- 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 8c121964d7286c2995d13b8143baeb717a7c366e..8f3eda0a6f9168c4148591dc175493d7b5df38cb 100644 (file)
--- 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