From: hackbard <hackbard@sage.physik.uni-augsburg.de>
Date: Tue, 20 May 2008 19:00:12 +0000 (+0200)
Subject: oups ...
X-Git-Url: https://hackdaworld.org/cgi-bin/gitweb.cgi?a=commitdiff_plain;h=bb20e9a1cf8917e5353683c00dd733016cfbbd3a;p=lectures%2Flatex.git

oups ...
---

diff --git a/solid_state_physics/tutorial/2_02.tex b/solid_state_physics/tutorial/2_02.tex
index 8787db5..ecafd2a 100644
--- a/solid_state_physics/tutorial/2_02.tex
+++ b/solid_state_physics/tutorial/2_02.tex
@@ -56,7 +56,7 @@ and $\lambda$ is the London penetration depth.
        of the wire. Assume, that the penetration depth $\lambda$ is much
        smaller than the radius $R$ of the cylinder.
        {\bf Hint:}
-       Use the relation $I_c=\int_0^R dr \int_0^{2\pi} d\phi \, j_c(r)$
+       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 $T=0K$.
        The critical current and penetration depth at $T=0K$  are