From bb20e9a1cf8917e5353683c00dd733016cfbbd3a Mon Sep 17 00:00:00 2001
From: hackbard <hackbard@sage.physik.uni-augsburg.de>
Date: Tue, 20 May 2008 21:00:12 +0200
Subject: [PATCH] oups ...
---
solid_state_physics/tutorial/2_02.tex | 2 +-
1 file changed, 1 insertion(+), 1 deletion(-)
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
--
2.39.5