From bb20e9a1cf8917e5353683c00dd733016cfbbd3a Mon Sep 17 00:00:00 2001 From: hackbard 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 -- 2.20.1