X-Git-Url: https://hackdaworld.org/gitweb/?a=blobdiff_plain;f=solid_state_physics%2Ftutorial%2F2_02.tex;h=b9b5c5a93af222bb4722c4d5382faceb861b7866;hb=fec7e52be07515b5dac392facfa0b2022d002c21;hp=8787db56333869653cc81eae8062045268410611;hpb=82e203acc6d8f125b12df53e5697d5df0875f0e3;p=lectures%2Flatex.git

diff --git a/solid_state_physics/tutorial/2_02.tex b/solid_state_physics/tutorial/2_02.tex
index 8787db5..b9b5c5a 100644
--- a/solid_state_physics/tutorial/2_02.tex
+++ b/solid_state_physics/tutorial/2_02.tex
@@ -56,9 +56,10 @@ 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 $T=0K$.
+ \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}