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final ice changes
[lectures/latex.git]
/
solid_state_physics
/
tutorial
/
2_02.tex
diff --git
a/solid_state_physics/tutorial/2_02.tex
b/solid_state_physics/tutorial/2_02.tex
index
8787db5
..
b9b5c5a
100644
(file)
--- 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:}
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.
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}
The critical current and penetration depth at $T=0K$ are
$I_c=75\, A$ and $\lambda =300\cdot 10^{-10}\, m$.
\end{enumerate}