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 \begin{document}
 
 \section{\ce{Nb3Sn} cylinder}
+Consider a cylinder of \ce{Nb3Sb}.
+From lecture 4, we have the following properties for \ce{Nb3Sn}:
+$T_c = \SI{18.2}{\kelvin}$,
+$\xi = \SI{3.6}{\nano\meter}$,
+$\lambda = \SI{124}{\nano\meter}$,
+$\kappa = \frac{\lambda}{\xi} = 34 > \frac{1}{\sqrt{2}}$,
+which means we are indeed dealing with a type-II superconductor.
+As $B_{c1} < B_E < B_{c2}$, the cylinder is in the vortex state.
+From the previous set of assignments, we know what the currents in the cylinder look like.
+
+The average field inside the cylinder is gives as
+\[
+	\langle \vec{B} \rangle = \frac{1}{V_{\text{cylinder}}} \int_{\text{cylinder}} \vec{B}(\vec{r}) d\vec{r} .
+\]
+
+To determine this $\vec{B}$ inside the material, we first need to know how many vortices there are.
+We assume that every vortex lets through only one flux quantum $\Phi_0$,
+and that the vortices will arange themselves as far as possible from eachother.
+If their distance then is large enough to assume there is no overlap between regions of finite $\vec{B}$ around them,
+we can calculate the average field by just summing over the quanta and lastly over the field that penetrates the material in the outside of the cylinder.
+For this latter calculation, we can use the field for a type-I superconductor.
 
 \section{Superconducting wire}