ass1: Exam question 2 is done!

superconductivity_assignment1_kvkempen.tex

@@ -116,16 +116,59 @@ The first and second London equations are
 \[
 	\frac{\partial}{\partial t}(\Lambda \vec{J_s}) = \vec{E}
 	\qquad
-	\nabla \times (\Lambda \vec{J_s}) = \vec{B},
+	- \nabla \times (\Lambda \vec{J_s}) = \vec{B},
 \]
 with $\Lambda = \frac{m_e}{n_se^2}$, $m_e$ the normal electron mass, $n_s$ the superfluid electron density.
 
 \begin{enumerate}
-	\item \textbf{Describe the idea behind the London model for the Meissner-Ochsenfeld effect.}
+\item Describe the idea behind the London model for the Meissner-Ochsenfeld effect.
 
-	\emph{The London model is based on the two fluid model that was proposed by Lev Landau a few years prior to the London model. The idea of the London model is the coexistence of two types of electrons in a superconductor, namely normal electrons and superelectrons. Both types can transfer electrical current, the normal electrons with resistance, the superelectrons without, and they do so in parallel. Heat can only be transferred by the normal electrons.\\\\ Electrons can change type. At high temperature, all the electrons are of the normal type. Below a critical temperature $T_c$, however, the density of superelectrons quickly emerges and a supercurrent can flow.}
+\begin{em}
+The London model is based on the two fluid model that was proposed by Lev Landau a few years prior to the London model.
+The idea of the London model is the coexistence of two types of electrons in a superconductor, namely normal electrons and superelectrons.
+Both types can transfer electrical current, the normal electrons with resistance, the superelectrons without, and they do so in parallel.
+Heat can only be transferred by the normal electrons.
+
+Electrons can change type.
+At high temperature, all the electrons are of the normal type.
+Below a critical temperature $T_c$, however, the density of superelectrons quickly emerges and a supercurrent can flow.
+\end{em}
+
+\item Using the second London equation, derive the screening equation
+	\[
+		\nabla^2\vec{B} = \frac{\vec{B}}{\lambda_L^2}
+	\]
+	with $\lambda_L^2 = \frac{\Lambda}{\mu_0}$.
+
+	Use the Maxwell-Amp\`ere law
+	\[
+		\nabla \times \vec{B} = \mu_0(\vec{J} + \frac{\partial \vec{E}}{\partial t})
+	\]
+
+\begin{em}
+We assume a steady-state, so starting from the Maxwell-Amp\`ere law, we can immediately cross off the derivative $\frac{\partial \vec{E}}{\partial t} = 0$.
+As we are in the superconducting state, the total current is mostly due to current of superelectrons (supercurrent), as these flow without resistance and thus flow way easier.
+So we set $\vec{J} = \vec{J_s}$.
+We now have
+\[
+	\nabla \times \vec{B} = \mu_0(\vec{J_s}).
+\]
+Taking the curl of both sides, we end up with
+\[
+	\nabla \times (\nabla \times \vec{B}) = \nabla^2\vec{B} = \mu_0\nabla \times \vec{J_s}.
+\]
+Assuming that $\Lambda$ is constant over the material, we have from the second London equation that
+\[
+	\nabla \times (\Lambda\vec{J_s}) = - \Lambda \nabla \times \vec{J_s} = \vec{B},
+\]
+which we can easily substitute into our early result.
+This way we find our result:
+\[
+	\nabla^2\vec{B} = \frac{\mu_0\vec{B}}{\Lambda} = \frac{\vec{B}}{\lambda_L^2}
+\]
+%If we now substitute in the second London equation, assuming that $\Lambda$ is constant over the material,
+\end{em}
 
-	\item ..
 \end{enumerate}
 
 \section{Difference between type-I and type-II superconductors}