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ass3: Solve 9b, add sub letters, add todo

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Kees van Kempen
2022-03-17 15:09:18 +01:00
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superconductivity_assignment3_kvkempen.tex
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@ -102,6 +102,7 @@ For the gradient we thus find
The size of the supercurrent density has the same relation, $J_S \propto 1/r$. The size of the supercurrent density has the same relation, $J_S \propto 1/r$.
\section{Superconducting wire} \section{Superconducting wire}
\textbf{(a)}
The voltage $U = \SI{1.5e-5}{\volt}$ across the wire of length $\ell = \SI{.08}{\meter}$ induces a current $J_t$. % through the resistive wire with unknown resistivity $\rho$ according to Ohm's law. The voltage $U = \SI{1.5e-5}{\volt}$ across the wire of length $\ell = \SI{.08}{\meter}$ induces a current $J_t$. % through the resistive wire with unknown resistivity $\rho$ according to Ohm's law.
Due to the presence of the magnetic field $B = \SI{5}{\tesla}$, if the vortices move with velocity $v_L$, a Lorentz force $f_L$ per vortex acts on the vortices. Due to the presence of the magnetic field $B = \SI{5}{\tesla}$, if the vortices move with velocity $v_L$, a Lorentz force $f_L$ per vortex acts on the vortices.
This results in a power input $P_L = f_Lv_L = J_tBv_L$ per vortex. This results in a power input $P_L = f_Lv_L = J_tBv_L$ per vortex.
@ -112,6 +113,18 @@ Equating these expressions and rewriting yields
v_L = \frac{U}{B\ell} = \SI{3.75e5}{\meter\per\second}. v_L = \frac{U}{B\ell} = \SI{3.75e5}{\meter\per\second}.
% https://www.wolframalpha.com/input?i=1.5*10%5E-5+%2F+%285*+.08%29 % https://www.wolframalpha.com/input?i=1.5*10%5E-5+%2F+%285*+.08%29
\] \]
\todo{Direction?}
\textbf{(b)}
The vortices are aranged in a lattice with separation $r_{sep} = \sqrt{\frac{\Phi_0}{B}}$.
They move along the wire with velocity $v_L$ as determined above.
The expected frequency is then given by their velocity over the separation, as that is the period of the changing fields due to the vortices:
\[
f = \frac{v_L}{r_{sep}} = \frac{U}{B\ell}\sqrt{\frac{B}{\Phi_0}} = \frac{U}{\ell\sqrt{B\Phi_0}} = \SI{1.84}{\kilo\hertz},
% https://www.wolframalpha.com/input?i=1.5*10%5E-5+%2F+%28.08%29+%2Fsqrt%285+*+2.067*10%5E%28-15%29%29
\]
where we used that $\Phi_0 = \SI{2.067e-15}{\volt\second}$.
This is very close to what is written in the assignment, but not precisely the same, so maybe I used a different value for $\Phi_0$.
\section{Fine type-II superconducting wire} \section{Fine type-II superconducting wire}