Benutzer:Alturand/Wirbelstrom

Es gilt:

${\displaystyle }$

${\displaystyle \nabla \times {\vec {E}}=-{\tfrac {d}{dt}}{\vec {B}}}$

für

${\displaystyle B(x,y,z)={\begin{pmatrix}0\\0\\B_{z}\end{pmatrix}}=B{\begin{pmatrix}0\\0\\1\end{pmatrix}}\delta (x-x_{0},y-y_{0},0)\Theta (t-t_{0})}$

ist eine Lösung:

${\displaystyle E(x,y,z)={\begin{pmatrix}E_{x}\\E_{y}\\0\end{pmatrix}}={\tfrac {E}{\sqrt {(x-x_{0})^{2}+(y-y_{0})^{2}}}}{\begin{pmatrix}-{\tfrac {(y-y_{0})}{\sqrt {(x-x_{0})^{2}+(y-y_{0})^{2}}}}\\{\tfrac {(x-x_{0})}{\sqrt {(x-x_{0})^{2}+(y-y_{0})^{2}}}}\\0\end{pmatrix}}\delta (t-t_{0})}$

denn

${\displaystyle {\begin{alignedat}{1}\nabla \times {\vec {E}}&=&{\begin{pmatrix}{\tfrac {d}{dx}}\\{\tfrac {d}{dy}}\\{\tfrac {d}{dz}}\\\end{pmatrix}}\times {\tfrac {E}{2\pi {\sqrt {(x-x_{0})^{2}+(y-y_{0})^{2}}}}}{\begin{pmatrix}-{\tfrac {(y-y_{0})}{\sqrt {(x-x_{0})^{2}+(y-y_{0})^{2}}}}\\{\tfrac {(x-x_{0})}{\sqrt {(x-x_{0})^{2}+(y-y_{0})^{2}}}}\\0\end{pmatrix}}\delta (t-t_{0})\\&=&E{\begin{pmatrix}0\\0\\{\tfrac {d}{dx}}E_{y}-{\tfrac {d}{dy}}E_{x}\end{pmatrix}}\delta (t-t_{0})\\{\dot {B_{z}}}&=&{\tfrac {d}{dx}}E_{y}-{\tfrac {d}{dy}}E_{x}&=&0\\\end{alignedat}}}$

Mit der hilfsweisen Annäherung der Deltafunktion über die Dreiecksfunktion

${\displaystyle \delta _{\epsilon }(x)=\lim _{\epsilon \rightarrow 0}{\begin{cases}{\frac {\epsilon +x}{\epsilon ^{2}}}&\ -\epsilon \leq x\leq 0\\{\frac {\epsilon -x}{\epsilon ^{2}}}&\ 0<x\leq \epsilon \\0&\ {\text{sonst}}\end{cases}}}$

erhält man die Lösung zu: