{\large \displaystyle \,\,-\overrightarrow{{{F}_{L}}}\,\,\,\,\,=\,\overrightarrow{{{F}_{el}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{F}_{L}}\,=\,e\cdot v\cdot B\,\,\,\,\,\,(\overrightarrow{v}\,\bot \,\overrightarrow{B})\,\,\,;\,\,\,{{F}_{el}}\,=e\cdot U }
{\large \displaystyle \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{F}_{L}}\,=\,e\cdot v\cdot B\,\cdot \,\sin \measuredangle \left( \overrightarrow{v},\,\overrightarrow{B} \right)}
Wir gehen zunächst davon aus, dass die Bewegung senkrecht zu den magnetischen Feldlinien erfolgt.
{\large \begin{array}{l}e\cdot v\cdot B\,=\,e\cdot E\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,E\,=\frac{{{U}_{ind}}}{d}\\\,\,\,\,\,v\,\cdot \,\,B\,=\,\frac{{{U}_{ind}}}{d}\\{{U}_{ind}}\,=\,B\,\cdot \,d\,\cdot \,v\end{array}}
Für die Induktionsspannung Uind in einer Leiterschleife gilt dann:
{\large \begin{array}{l}{{U}_{ind}}\,=\,B\,\cdot \,d\,\cdot \,v\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,v=\,\frac{\Delta s}{\Delta t}\\\\{{U}_{ind}}\,=\,B\,\cdot \,d\,\cdot \,\frac{\Delta s}{\Delta t}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta s\cdot d=\Delta A\\\\{{U}_{ind}}\,=\,B\,\cdot \,\frac{\Delta A}{\Delta t}\end{array} }