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  4. An electron accelerated by a potential difference V=1.0 kV An electron accelerated by a potential difference α=30° to the vector vecB whose modulus is B = 29 mT. Find the pitch (in cm) of the helical trajectory of the electron

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Q. An electron accelerated by a potential difference $V=1.0\,kV$ An electron accelerated by a potential difference $\alpha=30^{\circ}$ to the vector $\vec{B}$ whose modulus is $B = 29 \,mT$. Find the pitch (in cm) of the helical trajectory of the electron

Moving Charges and Magnetism

Solution:

If v is the velocity of the electron, then
$\frac{1}{2} mv^{2} =eV$
$\therefore v=\sqrt{\frac{2eV}{m}}$
Pitch $=v\,cos\,\alpha \times T =v\,cos\,\alpha \times\frac{2\pi m}{eB}$
$= \begin{matrix}2\pi m\\ eB\end{matrix} \sqrt{\begin{matrix}2e V\\ m\end{matrix}}cos\,\alpha $
$=2\pi \sqrt{\frac{2mV}{eB^{2}}}cos\,\alpha$