Question

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

Answer 1

Answer: 2

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