Question

A charged particle with a charge-to-mass ratio of |q|/m = 5.7x 10⁸ C/kg travels on a circular path that isperpendicular to a magnetic field whose magnitude is 0.76 T. How much time does it take for the particle to complete onerevolution? t = ?

Answer 1

The \(\mathrm{q} / \mathrm{m}\) ratio \(=5.7 \times 10^{8} \mathrm{C} / \mathrm{kg}\)

the magnitude of magnetic field \(\mathrm{B}=0.76 \mathrm{~T}\)

we know that in circular path

$$ \begin{aligned} \text { speed } v &=\frac{2 \pi r}{t} \\ \text { then } t=\frac{2 \pi r}{v} &=\frac{2 \pi r}{B q r / m}=\frac{2 \pi(m)}{B q}=\frac{2 \pi}{0.76}\left(\frac{1}{5.7 \times 10^{8}}\right)=1.5 \times 10^{-8} \mathrm{sec} \end{aligned} $$