Question

A particle moves in a circle of diameter $1.0cm$ under the action of a magnetic field of $0.40tesla$. An electric field of $200V{m}^{-1}$ makes the path straight. Find the charge/mass ratio of the particle.

Answer 1

As applying electric field makes path straight, it means electric force is being balanced by magnetic force.

$F_E=qE$

$F_B=q\vartheta B$

$F_E=F_B$

$\therefore qE=q\vartheta B$

$200=\vartheta \ast 0.4$

$\therefore \vartheta =\frac{200}{0.4}=500m/sec$

now

particle (without electric field) moves in acircle of diameter 1 cm

$\therefore \frac{m{v}^2}{r}=q\vartheta B$

$\frac{m\vartheta }{r}=qB$

$\frac{m}{q}=\frac{rB}{\vartheta }$

as $r=\frac{1}{2}\ast {10}^{-2}=0.5\ast {10}^{-2}metre$

$\therefore \frac{m}{q}=\frac{0.5\ast {10}^{-2}\ast 0.4}{500}$

$\therefore \frac{q}{m}=2500\ast {10}^2$

$=25\ast {10}^4$