Question

An electron experiences a magnetic force of magnitude $4.60\times {10}^{-15}N$, when moving at an angle of ${60.0}^o$ with respect to a magnetic field of magnitude $3.50\times {10}^{-3}T$.If $v$ is the speed of the electron, find $x$, such that $x=v/{10}^4$.

Answer 1

$F=Bqvsin\theta$
$v=\frac{F}{Bqsin\theta }$
$=\frac{4.6\times {10}^{-15}}{3.5\times {10}^{-3}\times 1.6\times {10}^{-19}\times sin{60}^o}$

$=9.47\times {10}^{6}m/s$