Question

An electron of mass $0.90\times {10}^{-30}kg$ under the action of a magnetic field moves in a circle of 2.0 $cm$ radius at a speed of $3.0\times {10}^{6}m{s}^{-1}$. If a proton of mass $1.8\times {10}^{-27}kg$ has to move in a circle of the same radius in the same magnetic field, then its speed will be $m{s}^{-1}$.

Answer 1

$Bqv=\frac{m{v}^2}{r}$ or $Bqr=mv$
For electron as well as proton $B$ is the same, $r$ is same and charge $q$ is same. Therfore $mv$ is constant
$m_e{v}_e$ or $m_p{v}_p$ or $v_p=\left(\frac{m_e}{m_p}\right)v_e$
or $v_p=\left(\frac{0.90\times {10}^{-30}}{1.8\times {10}^{-27}}\right)(3.0\times {10}^6)$
$=1.5\times {10}^{3}m{s}^{-1}$