Question

A uniform magnetic field $\vec{B}=B_{0} \hat{j}$ exists in space. A particle of mass $m$ and charge $q$ is projected towards negative $x$-axis with speed $v$ from the a point $(d, 0,0)$. If the maximum value of $v$ for which the particle does not hit $y-z$ plane is $\frac{4 B_{0} q d}{\alpha m}$, then calculate $\alpha$.

Answer 1

$\vec{B}=B_{0} \hat{j}$
The particle will not hit $y-z$ plane if radius of curvature
$ r \leq d $
$\therefore \frac{m v^{2}}{d}=q v B_{0} $
$\Rightarrow v=\frac{B_{0} q d}{m}$