Question

A charged particle (charge $q$, mass $m$) has velocity $v_0$ at the origin in +x direction. In space, there is a uniform magnetic field B in –z direction. Find the y coordinate of the particle when it crosses y axis.

• A. $\frac{2m{v}_0}{qB}$
• B. $\frac{m{v}_0}{qB}$
• C. $\frac{4m{v}_0}{qB}$
• D. None of these

Answer 1

The correct option is A $\frac{2m{v}_0}{qB}$

Force on a charged particle in the magnetic field
$F=q(\overrightarrow{v}\times \overrightarrow{B})$
$F=qvB\sin \theta$
$\because$ here $v$ and $B$ are perpendicular, therefore
$\theta ={90}^{\circ }$
$F=qvB\left(1\right)$
It is equal to centripetal force $\frac{m{v}^2}{R}...\left(2\right)$
$\therefore$ from equation (1) and (2)
$qvB=\frac{m{v}^2}{R}$ (here $v=v_0$)
$R=\frac{m{v}_0}{qB}$
Now y coordiante will be $(0,2R)$
i.e., $\frac{2m{v}_0}{qB}$