Question

A particle of charge $q$ and mass $m$ is projected with a velocity $v_0$ towards a circular region having uniform Magnetic field $B$ perpendicular and into the plane of paper, from point $P$ as shown in figure. $R$ is the radius and $O$ is the centre of the circular region. If the line $OP$ makes an angle $\theta$ with the direction of $v_0$ then the value of $v_0$ so that particle passes through $O$ is

• A. $\frac{qBR}{msin\theta }$
• B. $\frac{qBR}{2msin\theta }$
• C. $\frac{2qBR}{msin\theta }$
• D. $\frac{3qBR}{2msin\theta }$

Answer 1

Answer: (B)

Let $r$ be the radius of circular path. If the particle passes through $O$, then from the given figure,

$sin\theta =\frac{R/2}{r}=\frac{R}{2r}\dots (i)$
As $r=\frac{m{v}_0}{qB}$
$\therefore v_0=\frac{qBr}{m}$
$=\frac{qBR}{2msin\theta }$ (Using (i))