Question

A particle of charge $q$ and mass $m$ is projected with a velocity $v_0$ towards a circular region having a uniform magnetic field $B$ perpendicular and into the plane of paper from point $P$ as shown in the figure. $R$ is the radius and $O$ is the center 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. $\frac{qBR}{m\sin \theta }$
• B. $\frac{qBR}{2m\sin \theta }$
• C. $\frac{2qBR}{m\sin \theta }$
• D. $\frac{3qBR}{2m\sin \theta }$

Answer 1

Answer: (B)

Let $r$ be the radius of the circular path.

$\sin \theta =\frac{\left(\frac{R}{2}\right)}{r}=\frac{R}{2r}$

$r=\frac{mv}{qB}$

Substituting $r,$

$\frac{R}{2\sin \theta }=\frac{mv}{qB}$