Question

A charged particle moving along $+ve,x-$direction with a velocity $v$ enters in a region where there is a uniform magnetic field $B_0(-\hat{k})$, from $x=0$ to $x=d$. The particle gets deflected at an angle $\theta$ from its initial path. The specific charge of the particle is

• A. $\frac{v\cos \theta }{Bd}$
• B. $\frac{v\tan \theta }{Bd}$
• C. $\frac{v}{Bd}$
• D. $\frac{v\sin \theta }{Bd}$

Answer 1

The correct option is D $\frac{v\sin \theta }{Bd}$

As can be seen from the diagram, $\sin \theta$ is the angle of deviation for the charged particle.

$\sin \theta =\frac{CD}{OA}=\frac{d}{R}$ where $O$ is the center of the curved portion of the circular track of radius $R$.

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

$\therefore \frac{d}{\sin \theta }=\frac{mv}{qB}$

$\therefore \frac{q}{m}=\frac{v\sin \theta }{Bd}$