Question

A positive charged particle of charge $q$ and mass $m$ is projected with speed $v$ perpendicular to a uniform magnetic field $B$ as shown in the figure. Neglecting gravity, calculate the $x-$coordinate of the point on the screen at which the charged particle will hit.

Take, $d=\frac{R\sqrt{3}}{2};$ where $R=\frac{mv}{qB}$

• A. $\sqrt{3}R$
• B. $2R$
• C. $2.5R$
• D. $0.5R$

Answer 1

The correct option is B $2R$



From the figure,

$\tan \theta =\frac{PS}{PB}=\frac{PS}{d}$

$\Rightarrow PS=d\tan \theta$

Also,

$\sin \theta =\frac{d}{R}$

Given, $d=\frac{R\sqrt{3}}{2}$ and $R=\frac{mv}{Bq}$

$\sin \theta =\frac{\frac{R\sqrt{3}}{2}}{R}$

$\Rightarrow \theta ={60}^{\circ }$

Now, $AB=R-R\cos {60}^{\circ }=\frac{R}{2}$

Further, $x-$coordinate where it strikes is

$x=AB+PS$

$\Rightarrow x=\frac{R}{2}+d\tan {60}^{\circ }$

$\Rightarrow x=\frac{R}{2}+\frac{R\sqrt{3}}{2}\times \sqrt{3}=2R$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option $\left(B\right)$ is the correct answer.