Question

A non-relativistic positive charge particle of charge $q$ and mass $m$ is projected perpendicular to uniform magnetic field $B$ as shown. Neglecting gravity, calculate $x-$coordinate of point on screen at which the charge particle will hit : $d=\frac{R\sqrt{3}}{2}$, where $R=\frac{mV}{qB}$

• A. $\sqrt{3}R$
• B. $2R$
• C. The particle describes a helical path of radius $R=\frac{mv\sin \alpha }{qB}$
• D. The particle describes a circular path of radius $R=\frac{mv}{qB}$

Answer 1

Answer: (B), (D)

The particle describes a circular path of radius $R=\frac{mv}{qB}$

Since it is given that motion of the charged particle is perpendicular to the magnetic field, the particle describes a circular path of radius, $R=\frac{mV}{qB}$
From figure,
$\sin \theta =\frac{d}{R}=\frac{\sqrt{3}}{2}$
$\Rightarrow \theta ={60}^{\circ }$
$\therefore$ The required $x-$ coordinate where the charge will hit is
$x=(R-R\cos \theta )+d\cot {30}^{\circ }$
$\Rightarrow x=\frac{R}{2}+\frac{R\sqrt{3}}{2}\times \sqrt{3}=2R$