Question

A particle of charge per unit mass $\alpha$ is released from the origin with velocity $\overrightarrow{ v }= v _{0} \hat{ i }$ in the magnetic field
$\vec{ B }=- B _{0} \hat{ k }$ for $x \leq \frac{\sqrt{3}}{2} \frac{ v _{0}}{ B _{0} \alpha}$
and $\vec{ B }=0$ for $x >\frac{\sqrt{3}}{2} \frac{ v _{0}}{ B _{0} \alpha}$
The $x$-coordinate of the particle at time $t \left(>\frac{\pi}{3( B )_{0} \alpha}\right)$ would be

• A. $\frac{\sqrt{3}}{2} \frac{ v _{0}}{ B _{0} \alpha}+\frac{\sqrt{3}}{2} v _{0}\left( t -\frac{\pi}{( B )_{0} \alpha}\right)$
• B. $\frac{\sqrt{3}}{2} \frac{ v _{0}}{ B _{0} \alpha}+ v _{0}\left( t -\frac{\pi}{3( B )_{0} \alpha}\right)$
• C. $\frac{\sqrt{3}}{2} \frac{ v _{0}}{ B _{0} \alpha}+\frac{ v _{0}}{2}\left( t -\frac{\pi}{3( B )_{0} \alpha}\right)$
• D. $\frac{\sqrt{3}}{2} \frac{ v _{0}}{ B _{0} \alpha}+\frac{ v _{0} t }{2}$

Answer 1

Answer: (C)

$r =\frac{ mv _{0}}{ B _{0} q }=\frac{ v _{0}}{ B _{0} \alpha}, \frac{ x }{ r }=\frac{\sqrt{3}}{2}=\sin \theta$
$ \Rightarrow \theta=60^{\circ}$
$t _{ OA }=\frac{ T }{6}=\frac{\pi}{3 B _{0} \alpha}$

Therefore $x$-coordinate of particle at any time $t>\frac{\pi}{3 B _{0} \alpha}$ will be
$x =\frac{\sqrt{3}}{2} \frac{ v _{0}}{ B _{0} \alpha}+ v _{0}\left( t -\frac{\pi}{3( B )_{0} \alpha}\right) \cos 60^{\circ} $
$=\frac{\sqrt{3}}{2} \frac{ v _{0}}{ B _{0} \alpha}+\frac{ v _{0}}{2}\left( t -\frac{\pi}{3( B )_{0} \alpha}\right)$