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 $x\le \frac{\sqrt{3}}{2}\frac{v_0}{B_0\alpha }$
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)

$r=\frac{m{v}_0}{B_{0}q}=\frac{v_0}{B_0\alpha },\frac{x}{r}=\frac{\sqrt{3}}{2}=\sin \theta$
$\Rightarrow \theta =60^{\circ }$
$t_{OA}=\frac{T}{6}=\frac{\pi }{3B_0\alpha }$

Therefore $x$-coordinate of particle at any time $t>\frac{\pi }{3B_0\alpha }$ will be
$x=\frac{\sqrt{3}}{2}\frac{v_0}{B_0\alpha }+v_0\left(t-\frac{\pi }{3(B{)}_0\alpha }\right)\cos 60^{\circ }$
$=\frac{\sqrt{3}}{2}\frac{v_0}{B_0\alpha }+\frac{v_0}{2}\left(t-\frac{\pi }{3(B{)}_0\alpha }\right)$