Question

A particle of charge per unit mass $\alpha$ is released from origin with velocity $\overrightarrow{v}=v_0\hat{i}$ in a magnetic field $\overrightarrow{B}=-B_0\hat{k}$ for $x\le \frac{\sqrt{3}}{2}\frac{v_0}{B_0\alpha }$ and $\overrightarrow{B}=0$ for $x>\frac{\sqrt{3}}{2}\frac{v_0}{B_0\alpha }$. The x-coordinate of the particle at time $t(>\frac{\pi }{3B_0\alpha })$ would be

• A. $\frac{\sqrt{3}}{2}\frac{v_0}{B_0\alpha }+\frac{\sqrt{3}}{2}-v_0(t-\frac{\pi }{B_0\alpha })$
• B. $\frac{\sqrt{3}}{2}\frac{v_0}{B_0\alpha }+v_0(t-\frac{\pi }{3B_0\alpha })$
• C. $\frac{\sqrt{3}}{2}\frac{v_0}{B_0\alpha }+\frac{v_0}{2}(t-\frac{\pi }{3B_0\alpha })$
• D. $\frac{\sqrt{3}}{2}\frac{v_0}{B_0\alpha }+\frac{v_{0}t}{2}$

Answer 1

The correct option is C $\frac{\sqrt{3}}{2}\frac{v_0}{B_0\alpha }+\frac{v_0}{2}(t-\frac{\pi }{3B_0\alpha })$

$r=\frac{m{v}_0}{B_{0}q}=\frac{v_0}{B_0\alpha },\frac{x}{r}=\frac{\sqrt{3}}{2}=sin\theta$

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

$t_{OA}=\frac{T}{6}=\frac{\pi }{3B_0\alpha }$

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

$=\frac{\sqrt{3}}{2}\frac{v_0}{B_0\alpha }+\frac{v_0}{2}(t-\frac{\pi }{3B_0\alpha })$