A particle of charge per unit mass $α$ is released from origin with velocity $\vec{V} = v_{0} \hat{i}$ in a magnetic field $\vec{B} = - B_{0} \hat{k} for x \leq \frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} α} and \vec{B} = 0 for x > \frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} α}$ . Then x–coordinate of the particle at time $t > (\frac{π}{3 B_{0} α})$ would be

Moderate

A

$\frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} α} + \frac{\sqrt{3}}{2} v_{0} (t - \frac{π}{3 B_{0} α})$
B

$\frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} α} + v_{0} (t - \frac{π}{3 B_{0} α})$
C

$\frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} α} + \frac{v_{0}}{2} (t - \frac{π}{3 B_{0} α})$
D

$\frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} α} + \frac{v_{0} t}{2}$

Solution

[Step 1]:    $r = \frac{m V_{0}}{B_{0} q} = \frac{V_{0}}{B_{0} α}$
[Step 2]:    $\frac{x_{}}{r} = \frac{\sqrt{3}}{2} = \sin θ$
$∴ θ = 60^{0} - - - - (1)$
[Step 3]:    $t_{0 A} = \frac{T}{6} = \frac{\frac{2 π m}{q B_{0}}}{6} = \frac{π}{3 B_{0} α} (∵ \frac{q}{m} = α)$
[Concept]: Therefore x – coordinate of particle at any time
$\begin{array}{l} \begin{array}{l} t > \frac{π}{3 B_{0} α} w i l l b e \\ [s t e p 4] : x' = x_{0} + v_{0} \cos θ (t - t_{0 A}) \end{array} \\ \Rightarrow x' = \frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} α} + v_{0} (t - \frac{π}{3 B_{0} α}) \cos 60^{0} \\ \Rightarrow x' = \frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} α} + \frac{v_{0}}{2} (t - \frac{π}{3 B_{0} α}) \end{array}$

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