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The Circular Motion

A particle of...

Question

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

A

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

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B

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

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C

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

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D

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

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Solution

The correct option is C $\frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} α} + \frac{v_{0}}{2} (t - \frac{π}{3 B_{0} α})$
$r = \frac{m v_{0}}{B_{0} q} = \frac{v_{0}}{B_{0} α}, \frac{x}{r} = \frac{\sqrt{3}}{2} = s i n θ$
$\Rightarrow θ = 60^{o}$
$t_{O A} = \frac{T}{6} = \frac{π}{3 B_{0} α}$
Therefore, x-coordinate of particle at any time $t > \frac{π}{3 B_{0} α}$ will be
$x = \frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} α} + v_{0} (t - \frac{π}{3 B_{0} π}) c o s 60^{o}$
$= \frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} α} + \frac{v_{0}}{2} (t - \frac{π}{3 B_{0} α})$

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