A particle of charge per unit mass α is released from origin with velocity V→=v0i^ in a magnetic field B→=−B0k^ for x≤32v0B0α and B→=0 for x>32v0B0α. Then x–coordinate of the particle at time t>π3B0α would be

[Step 1]: r=mV0B0q=V0B0α[Step 2]: x r=32=sinθ∴θ=600----(1)[Step 3]: t0A=T6=2πmqB06=π3B0α∵qm=α[Concept]: Therefore x – coordinate of particle at any time t>π3B0αwill bestep 4:x'=x0+v0 cosθt−t0A⇒x'=32v0B0α+v0t−π3B0αcos600 ⇒x'=32v0B0α+v02t−π3B0α

A particle of charge per unit mass α is released from origin with velocity V→=v0i^ in a magnetic field B→=−B0k^ for x≤32v0B0α and B→=0 for x>32v0B0α. Then x–coordinate of the particle at time t>π3B0α would be

[Step 1]: r=mV0B0q=V0B0α[Step 2]: x r=32=sinθ∴θ=600----(1)[Step 3]: t0A=T6=2πmqB06=π3B0α∵qm=α[Concept]: Therefore x – coordinate of particle at any time t>π3B0αwill bestep 4:x'=x0+v0 cosθt−t0A⇒x'=32v0B0α+v0t−π3B0αcos600 ⇒x'=32v0B0α+v02t−π3B0α

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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

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

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

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

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

answer is C.

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Detailed 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
$\begin{array}{l} \begin{array}{l} t > \frac{π}{3 B_{0} α} w i l l b e \\ [s t e p 4] : x' = x_{0} + v_{0} \cos θ (t - t_{0 A}) \end{array} \\ \Rightarrow x' = \frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} α} + v_{0} (t - \frac{π}{3 B_{0} α}) \cos 60^{0} \\ \Rightarrow x' = \frac{\sqrt{3}}{2} \frac{v_{0}}{B_{0} α} + \frac{v_{0}}{2} (t - \frac{π}{3 B_{0} α}) \end{array}$