A charged particle is projected with velocity v0 along positive x-axis. The magnetic field B is directed along negative z-axis between x = 0 and x = L. The particle emerges out (at x = L) at an angle of 60º with the direction of projection. Find the velocity with which the same particle were projected (at x = 0) along positive x-axis so that when it emerges out (at x = L), the angle made by it is 30º with the direction of projection

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A charged particle is projected with velocity v₀ along positive x-axis. The magnetic field B is directed along negative z-axis between x = 0 and x = L. The particle emerges out (at x = L) at an angle of 60º with the direction of projection. Find the velocity with which the same particle were projected (at x = 0) along positive x-axis so that when it emerges out (at x = L), the angle made by it is 30º with the direction of projection

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$\frac{v_{0}}{\sqrt{3}}$

$\frac{v_{0}}{2}$

$V_{0} \sqrt{3}$

2v₀

answer is D.

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Detailed Solution

In $∆$ OAB
AB = L = constant
$\begin{array}{l} OB = r = \frac{mv}{qB} \\ \sin θ = \frac{L}{r} \Rightarrow L = rsin θ = constant \\ or vsin θ = constant \\ v_{0} \sin 60^{\circ} = v^{'} \sin 30^{\circ} \\ v^{'} = \sqrt{3} v_{0} \end{array}$