A charged particle is projected with velocity υ₀ along positive $x -$ axis. The magnetic field $B$ is directed along negative $z -$ axis between $x = 0$ and $x = L$ . The particle direction of projection. Find the velocity with which the same particle is projected (at $x = 0$ ) along positive $x -$ axis so that when it emerges out (at $x = L$ ), the angle made by it is $30^{\circ}$ with the direction of projection

$2 v_{0}$

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

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

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$\sqrt{3} v_{0}$

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Solution

The correct option is D
$\sqrt{3} v_{0}$

$sin θ = \frac{L}{r} = \frac{L}{m v / B q}$
$∴ v \propto \frac{1}{sin θ}$
or $\frac{v}{v_{0}} = \frac{sin 60 °}{sin 30 °}$
$∴ v = \sqrt{3} v_{0}$

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The correct option is D √3v0 sinθ=Lr=Lmv/Bq ∴v∝1sinθ or vv0=sin60°sin30° ∴v=√3v0

The correct option is D
$\sqrt{3} v_{0}$

$sin θ = \frac{L}{r} = \frac{L}{m v / B q}$
$∴ v \propto \frac{1}{sin θ}$
or $\frac{v}{v_{0}} = \frac{sin 60 °}{sin 30 °}$
$∴ v = \sqrt{3} v_{0}$