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Q.

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

2v0

b

v02

c

v03

d

V03

answer is D.

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Detailed Solution

In ∆OABAB = L = constantOB=r=mvqBsin⁡θ=Lr⇒L=rsin⁡θ=constant or vsin⁡θ= constant v0sin⁡60∘=v′sin⁡30∘v′=3v0
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