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Two particles of charges $+Q$ and $-Q$ are projected from the same point with a velocity $v$ in a region of uniform magnetic field $B$ such that the velocity vector makes an angle $q$ with the magnetic field. Their masses are $M$ and $2M,$ respectively. Then, they will meet again for the first time at a point whose distance from the point of projection is
$2\pi Mv \,\,cos\theta\, Q/B$
$8\pi Mv \,\,cos\theta \,Q/B$
$\pi Mv\,\, cos\theta \,Q/B$
$4\pi Mv\,\, cos\theta \,/QB$
Solution
$T_{1}=\frac{2 \pi M}{q B} T_{2}=\frac{2 \pi(2 M)}{q B} \Rightarrow T_{2}=\frac{4 \pi M}{q B}$
they will meet again when first will complete two revolution and second will complete one revoluation so direction
$d=v \cos \theta t=v \cos \theta .2 \frac{2 \pi M}{q B}=\frac{4 \pi M v \cos \theta}{q B}$
