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A particle of mass $m$ and charge $q$ enters a region of magnetic field (as shown) with speed $v$ . There is a region in which the magnetic field is absent, as shown. The particle after entering the region collides elastically with a rigid wall. Time after which the velocity of particle becomes antiparallel to its initial velocity is

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$\frac{m}{4 q B} (π + 2)$

$\frac{m}{q B} (π + 2)$

$\frac{m}{4 q B} (2 π + 3)$

$\frac{m}{2 q B} (π + 4)$

answer is D.

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Detailed Solution

$r = \frac{m v}{q B} \sin θ = \frac{x}{r} = \frac{\frac{m v}{\sqrt{2} q B}}{\frac{m v}{q B}} = \frac{1}{\sqrt{2}}$

or $θ = \frac{π}{4}$

Time to complete the circle $(2 π)$ , $T = \frac{2 π m}{q B}$

time taken to traverses $\frac{π}{4}$ , $t = \frac{π m}{4 q B}$

Time taken to travel horizontal distance

$t_{1} = \frac{\frac{m v}{\sqrt{2} q B}}{\frac{v}{\sqrt{2}}} = \frac{m}{q B}$

Total time taken $= 2 t + 2 t_{1} = \frac{m}{2 q B} (π + 4)$

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