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A particle of mass $m$ and charge $q$ , accelerated by a potential difference $V$ enters a region of a uniform transverse magnetic field $B$ . If $d$ is the thickness of the region of $B$ , the angle $θ$ through which the particle deviates from the initial direction on leaving the region is given by:

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$tanθ = Bd {(\frac{q}{2 mV})}^{1 / 2}$

$cotθ = Bd {(\frac{q}{2 mV})}^{1 / 2}$

$sinθ = Bd {(\frac{q}{2 mV})}^{1 / 2}$

$cosθ = Bd {(\frac{q}{2 mV})}^{1 / 2}$

answer is A.

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Detailed Solution

Let $v$ be the velocity of the particle. Its kinetic energy is:

$\frac{1}{2} {mv}^{2} = qV \Rightarrow v = \sqrt{\frac{2 qV}{m}} . . . . . . (1)$

The particle follows a circular path from $A$ to $B$ of radius r which is given by

$\frac{{mv}^{2}}{r} = qvB \Rightarrow r = \frac{mv}{qB} . . . . . . (2)$

using $(1)$ and $(2)$ , we get:

$r = \frac{m}{qB} {(\frac{2 qV}{m})}^{1 / 2}$

$= \frac{1}{B} {(\frac{2 mV}{q})}^{1 / 2}$

in triangle $BCD, sinθ = \frac{BD}{BC}$ $= \frac{d}{r}$

Therefore, $sinθ = Bd {(\frac{q}{2 mV})}^{\frac{1}{2}}$

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