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A positively charged particle of charge ‘q’ enters in a uniform magnetic field $\vec{B}$ directed inward and is deflected a distance x after traveling a distance Y. The magnitude of linear momentum of the particle is

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$\frac{q B}{2 x} (x^{2} + Y^{2})$

$\frac{q B}{2} (\frac{x^{2}}{Y} + x)$

$\frac{q B Y}{2}$

$\frac{q B}{2 Y} (x^{2} + Y^{2})$

answer is C.

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Detailed Solution

$x = R - R \cos θ y = R \sin θ$

Eliminating ‘ $θ$ ’
$x^{2} + y^{2} - 2 R x = 0$ --- (1)
Linear momentum $P = m v = q B r = q B (\frac{x^{2} + y^{2}}{2 x})$

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