A constant horizontal uniform magnetic field $B = 5 \times 10^{6}$ tesla exist in a uniform vertical gravitational field $g = 10 m / s^{2}$ . Both fields are perpendicular to each other. A particle of mass $m = 50 gm$ and charge $q = 2 μC$ is released in the space. Find the displacement (in cm) of particle along y-axis when velocity particle is along x-axis first time.

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answer is 0.05.

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Detailed Solution

$mg - {qBv}_{x} = m \frac{{dv}_{y}}{dt} {qBv}_{y} = m \frac{{dv}_{x}}{dt}$
Differentiate equation (i) with respect to time
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$0 - qB \frac{{dv}_{x}}{dt} = m \frac{d^{2} v_{y}}{{dt}^{2}} \frac{d^{2} v_{y}}{{dt}^{2}} = - \frac{q^{2} B^{2}}{m^{2}} v_{y} = - ω^{2} v_{y} Solution of equation v_{y} = {(v_{y})}_{0} sinωt \frac{{dv}_{y}}{dt} = {(v_{y})}_{0} ωcosωt at = 0, \frac{{dv}_{y}}{dt} = g, so {(v_{y})}_{0} = \frac{g}{ω} Now form equation (i) v_{x} = \frac{mg}{q B} (1 - cosωt) So, v_{y} = 0, at ωt = π, t = \frac{π}{ω}$
$v_{x} = \frac{mg}{qB} [1 - (- 1)] = \frac{2 mg}{qB} From work energy theorem mgy = \frac{1}{2} {mv}_{x}^{2} y = \frac{2 m^{2} g}{q^{2} B^{2}} = 0.05 cm$

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