A charged particle (q,m) initially at rest, is released in uniform gravitational field. If uniform magnetic field B0(−k^) is also present, then maximum speed acquired by the charged particle, is xmgqB0. Find x

Resolve the velocity in x and y and write down the equation of motion.At lowest position, vx=2gyF→=−mgj^+q−vyj^+vxi^×B0(−k^)mdvxdt=qvyB0mvx=qB0ym2gy=qyB0y=2m2gq2B02Maximum speed, vx=qB0m×2m2gq2B02=2mgqB0

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A charged particle $(q, m)$ initially at rest, is released in uniform gravitational field. If uniform magnetic field $B_{0} (- \hat{k})$ is also present, then maximum speed acquired by the charged particle, is $\frac{x m g}{q B_{0}}$ . Find x

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Detailed Solution

Resolve the velocity in $x$ and $y$ and write down the equation of motion.
At lowest position, $v_{x} = \sqrt{2 g y}$
$\begin{array}{l} \vec{F} = - m g \hat{j} + q (- v_{y} \hat{j} + v_{x} \hat{i}) \times B_{0} (- \hat{k}) \\ m \frac{d v_{x}}{d t} = q v_{y} B_{0} \\ m v_{x} = q B_{0} y \\ m \sqrt{2 g y} = q y B_{0} \\ y = \frac{2 m^{2} g}{q^{2} B_{0}^{2}} \end{array}$