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(D) The time period of a charged particle in a magnetic field is given by $T = \frac{2\pi m}{QB}$.For the first particle with mass $M$ and charge $Q$,

Vedclass · Accepted Answer

$4\pi Mv \cos	heta / QB$

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(D) The time period of a charged particle in a magnetic field is given by $T = \frac{2\pi m}{QB}$.For the first particle with mass $M$ and charge $Q$,the time period is $T_1 = \frac{2\pi M}{QB}$.For the second particle with mass $2M$ and charge $Q$,the time period is $T_2 = \frac{2\pi(2M)}{QB} = \frac{4\pi M}{QB}$.The particles will meet again when the time elapsed is a common multiple of their time periods. The least common multiple of $T_1$ and $T_2$ is $T = 4\pi M / QB$.At this time,the first particle completes $n_1 = T/T_1 = 2$ revolutions,and the second particle completes $n_2 = T/T_2 = 1$ revolution.The distance traveled along the direction of the magnetic field is given by $d = v \cos	heta 	imes t$.Substituting $t = T = \frac{4\pi M}{QB}$,we get $d = v \cos	heta 	imes \frac{4\pi M}{QB} = \frac{4\pi Mv \cos	heta}{QB}$.

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