A particle of specific charge (q/m) is projec | vedclass

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(C) The electric field $\vec{E}$ is along the $+y$ direction,so the acceleration of the particle is $a_y = (qE/m)$.For the particle to return to the o

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$[vB / \pi E]$ is an integer

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(C) The electric field $\vec{E}$ is along the $+y$ direction,so the acceleration of the particle is $a_y = (qE/m)$.For the particle to return to the origin,its displacement along the $y$-axis must be zero at some time $t > 0$.Using the equation of motion along the $y$-axis: $y = u_y t + \frac{1}{2} a_y t^2$.Setting $y = 0$ and $u_y = -v$,we get: $0 = -vt + \frac{1}{2} (qE/m) t^2$.Solving for $t$,we find: $t = \frac{2mv}{qE}$.In this time $t$,the particle must complete an integer number of revolutions in the $x$-$z$ plane due to the magnetic field $\vec{B}$ along the $y$-axis.The time period of revolution in the magnetic field is $T = \frac{2\pi m}{qB}$.For the particle to return to the origin,the time $t$ must be an integer multiple of the time period $T$,i.e.,$t = nT$ where $n = 1, 2, 3, \dots$.Substituting the expressions for $t$ and $T$: $\frac{2mv}{qE} = n \left( \frac{2\pi m}{qB} \right)$.Simplifying this,we get: $\frac{v}{E} = n \frac{\pi}{B}$,which implies $\frac{vB}{\pi E} = n$.Thus,$[vB / \pi E]$ must be an integer.

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