Question

A potential difference of V volt is applied between two parallel plates $P_1$ and $P_2$, separated by a distance d, with the planes of plates being parallel to xOz plane as shown in the Figure. A uniform magnetic field $B\hat{k}$ is applied out of the cross- section, perpendicular to it. A particle of charge q and mass m is released at rest from O$(0,0,0)$ on the plale $P_1$. it is found that the particle never touches the plate $P_2$ if $d\ge \frac{1}{B}\sqrt{\frac{nVm}{q}}$ What is the value of $n$?

Answer 1

$\overrightarrow{E}=\frac{V}{d}\hat{j},\overrightarrow{B}=B\hat{k}$

The motion of the particle is confined to x-y plane (since initially it is at rest and forces due to E and B are in x-y plane)

$\therefore \overrightarrow{v}=x\hat{i}+y\hat{j}$
$m\overrightarrow{a}=q[\frac{V}{d}\hat{j}+(-xB\hat{j}+yB\hat{i}\left)\right]$
$=[(\frac{qV}{d}-qBx)\hat{j}+qBy\hat{i}]$
Decomposing,
$mx=qBy$ and $my=-qBx+\frac{qV}{d}$
$\therefore m\frac{d}{dt}\left(y\right)=-qBx=-\frac{q^2{B}^2}{m}y$
$\therefore \frac{d^2}{d{t}^2}\left(y\right)=-{\left(\frac{qB}{m}\right)}^{2}y$ or $\frac{d^2}{d{t}^2}\left(y\right)+{\omega }^{2}y=0$

Let $y=A\sin \omega t+B\cos \omega t$
$y=\frac{1}{\omega }[-Acos\omega t+B\sin \omega t]+C$
$y=\omega [A\cos \omega t-B\sin \omega t]$

At $t=0,y=0,\dot{y}=0,\overline{y}=\frac{qV}{md}$
$y=0\Rightarrow B=0,y=0\Rightarrow C=\frac{A}{\omega },\overline{y}=\frac{qV}{md}=\omega A$

$\therefore A=\frac{qV}{md.\omega }=\frac{qV}{md}.\frac{m}{qB}=\frac{V}{dB}=\frac{E}{B}$
$C=\frac{V}{dB\omega }$

Hence $y=\frac{A}{\omega }(1-cos\omega t)=\frac{V}{\omega dB}(1-cos\omega t)$
$y_{max}=\frac{2V}{\omega dB}\le d$

or $d^2\ge \frac{2V}{\omega B}(=\frac{2Vm}{q{B}^2})$

or $d^2\ge \frac{2Vm}{q{B}^2}$
$\therefore d\ge \frac{1}{B}\sqrt{\frac{2Vm}{q}}\Rightarrow n=2$