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A rectangular region of dimensions ( $\omega \times l(\omega) \ll l$ ) has a constant magnetic field into the plane of the paper as shown in the figure below. On one side, the region is bounded by a screen. On the other side, positive ions of mass $m$ and charge $q$ are accelerated from rest and towards the screen by a parallel plate capacitor at constant potential difference $V < 0$ and come out through a small hole in the upper plate. Which one of the following statements is correct regarding the charge on the ions that hit the screen?
Ions with $q > \frac{2 V m}{B^{2} \omega^{2}}$ will hit the screen
Ions with $q < \frac{2 V m}{B^{2} \omega^{2}}$ will hit the screen
All ions will hit the screen
Only ions with $q=\frac{2 V m}{B^{2} \omega^{2}}$ will hit the screen
Solution
$(b)$ Trajectory of charged particle in region of perpendicular magnetic field to its velocity is a circle.
So, path of charged particle will be as shown below in the figure,
Clearly, charged particle will hit the screen, if $R > \omega$.
$\Rightarrow \left.\frac{m v}{B q} > \omega \text { [from, } B q v=\frac{m v^{2}}{R}\right]$
$\Rightarrow \frac{\sqrt{2 m K}}{B q} > \omega$
${[\therefore \text { momentum, } p=m v=\sqrt{2 m K}]}$
$\Rightarrow \frac{\sqrt{2 m q V}}{B q} > \omega$
$\text { or } \frac{2 m V}{B^{2} q} > \omega^{2} \text { or } \quad q < \frac{2 m V}{B^{2} \omega^{2}}$
