A particle moving with velocity v having specific charge $(q/m)$ enters a region of magnetic field $B$ having width $d=\frac{{3mv}}{{5qB}}$ at angle $53^o$ to the boundary of magnetic field. Find the angle $\theta$ in the diagram......$^o$
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angle $\theta$ in the diagram is $90^{\circ}$.
let angle of deviation is a.
from figure, sina $=\mathrm{d} / \mathrm{R}$
where, $\mathrm{d}=3 \mathrm{mv} / 5 \mathrm{qb}$ and we know radius of circular path attained by charged particle $, \mathrm{R}=\mathrm{mv} / \mathrm{qb}$
now, sina $=(3 m v / 5 q b) /(m v / q b)=3 / 5=\sin 37^{\circ}$
$\Rightarrow a=37^{\circ}$
we know, angle of deviation = emergent angle - incidence angle
$=\theta+53^{\circ}$
or, $37^{\circ}=\theta+53^{\circ}$
or, $\theta=90^{\circ}$
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