A particle moving with velocity v having spec | vedclass

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Question

angle $	heta$ in the diagram is $90^{\circ}$.

let angle of deviation is a.

from figure, sina $=\mathrm{d} / \mathrm{R}$

where, $\mathrm{d}=3

Vedclass · Accepted Answer

$90$

Vedclass · Answer

angle $	heta$ 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

$=	heta+53^{\circ}$

or, $37^{\circ}=	heta+53^{\circ}$

or, $	heta=90^{\circ}$

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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