An electron, moving along the $x-$ axis with an initial energy of $100\, eV$, enters a region of magnetic field $\vec B = (1.5\times10^{-3}T)\hat k$ at $S$ (See figure). The field extends between $x = 0$ and $x = 2\, cm$. The electron is detected at the point $Q$ on a screen placed $8\, cm$ away from the point $S$. The distance $d$ between $P$ and $Q$ (on the screen) is :......$cm$ (electron's charge $= 1.6\times10^{-19}\, C$, mass of electron $= 9.1\times10^{-31}\, kg$)
JEE MAIN 2019, Diffcult
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${\text{R}} = \frac{{{\text{mv}}}}{{{\text{qB}}}}$
$ = \frac{{\sqrt {2{\text{m}}({\text{KE}})} }}{{{\text{qB}}}}$
${\text{R}} = \frac{{\sqrt {2 \times 9.1 \times {{10}^{ - 31}} \times \left( {100 \times 1.6 \times {{10}^{ - 18}}} \right)} }}{{1.6 \times {{10}^{ - 10}} \times 1.5 \times {{10}^{ - 3}}}}$
${\text{R}} = 2.248\,{\text{cm}}$
$\sin \theta = \frac{2}{{2.248}};\,\,\,\,\,\tan \,\theta = \frac{{QU}}{{TU}}$
$\frac{2}{{1.026}} = \frac{{QU}}{6}$
$QU = 11.69$
${\text{PU}} = {\text{R}}(1 - \cos \,\theta )$
$ = 1.22$
${\text{d}} = {\text{QU}} + {\text{PU}}$
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