Question 158 of 276

An electron, moving along the $x-$ axis with an initial energy of $100eV,$ enters a region of magnetic field $\vec{B}=\left(1.5\times {10}^{-3}T\right)\hat{k}$ at S (see figure). The field extends between $x=0$ and $x=2cm.$ The electron is detected at the point $Q$ on a screen placed $8cm$ away from the point $S.$ The distance d between $P$ and $Q$ (on the screen) is:
(electron’s charge $1.6\times {10}^{-19}C$ , mass of electron $=9.1\times {10}^{-31}kg$ )

A$1.22cm$B$12.87cm$C$11.65cm$D$2.25cm$

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Answer: (B) $12.87cm$

$R=\dfrac{mV}{qB}$
$=\dfrac{\sqrt{2m(K.E)}}{qB}$
$\left\{mV=\sqrt{2m K.E}\right\}$
$=\dfrac{\sqrt{2\times 9.1\times {10}^{-31}\times 100\times 1.6\times {10}^{-19}}}{1.6\times {10}^{-19}\times 1.5\times {10}^{-3}}$
$=2.248 cm\approx 2.25 cm$

$\sin ⁡\theta =\dfrac{2}{2.248}=0.89$
$PA=R\left(1-\cos ⁡\theta \right)=1.22cm$
$AB=6\tan ⁡\theta =11.71cm$
$d=PA+AB=12.93cm$