Question

An electron, moving along the x-axis with an initial energy of $100eV$, enters a region of magnetic field $\vec{B}=(1.5\times 10^{-3}T)\hat{k}$ 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. 12.87 cm
• B. 1.22 cm
• C. 11.65 cm
• D. 2.25 cm

Answer 1

Answer: (A)

$R=\frac{mv}{qB}$
$=\frac{\sqrt{2m(K.E.)}}{qB}$
$R=\frac{\sqrt{2\times 9.1\times 10^{-31}\times (100\times 1.6\times 10^{-19}}}{1.6\times 10^{-19}\times 1.5\times 10^{-3}}$
$R=2.248$ cm
$sin\theta =\frac{2}{2.248}$
$tan\theta =\frac{QU}{TU}$
$\frac{2}{1.026}=\frac{QU}{6}$
$QU=11.69$
$PU=R(1-cos\theta )$
= $1.22$
$d=QU+PU$