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  4. An electron, moving along the x-axis with an initial energy of 100 eV, enters a region of magnetic field vecB = (1.5 × 10-3 T) hatk 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: (electron's charge = 1.6 × 10-19C. mass of electron = 9.1 × 10-31kg)

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Q. An electron, moving along the x-axis with an initial energy of $100\, eV$, 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 $x = 2\, cm$. 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 : (electron's charge = $1.6 \times 10^{-19}C.$ mass of electron = $9.1 \times 10^{-31}kg)$Physics Question Image

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Solution:

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

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