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A monoenergetic (18 keV) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of $0.04 \mathrm{G}$ normal to the initial direction. Estimate the up or down deflection of the beam over a distance of $30 \mathrm{~cm}\left(\mathrm{~m}_{\mathrm{e}}=9.11 \times 10^{-19} \mathrm{C}\right)$. [Note: Data in this exercise are so chosen that the answer will give you an idea of the effect of earth's magnetic field on the motion of the electron beam from the electron gun to the screen in a TV set.]
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Kinetic energy of electron $=18 \mathrm{keV}$
$\frac{1 \mathrm{~m}_e v^2}{2}=18 \times 10^3 \times 1.6 \times 10^{-19}$ joule
velocity $\mathrm{v}=8 \times 10^7 \mathrm{~ms}^{-1}$
velocity in $\mathrm{x}$ direction remain constant, hence time to cross $30 \mathrm{~cm}$
$t=\frac{30 \times 10^{-2}}{8 \times 10^7}$
$t=\frac{3}{8} \times 10^{-9} \mathrm{sec}$
A magnetic force $F=e v B$ is acting in vertical direction, which provides acceleration in vertical direction $a_y=\frac{e v B}{m_e}=5.63 \times 10^{14} \mathrm{~ms}^{-2}$
Vertical deflection $S_y=u_y t+\frac{1}{2} a_y t^2$
$$
\begin{aligned}
S_y &=0+\frac{1}{2} \times 5.63 \times 10^{14} \times\left[\frac{3}{8} \times 10^{-9}\right]^2 \\
S_y &=\frac{1}{2} \times \frac{9}{64} \times 5.63 \times 10^{-4} \\
&=395.85 \times 10^{-7} \text { meter. }
\end{aligned}
$$
$\frac{1 \mathrm{~m}_e v^2}{2}=18 \times 10^3 \times 1.6 \times 10^{-19}$ joule
velocity $\mathrm{v}=8 \times 10^7 \mathrm{~ms}^{-1}$
velocity in $\mathrm{x}$ direction remain constant, hence time to cross $30 \mathrm{~cm}$
$t=\frac{30 \times 10^{-2}}{8 \times 10^7}$
$t=\frac{3}{8} \times 10^{-9} \mathrm{sec}$
A magnetic force $F=e v B$ is acting in vertical direction, which provides acceleration in vertical direction $a_y=\frac{e v B}{m_e}=5.63 \times 10^{14} \mathrm{~ms}^{-2}$
Vertical deflection $S_y=u_y t+\frac{1}{2} a_y t^2$
$$
\begin{aligned}
S_y &=0+\frac{1}{2} \times 5.63 \times 10^{14} \times\left[\frac{3}{8} \times 10^{-9}\right]^2 \\
S_y &=\frac{1}{2} \times \frac{9}{64} \times 5.63 \times 10^{-4} \\
&=395.85 \times 10^{-7} \text { meter. }
\end{aligned}
$$
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