A beam of cathode rays is subjected to crossed Electric ( $E$ ) and Magnetic fields $(B)$. The fields are…
A beam of cathode rays is subjected to crossed Electric ( $E$ ) and Magnetic fields $(B)$. The fields are adjusted such that the beam is not deflected. The specific charge of the cathode rays is given by
(where $V$ is the potential difference between cathod and anode)
$\frac{\mathrm{B}^2}{2 \mathrm{VE}^2}$
$\frac{2 \mathrm{VB}^2}{\mathrm{E}^2}$
$\frac{2 \mathrm{VE}^2}{\mathrm{~B}^2}$
$\frac{\mathrm{E}^2}{2 \mathrm{VB}^2}$
Solution
As the electron beam is not deflected, then or
$\begin{aligned}
\mathrm{F}_{\mathrm{m}} & =\mathrm{F}_{\mathrm{e}} \\
\mathrm{Bev} & =\mathrm{Ee} \\
\mathrm{v} & =\frac{\mathrm{E}}{\mathrm{B}}
\end{aligned}$
As the electron moves from cathode to anode, its potential energy at the cathode appears as its kinetic energy at the anode. If $V$ is the potential difference between the anode and cathode, then potential energy of the electron at cathode $=\mathrm{eV}$. Also, kinetic energy of the electron at anode $=\frac{1}{2} \mathrm{mv}^2$. According to law of conservation of energy
$\begin{aligned}
\frac{1}{2} \mathrm{mv}^2 & =\mathrm{eV} \\
\mathrm{v} & =\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}
\end{aligned}$
From Eqs. (i) and (ii), we have
$\begin{aligned}
\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}} & =\frac{E}{\mathrm{~B}} \\
\frac{\mathrm{e}}{\mathrm{m}} & =\frac{\mathrm{E}^2}{2 \mathrm{VB}^2}
\end{aligned}$