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An electron gun with its collector at a potential of $100 \mathrm{~V}$ fires out electrons in a spherical bulb containing hydrogen gas at low pressure $\left(\sim 10^{-2} \mathrm{~mm}\right.$ of $\left.\mathrm{Hg}\right)$. A magnetic field of $2.83 \times 10^{-4} \mathrm{~T}$ curves the path of the electrons in a circular orbit of radius $12.0 \mathrm{~cm}$.
(The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known as the fine beam tube' method.)
(The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known as the fine beam tube' method.)
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Verified Answer
Given $\mathrm{V}=100 \mathrm{~V}, \mathrm{~B}=2.83 \times 10^{-4} \mathrm{~T}$,
$\mathrm{r}=12.0 \mathrm{~cm}=12 \times 10^{-2} \mathrm{~m}$
k.E. of accelerated electron $\frac{1}{2} \mathrm{mv}^2=\mathrm{eV}$
$\Rightarrow \mathrm{v}^2=\frac{2 \mathrm{eV}}{\mathrm{m}}$
on application of magnetic field B normally to electron beam, the radius of circular path is
$\begin{aligned} & \operatorname{Bev}=\frac{\mathrm{mv}^2}{\mathrm{r}} \\ \Rightarrow & \mathrm{v}=\frac{\mathrm{Ber}}{\mathrm{m}} \Rightarrow \frac{\mathrm{B}^2 \mathrm{e}^2 \mathrm{r}^2}{\mathrm{~m}^2}=\mathrm{v}^2 \\ \text { Equating eqs. (1) and (2), } \end{aligned}$
$$
\frac{2 \mathrm{eV}}{\mathrm{m}}=\frac{\mathrm{B}^2 \mathrm{e}^2 \mathrm{r}^2}{\mathrm{~m}^2} \Rightarrow \frac{\mathrm{e}}{\mathrm{m}}=\frac{2 \mathrm{~V}}{\mathrm{~B}^2 \mathrm{r}^2}
$$
$$
\begin{aligned}
\Rightarrow \frac{\mathrm{e}}{\mathrm{m}} &=\frac{2 \times 100}{\left(2.83 \times 10^{-4}\right)^2 \times\left(12 \times 10^{-2}\right)^2} \\
&=\frac{200}{8.01 \times 10^{-8} \times 144 \times 10^{-4}}
\end{aligned}
$$
$$
=\frac{200 \times 10^{12}}{8 \times 144}=1.734 \times 10^{11} \mathrm{C} / \mathrm{kg}
$$
$\mathrm{r}=12.0 \mathrm{~cm}=12 \times 10^{-2} \mathrm{~m}$
k.E. of accelerated electron $\frac{1}{2} \mathrm{mv}^2=\mathrm{eV}$
$\Rightarrow \mathrm{v}^2=\frac{2 \mathrm{eV}}{\mathrm{m}}$
on application of magnetic field B normally to electron beam, the radius of circular path is
$\begin{aligned} & \operatorname{Bev}=\frac{\mathrm{mv}^2}{\mathrm{r}} \\ \Rightarrow & \mathrm{v}=\frac{\mathrm{Ber}}{\mathrm{m}} \Rightarrow \frac{\mathrm{B}^2 \mathrm{e}^2 \mathrm{r}^2}{\mathrm{~m}^2}=\mathrm{v}^2 \\ \text { Equating eqs. (1) and (2), } \end{aligned}$
$$
\frac{2 \mathrm{eV}}{\mathrm{m}}=\frac{\mathrm{B}^2 \mathrm{e}^2 \mathrm{r}^2}{\mathrm{~m}^2} \Rightarrow \frac{\mathrm{e}}{\mathrm{m}}=\frac{2 \mathrm{~V}}{\mathrm{~B}^2 \mathrm{r}^2}
$$
$$
\begin{aligned}
\Rightarrow \frac{\mathrm{e}}{\mathrm{m}} &=\frac{2 \times 100}{\left(2.83 \times 10^{-4}\right)^2 \times\left(12 \times 10^{-2}\right)^2} \\
&=\frac{200}{8.01 \times 10^{-8} \times 144 \times 10^{-4}}
\end{aligned}
$$
$$
=\frac{200 \times 10^{12}}{8 \times 144}=1.734 \times 10^{11} \mathrm{C} / \mathrm{kg}
$$
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