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An electron moves at right angle to a magnetic field of $1.5 \times 10^{-2} \mathrm{~T}$ with a speed of $6 \times 10^7 \mathrm{~m} / \mathrm{s}$. If the specific charge of the electron is $1.7 \times 10^{11} \mathrm{C} / \mathrm{kg}$. The radius of the circular path will be
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The correct answer is:
2.35 cm
The formula for radius of circular path is
$r=\frac{m v}{e B}=\frac{v}{\left(\frac{e}{m}\right) B}$ ...(1)
Given $: e / m$ of electron
$\begin{aligned}
& =1.7 \times 10^{11} \mathrm{C} / \mathrm{kg} \text { and } v=6 \times 10^7 \mathrm{~m} / \mathrm{s} \\
B & =1.5 \times 10^{-2} \mathrm{~T}
\end{aligned}$
$\quad \begin{aligned} So, r & =\frac{6 \times 10^7}{1.7 \times 10^{11} \times 1.5 \times 10^{-2}} \\ & =2.35 \times 10^{-2} \mathrm{~m}=2.35 \mathrm{~cm}\end{aligned}$
$r=\frac{m v}{e B}=\frac{v}{\left(\frac{e}{m}\right) B}$ ...(1)
Given $: e / m$ of electron
$\begin{aligned}
& =1.7 \times 10^{11} \mathrm{C} / \mathrm{kg} \text { and } v=6 \times 10^7 \mathrm{~m} / \mathrm{s} \\
B & =1.5 \times 10^{-2} \mathrm{~T}
\end{aligned}$
$\quad \begin{aligned} So, r & =\frac{6 \times 10^7}{1.7 \times 10^{11} \times 1.5 \times 10^{-2}} \\ & =2.35 \times 10^{-2} \mathrm{~m}=2.35 \mathrm{~cm}\end{aligned}$
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