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In a chamber, a uniform magnetic field of $6.5 \mathrm{G}(1 \mathrm{G}$ $\left.=10^{-4} \mathrm{~T}\right)$ is maintained. An electron is shot into the field with a speed of $4.8 \times 10^6 \mathrm{~m} \mathrm{~s}^{-1}$ normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. $\left(e=1.6 \times 10^{-19} \mathrm{C}, \mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-31} \mathrm{~kg}\right)$
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Given, $\mathrm{B}=6.5 \mathrm{G}=6.5 \times 10^{-4} \mathrm{~T}$
$\mathrm{v}=4.8 \times 10^6 \mathrm{~m} / \mathrm{s}, \mathrm{q}=\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$
$\mathrm{m}=9.1 \times 10^{-31} \mathrm{~kg}$.
$\because \frac{\mathrm{mv}^2}{\mathrm{r}}=\mathrm{qvB} \Rightarrow \mathrm{r}=\frac{\mathrm{mv}}{\mathrm{Bq}}=\frac{9.1 \times 10^{-31} \times 4.8 \times 10^6}{1.6 \times 10^{-19} \times 6.5 \times 10^{-4}}$
$=\frac{91 \times 48 \times 10^{-27}}{16 \times 65 \times 10^{-25}}=4.2 \times 10^{-2} \mathrm{~m}=4.2 \mathrm{~cm}$
$\mathrm{v}=4.8 \times 10^6 \mathrm{~m} / \mathrm{s}, \mathrm{q}=\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$
$\mathrm{m}=9.1 \times 10^{-31} \mathrm{~kg}$.
$\because \frac{\mathrm{mv}^2}{\mathrm{r}}=\mathrm{qvB} \Rightarrow \mathrm{r}=\frac{\mathrm{mv}}{\mathrm{Bq}}=\frac{9.1 \times 10^{-31} \times 4.8 \times 10^6}{1.6 \times 10^{-19} \times 6.5 \times 10^{-4}}$
$=\frac{91 \times 48 \times 10^{-27}}{16 \times 65 \times 10^{-25}}=4.2 \times 10^{-2} \mathrm{~m}=4.2 \mathrm{~cm}$
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