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A 10 eV electron is circulating in a plane at right angles to a uniform field at magnetic induction 10^{-4} \mathrm{Wb} / \mathrm{m}^2(=1.0 \mathrm{~gauss}). The orbital radius of the electron is.

Option: 1

12 cm


Option: 2

16 cm


Option: 3

11 cm


Option: 4

18 cm


Answers (1)

best_answer

\text { K.E. of electron }=10 \mathrm{eV} \\
\Rightarrow \frac{1}{2} \mathrm{mv}^2=10 \mathrm{eV} \\

\Rightarrow \frac{1}{2}\left(9.1 \times 10^{-31}\right) \mathrm{v}^2=10 \times 1.6 \times 10^{-19} \\

\Rightarrow \mathrm{v}^2=\frac{2 \times 10 . \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}} \\


Also, we know that for circular motion
\mathrm{ \frac{\mathrm{mv}^2}{\mathrm{r}}=\mathrm{Bev} \Rightarrow \mathrm{r~~~r}=\frac{\mathrm{mv}}{\mathrm{Be}}=11 \mathrm{~cm} }

Posted by

Anam Khan

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