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A $10 \;eV$ electron is circulating in a plane at right angles to a uniform field at magnetic induction $10^{-4} \;W b / m^{2}(=1.0$ gauss), the orbital radius of electron is ........ $cm$
$11$
$18$
$12$
$16$
Solution
Kinetic energy of electron $\frac{1}{2} \times m v^{2}=10 eV$
and magnetic induction $( B )=10^{-4} Wb / m ^{2}$
Therefore $\frac{1}{2}\left(9.1 \times 10^{-31}\right) v^{2}=10 \times\left(1.6 \times 10^{-19}\right)$
$v^{2}=\frac{2 \times 10 \times\left(1.6 \times 10^{-19}\right)}{9.1 \times 10^{-31}}=3.52 \times 10^{12}$
$v=1.876 \times 10^{6} m$.
Centripetal force $=\frac{m v^{2}}{r}=B e v$.
$r=\frac{m v}{B e}=\frac{\left(9.1 \times 10^{-31}\right) \times\left(1.876 \times 10^{6}\right)}{10^{-4} \times\left(1.6 \times 10^{-19}\right)}$
$=11 \times 10^{-2} m=11 cm$
