An electron with energy \( 8 \) keV moves at right angle in a uniform magnetic field of \( 1 \) Gauss. The radius of the circular path of the electron is, approximately
(Given: Mass of electron, \( m = 9\times10^{-31} \) kg)
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 3 m
A charge moving perpendicular to a uniform field moves in a circle; the magnetic force provides the centripetal force.
\( \displaystyle qvB = \frac{mv^{2}}{r} \Rightarrow r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{eB} \)
\( K = 8\,\mathrm{keV} = 8\times10^{3}\times1.6\times10^{-19}\,\mathrm{J} = 1.28\times10^{-15}\,\mathrm{J} \)
\( B = 1\,\mathrm{G} = 10^{-4}\,\mathrm{T} \)
\( \displaystyle \sqrt{2mK} = \sqrt{2\times9\times10^{-31}\times1.28\times10^{-15}} = \sqrt{2.304\times10^{-45}} = 4.8\times10^{-23}\,\mathrm{kg\,m\,s^{-1}} \)
\( \displaystyle r = \frac{4.8\times10^{-23}}{(1.6\times10^{-19})(10^{-4})}\,\mathrm{m} = \frac{4.8\times10^{-23}}{1.6\times10^{-23}}\,\mathrm{m} = 3\,\mathrm{m} \)