An electron placed in a magnetic field of \( 3.14\times10^{-4}\,\mathrm{T} \) moves in a circle. The frequency of the revolution of the electron is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 8.79\times10^{6}\,\mathrm{Hz} \)
The magnetic force supplies the centripetal force on the electron, \( evB = mv^{2}/r \). This gives the cyclotron frequency, which does not depend on the speed or the radius.
\( \displaystyle f = \frac{v}{2\pi r} = \frac{eB}{2\pi m} \)
\( e = 1.6\times10^{-19}\,\mathrm{C} \)
\( m = 9.1\times10^{-31}\,\mathrm{kg} \)
\( B = 3.14\times10^{-4}\,\mathrm{T} = \pi\times10^{-4}\,\mathrm{T} \)
\( \displaystyle f = \frac{1.6\times10^{-19} \times \pi\times10^{-4}}{2\pi \times 9.1\times10^{-31}} = \frac{1.6\times10^{-23}}{1.82\times10^{-30}} = 8.79\times10^{6}\,\mathrm{Hz} \)