An electron moves with a speed of \( 4.4\times10^{7} \) m/s along circular path of radius \( 28 \) cm in a uniform magnetic field. The frequency of revolution of the electron is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 25 \) MHz
In a uniform magnetic field the electron moves in a circle at constant speed, so the frequency of revolution is the speed divided by the circumference: \( f = v/2\pi r \).
\( v = 4.4\times10^{7}\,\mathrm{m/s} \)
\( r = 28\,\mathrm{cm} = 0.28\,\mathrm{m} \)
\( \displaystyle f = \frac{v}{2\pi r} = \frac{4.4\times10^{7}}{2\times\frac{22}{7}\times0.28} \)
\( \displaystyle f = \frac{4.4\times10^{7}}{1.76} = 2.5\times10^{7}\,\mathrm{Hz} = 25\,\mathrm{MHz} \)