Q. An electron is moving in a circular path under the influence of a transverse magnetic field of $3.57 \times 10^{-2}$ T. If the value of e/m is $1.76 \times 10^{11}$ C/kg, the frequency of revolution of the electron is
A) 1 GHz
B) 100 MHz
C) 62.8 MHz
D) 6.28 MHz
Explanation
[Solution] $f = \frac{eB}{2\pi m} = \frac{1.76 \times 10^{11} \times 3.57 \times 10^{-2}}{2\pi} \approx 10^9$ Hz. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.