Q. An electron $\left( mass = 9.0 \times 10^{- 31} \right)\ kg$ and charge $(1.6 \times 10^{- 19}\ C)$ is moving in a circular orbit in a magnetic field of $1.0 \times 10^{- 4}\ {Wbm}^{- 2}.$ Its period of revolution is
A) $2.1 \times 10^{- 6}\ s$
B) $1.05 \times 10^{- 6}\ s$
C) $7 \times 10^{- 7}\ s$
D) $3.5 \times 10^{- 7}\ s$
Explanation
(d) The time period of revolution of the electrion is
$T = \frac{2\pi m}{qB} = \frac{2 \times 3.14 \times 9.0 \times 10^{- 31}}{1.6 \times 10^{- 19} \times 1 \times 10^{- 4}}$
$\ \ \ \ = 3.5 \times 10^{- 7}\ s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.