Q. An electron beam travels with a velocity of $1.6 \times 10^{7}\ {ms}^{- 1}$ perpendicularly to magnetic field of intensity 0.1 T. The radius of the path of the electron beam $(m_{e} = 9 \times 10^{- 31}\ kg)$
A) $9 \times 10^{- 5}\ m$
B) $9 \times 10^{- 2}\ m$
C) $9 \times 10^{- 4}\ m$
D) $9 \times 10^{- 3}\ m$
Explanation
(c) In a perpendicular magnetic field, the radius of circular path travelled by electron beam is
$\ \ \ \ \ \ r = \frac{mv}{eB}$
$\therefore\ \ r = \frac{9 \times 10^{- 31} \times 1.6 \times 10^{7}}{1.6 \times 10^{- 19} \times 0.1}$
$\ \ \ \ \ \ \ \ \ = 9 \times 10^{- 4}\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.