Q. A electron moving with kinetic energy $6 \times 10^{- 16}$ J enters a field of magnetic induction $6 \times 10^{- 3}\ Wbm^{- 2}$ at right angle to its motion. The radius of its path is
A) 3.42 cm
B) 4.23 cm
C) 6.17 cm
D) 7.7 cm
Explanation
(a) $r = \frac{mv}{Bq} = \frac{\sqrt{2E_{k}m}}{Bq\ }$
$= \ \frac{\sqrt{2 \times 6 \times 10^{- 16} \times 9 \times 10^{- 31}}}{6 \times 10^{- 3} \times 1.6 \times 10^{- 19}}$
On solving $r = \ $3.42 cm.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.