Q. Electrons move at right angles to a magnetic field of $1.5 \times 10^{- 2}tesla$ with a speed of $6 \times 10^{7}m/s$. If the specific charge of the electron is $1.7 \times 10^{11}C/kg$, the radius of the circular path will be
A) 2.9 $cm$
B) 3.9 $cm$
C) 2.35 $cm$
D) 3 $cm$
Explanation
(c) $r = \frac{mv}{qB} = \frac{6 \times 10^{7}}{1.7 \times 10^{11} \times 1.5 \times 10^{- 2}} = 2.35\ cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.