Q. A beam of ions with velocity $2 \times 10^{5}m/s$ enters normally into a uniform magnetic field of $4 \times 10^{- 2}tesla$. If the specific charge of the ion is $5 \times 10^{7}C/kg$, then the radius of the circular path described will be
A) 0.10 $m$
B) 0.16 $m$
C) $0.20\ m$
D) 0.25 $m$
Explanation
(a) $r = \frac{mv}{Bq} = \frac{v}{(q/m)B} = \frac{2 \times 10^{5}}{5 \times 10^{7} \times 4 \times 10^{- 2}} = 0.1m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.