Q. In an ionisation experiment it is found that a doubly ionised particle enters a magnetic field of 1 T and moves in a circular path of radius 1 m with a speed of $1.6 \times 10^{7}\ ms^{- 1}$. The particle must be
A) $C^{+ +}$
B) $Be^{+ +}$
C) $Li^{+ +}$
D) $He^{+ +}$
Explanation
(a) As $q\ v\ B = mv^{2}/r$
or $m = \frac{qBr}{v} = \frac{(2 \times 1.6 \times 10^{- 19}) \times 1 \times 1}{1.6 \times 10^{7}}$
$= 2 \times 10^{- 26}kg = \frac{2 \times 10^{26}}{1.66 \times 10^{- 27}} = 12$
Therefore, particle must be $c^{+ +}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.