Q. A proton, a deutron and an $\alpha$-particle with the same KE enter a region of uniform magnetic field, moving at right angle to B. What is the ratio of the radius of their circular paths ?
A) $1\ :\ \sqrt{2}\ :1$
B) $1\ :\ \sqrt{2}\ :\ \sqrt{2}$
C) $\sqrt{2}\ :1\ :1$
D) $\sqrt{2}\ :\ \sqrt{2}\ :1$
Explanation
(a) In a magnetic field, perpendicular to velocity of particle
$\frac{mv^{2}}{r} = Bq\sqrt{2m\ E_{k}}v\ \ \ \ \ \ or\ \ \ r = \frac{mv}{Bq}\ \ \ \ \ and\ E_{k} = \frac{1}{2}\ mv^{2}$
$mv\ = \sqrt{2m\ E_{k}}v\ \ \ \ \ \ \ \ \ \ \ so,\ \ r = \frac{\sqrt{2m\ E_{k}}v}{Bq}\ \ \ or\ \ r \propto \frac{\sqrt{m}}{q}$
For same value of $E_{k}\ \ and\ B.$
$r_{P}\ :r_{d}\ :r_{\alpha} = \frac{\sqrt{m}}{l}\ :\frac{\sqrt{2m}}{l}\ :\frac{\sqrt{4m}}{2l}\ $
$\ = 1\ :\ \sqrt{2\ }\ :1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.