Q. An electron and a proton of equal linear momentum enter in the direction perpendicular to uniform magnetic field. If the radii of their circular paths be $r_{e}\ and\ r_{p}$ respectively, then $\frac{r_{e}}{r_{p}},$ is equal to $m_{e} =$ mass of electron, $m_{p} =$ mass of proton.
A) $\left( \frac{m_{p}}{m_{e}} \right)^{1/2}$
B) $\frac{m_{p}}{m_{e}}$
C) $\left( \frac{m_{e}}{m_{p}} \right)^{1/2}$
D) 1
Explanation
(d) Given, linear momentum of electron = linear momentum of proton.
or $m_{e}v_{e} = m_{p}v_{p}$
The radius of circular path is
$r = \frac{mv}{qB}$
For an electron the radius of circular path is
$r_{e} = \frac{m_{e}v_{e}}{qB}$
For a proton the radius of circular path is
$r_{p} = \frac{m_{p}v_{p}}{qB}$
Hence, $\frac{r_{e}}{r_{p}} = 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.