Q. A charged particle moving in a uniform magnetic field penetrates layer of lead and there by loss one-half of its kinetic energy. How does the radius of curvature of its path change?
A) The radius reduces to $r\sqrt{2}$
B) The radius reduces to $\frac{r}{\sqrt{2}}\ $
C) The radius remains the same
D) The radius becomes $r$/2
Explanation
(b) $r = \ \sqrt{2mE}$/ $Bq$ and $r_{1} = \ \sqrt{2m(E/2}$) / $Bq;$
So, $r_{1} = r/\sqrt{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.