Question

A charged particle moving in a uniform magnetic field penetrates the layer of lead and thereby losses one-half of its kinetic energy. The radius of curvature of its path

• 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 $r/2$

Answer 1

Answer: (B)

$r= \, \frac{\sqrt{2 m E}}{q B}$ and $r _{1}=\sqrt{\frac{2 \textit{m} \left(E ⁡ / 2\right)}{q B}}$
So, $r _{1} = \frac{r ⁡}{\sqrt{2}}$