A charged particle moves in a uniform magnetic field perpendicular to it, with a radius of curvature 4cm. On passing through a metallic sheet, it loses half of its kinetic energy. Then, the radius of curvature of the particle is

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$2 \sqrt{2} cm$

8 cm

2 cm

4 cm

answer is D.

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Detailed Solution

$r = \frac{m v}{q B} \Rightarrow r \propto p$ , where $p =$ momentum
Also, $p = \sqrt{2 m K}$

$\Rightarrow r \propto p \propto \sqrt{K}$
When $K \to \frac{K}{2}, p \to \frac{p}{\sqrt{2}} . So, r \to \frac{r}{\sqrt{2}}$

$r^{'} = \frac{r}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2 \sqrt{2} cm$

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