Question

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

• A. 2 cm
• B. 4 cm
• C. 8 cm
• D. $2\sqrt{2}cm$

Answer 1

The correct option is D $2\sqrt{2}cm$

$r=\frac{mv}{qB}$
$K=\frac{1}{2}m{v}^2$
$\therefore v=\sqrt{\frac{2K}{m}}$
$\therefore r=\frac{m\sqrt{\frac{2K}{m}}}{qB}=\frac{\sqrt{\frac{2K}{m}}}{qB}$
$\frac{r_1}{r_2}=\frac{\frac{\sqrt{\frac{2K}{m}}}{qB}}{\frac{\sqrt{\frac{2\frac{K}{2}}{m}}}{qB}}$
$\Rightarrow r_2=\frac{r_1}{\sqrt{2}}$

$\Rightarrow r_2=\frac{4}{\sqrt{2}}=2\sqrt{2}cm$