Question

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

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

Answer 1

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

Let the magnetic field be perpendicular to the plane (-z direction) and initial particle velocity be $\overrightarrow{v}$ along x direction. Then the particle will execute uniform circular motion in xy plane.


Radius of the circular path of the particle $r=\frac{mv}{qB}--\left(1\right)$
$mv=p$ (momentum) $=\sqrt{2mK}$
where $K$ is the kinetic energy of the particle.
$\therefore r=\frac{\sqrt{2mK}}{qB}$

After passing through the metal sheet, let the new radius of curvature be $r^{\prime }$.
Then $r^{\prime }=\frac{\sqrt{2m{K}^{\prime }}}{qB}--\left(2\right)$

(1) $÷$ (2) $\Rightarrow \frac{r}{r^{\prime }}=\frac{1}{\sqrt{2}}$
$\Rightarrow r^{\prime }=\frac{r}{\sqrt{2}}=\frac{4}{\sqrt{2}}$
$=2\sqrt{2}cm$