Question

A proton, a deuteron and an $\alpha -particle$ with the same kinetic energy enters a region of uniform magnetic field moving at right angles to $B$. What is the ratio of the radii of their circular paths :

• A. $1:\sqrt{2}:2$
• B. $\sqrt{2}:2:1$
• C. $1:\sqrt{2}:1$
• D. $\sqrt{2}:1:1$

Answer 1

The correct option is C $1:\sqrt{2}:1$

When a charge particle enters a magnetic field moving at right angle, then it follows a circular path & its radius is given by,

$R=\frac{mv}{qB}$

Where,

$m=$ Mass of the charged particle
$q=$ Charge of particle
$B=$ Magnetic field
$v=$ Velocity of particle

Realtion between momentum & $KE$ is given by,

$P=\sqrt{2mK}$, $R=\frac{\sqrt{2mK}}{qB}$

For Proton, $R_p=\frac{\sqrt{\left(2m\right)K}}{eB}$

For Deuteron, $R_p=\frac{\sqrt{2\left(2m\right)K}}{eB}$

For $\alpha -partilce$, $R_{\alpha }=\frac{\sqrt{2\left(4m\right)K}}{2eB}$

$\therefore R_p:R_d:R_{\alpha }=\frac{\sqrt{2mK}}{eB}:\frac{2\sqrt{mK}}{eB}:\frac{\sqrt{2mK}}{eB}$

$\therefore R_P:R_D:R_{\alpha }=1:\sqrt{2}:1$

Hence, option $\left(C\right)$ is the correct answer.