First slide

Force on a charged particle moving in a magnetic field

Question

An $α -$ particle and a deutron enter a uniform magnetic field with same kinetic energy in a direction perpendicular to the direction of magnetic field. Then ratio of the radii of their paths $r_{α} : r_{d}$ will be

Easy

A

$\sqrt{2} : 1$
B

$1 : 2 \sqrt{2}$
C

$1 : \sqrt{2}$
D

$2 \sqrt{2} : 1$

Solution

$ω = \frac{Q B}{m} ∴ r = \frac{V}{ω}$
Again $K = \frac{1}{2} m V^{2} \Rightarrow V = \sqrt{\frac{2 K}{m}}$
$∴ r = \frac{m}{B} \sqrt{\frac{2 K}{m}} = \frac{\sqrt{2 K m}}{Q B}$
$∴ \frac{r_{α}}{r_{d}} = \sqrt{\frac{m_{α}}{m_{d}}} . (\frac{Q_{d}}{Q_{α}}) = \sqrt{\frac{4}{2}} \frac{e}{2 e} = \frac{1}{\sqrt{2}}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App