First slide

Force on a charged particle moving in a magnetic field

Question

A proton, a deuteron and an a-particle having the same kinetic energy are moving in circular trajectories in a constant magnetic field. If r_p, r_d and $r_{α}$ denote, respectively, the radii of the trajectories of these particles, then

Moderate

A

$r_{α} = r_{p} < r_{d}$
B

$r_{α} > r_{d} > r_{p}$
C

$r_{α} = r_{d} > r_{p}$
D

$r_{p} = r_{d} = r_{α}$

Solution

Given that $K_{p} = K_{d} = K_{α} = K$ (say). We know that
$\begin{matrix} m_{p} = m, m_{d} = 2 m and m_{α} = 4 m and q_{p} = e, \\ q_{d} = e and q_{α} = 2 e \\ Further, r = \frac{\sqrt{2 mK}}{qB} \Rightarrow r_{p} = \frac{\sqrt{2 mK}}{eB} \\ r_{d} = \frac{\sqrt{2 (2 m) K}}{eB} = \sqrt{2} r_{p} \\ and r_{α} = \frac{\sqrt{2 (4 m) K_{1}}}{(2 e) B} = r_{p} . Hence, r_{α} = r_{p} < r_{d} \end{matrix}$

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