A proton and a deutron ( q =+ e , m=2.0 u ) having same kinetic energies enter a region of uniform magnetic field overrightarrow B , moving perpendicular to overrightarrow B . The ratio of the radius r d of deutron path to the radius r p of the proton path is :

Physics · Moving Charges and Magnetic Field

JEE Main 2024 — 9 April, Shift 2 — Question 41

A proton and a deutron $(\mathrm{q}=+\mathrm{e}, m=2.0 \mathrm{u})$ having same kinetic energies enter a region of uniform magnetic field $\overrightarrow{\mathrm{B}}$ , moving perpendicular to $\overrightarrow{\mathrm{B}}$ . The ratio of the radius $r_{d}$ of deutron path to the radius $r_{p}$ of the proton path is :

Option A:
$1: 1$
Option B:
$1: \sqrt{2}$
Option C:
$\sqrt{2}: 1$
Correct
Option D:
$1: 2$

Answer: C

Step-by-step solution

In uniform magnetic field,

$\mathrm{R}=\frac{\mathrm{m} v}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{~m}(\mathrm{~K} \cdot \mathrm{E})}}{\mathrm{qB}}$ Since same K.E $R \propto \frac{\sqrt{m}}{q}$

$\therefore \frac{\mathrm{R}_{\text {deutron }}}{\mathrm{R}_{\text {proton }}}=\sqrt{\frac{\mathrm{m}_{\mathrm{d}}}{\mathrm{m}_{\mathrm{p}}}} \times \frac{\mathrm{q}_{\mathrm{p}}}{\mathrm{q}_{\mathrm{d}}}$

$=\sqrt{2} \times 1$

$\therefore \gamma_{\mathrm{d}}: \gamma_{\mathrm{p}}=\sqrt{2}: 1$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 9 April, Shift 2
Subject: Physics
Chapter: Moving Charges and Magnetic Field
Topic: Motion of Charged Particles in magnetic Fields