Q62 / 90 · JEE Main 2024 (09 Apr Shift 2) · Physics
A proton and a deutron ( $q=+\mathrm{e}, m=2.0 \mathrm{u})$ having same kinetic energies enter a region of uniform magnetic field $\vec{B}$, moving perpendicular to $\vec{B}$. The ratio of the radius $r_d$ of deutron path to the radius $r_p$ of the proton path is:
- Correct answerA$\sqrt{2}: 1$
- B$1: 1$
- C$1: \sqrt{2}$
- D$1: 2$
Solution
In uniform magnetic field,
$\mathrm{R}=\frac{\mathrm{m} \nu}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{~m}(\mathrm{~K} . \mathrm{E})}}{\mathrm{qB}}$
Since same K.E
$\begin{aligned}& \mathrm{R} \propto \frac{\sqrt{\mathrm{m}}}{\mathrm{q}} \\& \therefore \frac{\mathrm{R}_{\text {deutron }}}{\mathrm{R}_{\text {proton }}}=\sqrt{\frac{\mathrm{m}_{\mathrm{d}}}{\mathrm{m}_{\mathrm{p}}}} \times \frac{\mathrm{q}_{\mathrm{p}}}{\mathrm{q}_{\mathrm{d}}} \\& =\sqrt{2} \times 1 \\& \therefore \gamma_{\mathrm{d}}: \gamma_{\mathrm{p}}=\sqrt{2}: 1\end{aligned}$
Hence, the correct answer is option: 'A'.