4.Moving Charges and Magnetism
hard

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 $\vec{B}$, 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:

A

$1: 1$

B

$1: \sqrt{2}$

C

$\sqrt{2}: 1$

D

$1: 2$

(JEE MAIN-2024)

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 $K.E$

$\mathrm{R} \propto \frac{\sqrt{\mathrm{m}}}{\mathrm{q}}$

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

$=\sqrt{2} \times 1$

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

Standard 12
Physics

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