Magnetic Effects & Magnetism

JEE Main level, single correct, medium

119 of 276

Question 119 of 276

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

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Answer: (A) 2:1\sqrt{2}: 1

In uniform magnetic field,
R=mνqB=2 m( K.E)qB\mathrm{R}=\frac{\mathrm{m} \nu}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{~m}(\mathrm{~K} . \mathrm{E})}}{\mathrm{qB}}
Since same K.E

R∝mq∴Rdeutron Rproton =mdmp×qpqd=2×1∴γd:γp=2:1\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}

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