• question_answer A proton, a deuteron and an $\alpha$ particle having the same kinetic energy are moving in circular trajectories in a constant magnetic field. If ${{r}_{p}},{{r}_{d}}$and${{r}_{a}}$denote respectively the radii of the trajectories of these particles, then A) ${{r}_{a}}={{r}_{P}}<{{r}_{d}}$                    B) ${{r}_{a}}>{{r}_{d}}<{{r}_{P}}$        C)   ${{r}_{a}}={{r}_{d}}>{{r}_{P}}$        D)   ${{r}_{P}}={{r}_{d}}<{{r}_{a}}$

Given that ${{K}_{P}}={{K}_{d}}={{K}_{a}}=K$       (say) We know that ${{q}_{P}}=e,\,{{q}_{d}}=e$and ${{q}_{a}}=2e$and ${{m}_{P}}=m,$${{m}_{d}}=2\,m$and ${{m}_{a}}=4\,m$ Further, $r=\frac{mv}{qB}=\frac{\sqrt{2mK}}{qB}$ Hence, ${{r}_{p}}=\frac{\sqrt{2mK}}{eB}$ ${{r}_{d}}=\frac{\sqrt{2(2m)K}}{eB}=\sqrt{2}\,{{r}_{P}}$ and ${{r}_{a}}=\sqrt{\frac{2(4m)K}{(2e)B}}={{r}_{P}}$ Hence, the correction option is (a).