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4.Moving Charges and Magnetism
medium
A deuteron and an alpha particle having equal kinetic energy enter perpendicular into a magnetic field. Let $r_{d}$ and $r_{\alpha}$ be their respective radii of circular path. The value of $\frac{r_{d}}{r_{\alpha}}$ is equal to
A
$\sqrt{2}$
B
$1$
C
$2$
D
$\frac{1}{\sqrt{2}}$
(JEE MAIN-2021)
Solution
${I}=\frac{{m} v}{{qB}}=\frac{\sqrt{2 {mk}}}{{qB}}$
$\frac{{r}_{{d}}}{{r}_{\alpha}}=\sqrt{\frac{{m}_{{d}}}{{m}_{\alpha}}} \frac{{q}_{a}}{{q}_{\hat{\imath}}}=\sqrt{\frac{2}{4}}\left(\frac{2}{1}\right)=\sqrt{2}$
Standard 12
Physics
