JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
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 \(\frac{1}{\sqrt{2}}\)
Answer & Solution
Correct Answer
(A) \(\sqrt{2}\)
Step-by-step Solution
Detailed explanation
\({I}=\frac{{m} v}{{qB}}=\frac{\sqrt{2 {mk}}}{{qB}}\) \(\frac{{r}_{{d}}}{{r}_{\alpha}}=\sqrt{\frac{{m}_{{d}}}{{m}_{\alpha}}} \frac{{q}_{a}}{{q}_{\hat{\imath}}}=\sqrt{\frac{2}{4}}\left(\frac{2}{1}\right)=\sqrt{2}\)
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