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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

  1. A \(\sqrt{2}\)
  2. B \(1\)
  3. C \(2\)
  4. D \(\frac{1}{\sqrt{2}}\)
Verified Solution

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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