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JEE Mains · Physics · STD 12 - Moving charges and magnetism
  • JEE Mains 2019
  • Medium
  • MCQ

A proton and an \(\alpha -\) particle (with their masses in the ratio of \(1 : 4\) and charges in the ratio of \(1:2\) are accelerated from rest through a potential difference \(V\). If a uniform magnetic field \((B)\) is set up perpendicular to their velocities, the ratio of the radii \(r_p : r_{\alpha }\) of the circular paths described by them will be

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Answer & Solution

Correct Answer

(A) \(1: \sqrt 2\)

Step-by-step Solution

Detailed explanation

\({{\text{m}}_p} = {\text{m}}\)               \({{\text{q}}_{\text{p}}} = {\text{q}}\)               \({{\text{k}}_{\text{p}}} = {\text{qV}} = {\text{k}}\)

\({{\text{m}}_\alpha } = {\text{4m}}\)               \({q_\alpha } = 2q\)               \({k_\alpha } = 2qV = 2k\)

Radius of circular path,

\(r=\frac{m v}{q B}=\frac{\sqrt{2 k m}}{q B}\)

\( \Rightarrow \) \(\frac{{{r_p}}}{{{r_\alpha }}} = \frac{1}{{\sqrt 2 }}\)

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