JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
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
- A \(1: \sqrt 2\)
- B \(1 : 2\)
- C \(1 : 3\)
- D \(1: \sqrt 3\)
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,…
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