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JEE Main 2019 (12 Jan, Shift 1) · Moving Charge in a Magnetic Field

Moving Charge in a Magnetic Field question from JEE Main 2019 (12 Jan, Shift 1)

Q88JEE Main 2019 · 12 JAN · SHIFT 1Medium

Moving Charge in a Magnetic Field

A proton and an α\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 rp: rα\alpha of the circular paths described by them will be;

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Answer: (c)

  1. After acceleration K=qVK = qV: r=2mqVqB=1B2mVq⇒r∝mqr = \dfrac{\sqrt{2mqV}}{qB} = \dfrac{1}{B}\sqrt{\dfrac{2mV}{q}} \Rightarrow r \propto \sqrt{\dfrac{m}{q}}
  2. rprα=11×24=12\dfrac{r_p}{r_\alpha} = \sqrt{\dfrac{1}{1} \times \dfrac{2}{4}} = \dfrac{1}{\sqrt{2}}
  3. So the answer is (c) 1:21 : \sqrt{2}.
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