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JEE Main 2023 · Physics · Magnetic Effects of Current · medium

A proton and an α-particle, both starting from rest, are accelerated through the same potential difference and enter a region of uniform magnetic…

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A proton and an α\alpha-particle, both starting from rest, are accelerated through the same potential difference and enter a region of uniform magnetic field, perpendicular to their velocity, that extends over a width dd. The proton leaves the region deflected by exactly 30∘30^\circ from its original direction. The sine of the deflection angle of the α\alpha-particle is
  1. A. 14\dfrac14
  2. B. 24\dfrac{\sqrt2}{4}
  3. C. 123\dfrac{1}{2\sqrt3}
  4. D. 34\dfrac{\sqrt3}{4}
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Answer: (B) 24\dfrac{\sqrt2}{4}

Radius r=mvqB=2mqVqB∝m/qr=\dfrac{mv}{qB}=\dfrac{\sqrt{2mqV}}{qB}\propto\sqrt{m/q}. For the α\alpha-particle (m→4mm\to4m, q→2qq\to2q): rα/rp=4/2=2r_\alpha/r_p=\sqrt{4/2}=\sqrt2. A particle entering perpendicular to the boundary and deflected by θ\theta in a slab of width dd obeys sin⁡θ=d/r\sin\theta=d/r. For the proton, sin⁡30∘=12=d/rp\sin30^\circ=\tfrac12=d/r_p, so rp=2dr_p=2d and rα=22 dr_\alpha=2\sqrt2\,d. Then sin⁡θα=d22 d=24≈0.354\sin\theta_\alpha=\dfrac{d}{2\sqrt2\,d}=\dfrac{\sqrt2}{4}\approx0.354 (about 20.7∘20.7^\circ).
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