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JEE Advanced 2016 Paper 2 · Physics · Magnetic Effects of Current · medium

A particle of mass m and charge q enters, perpendicular to its boundary, a region of width d in which a uniform magnetic field B acts perpendicular…

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A particle of mass mm and charge qq enters, perpendicular to its boundary, a region of width dd in which a uniform magnetic field BB acts perpendicular to the velocity. The speed is vv and the radius of its circular path inside the field is R=mv/(qB)R=mv/(qB). Choose the correct statement(s).
  1. A. If R=2dR=2d, the particle leaves the far boundary with its velocity deviated by 30∘30^\circ.
  2. B. If R=d/2R=d/2, the particle returns to the entry boundary at a distance dd from the entry point after a time πm/(qB)\pi m/(qB).
  3. C. If R=2dR=2d, the time spent inside the field is one sixth of the period of circular motion.
  4. D. The particle can emerge through the far boundary only if v>qBd/mv>qBd/m.
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Answer: A, B, D

Inside the field the particle moves on a circle whose centre lies on the entry boundary. It reaches the far boundary when the horizontal extent satisfies Rsin⁡θ=dR\sin\theta=d, where θ\theta is the angle turned. For R=2dR=2d, sin⁡θ=1/2\sin\theta=1/2, so θ=30∘\theta=30^\circ and the deviation is 30∘30^\circ; A is true. The time is (30/360)T=T/12(30/360)T=T/12, not T/6T/6; C is false. For R=d/2R=d/2 the particle completes a semicircle, returning to the entry boundary a distance 2R=d2R=d away after T/2=πm/(qB)T/2=\pi m/(qB); B is true. Crossing needs R>dR>d, i.e. mv/(qB)>dmv/(qB)>d, so v>qBd/mv>qBd/m; D is true.
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