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JEE Main 2021 · Physics · Electrostatics · medium

A particle of mass m and charge q enters a region of uniform magnetic field B (perpendicular to its velocity) normal to the plane boundary of the…

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A particle of mass mm and charge qq enters a region of uniform magnetic field BB (perpendicular to its velocity) normal to the plane boundary of the region, with speed vv. The region has width d=mv2qBd=\dfrac{mv}{2qB}. The time the particle spends inside the field is
  1. A. πm3qB\dfrac{\pi m}{3qB}
  2. B. πm6qB\dfrac{\pi m}{6qB}
  3. C. πm2qB\dfrac{\pi m}{2qB}
  4. D. 5πm6qB\dfrac{5\pi m}{6qB}
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Answer: (B) πm6qB\dfrac{\pi m}{6qB}

The radius of the circular path is r=mvqBr=\dfrac{mv}{qB}, so d=r/2d=r/2. The particle enters along the normal; it emerges through the opposite boundary when the horizontal displacement is dd, i.e. rsin⁡θ=d⇒sin⁡θ=12r\sin\theta=d\Rightarrow \sin\theta=\dfrac12, θ=30∘=π6\theta=30^\circ=\dfrac\pi6. Time =θω=π6⋅mqB=\dfrac\theta\omega=\dfrac{\pi}{6}\cdot\dfrac{m}{qB}.
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Original question written in the JEE Main 2021 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake