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JEE Main 2025 · Physics · Magnetic Effects of Current · medium

A particle of charge q and mass m enters, with speed v perpendicular to its boundary, a region of width d containing a uniform magnetic field…

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A particle of charge qq and mass mm enters, with speed vv perpendicular to its boundary, a region of width dd containing a uniform magnetic field perpendicular to its velocity. It leaves the region through the opposite boundary with its velocity turned by 30∘30^\circ from the original direction. The time spent by the particle in the field is
  1. A. πd6v\dfrac{\pi d}{6v}
  2. B. 2πd3v\dfrac{2\pi d}{3v}
  3. C. dv\dfrac{d}{v}
  4. D. πd3v\dfrac{\pi d}{3v}
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Answer: (D) πd3v\dfrac{\pi d}{3v}

The particle moves on a circular arc of radius RR. Having entered normally, the width of the region equals Rsin⁡θR\sin\theta for deflection θ\theta: d=Rsin⁡30∘d = R\sin30^\circ, so R=2dR = 2d. The arc subtends θ=π/6\theta = \pi/6 at the centre, so the path length is Rθ=2d⋅π/6=πd/3R\theta = 2d\cdot\pi/6 = \pi d/3. The speed is constant, so t=πd3vt = \dfrac{\pi d}{3v}. (Equivalently B=mv/2qdB = mv/2qd and t=θm/qBt = \theta m/qB.) The value d/vd/v would be the time along a straight path.
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