PrepRung
Sign inSign up free

JEE Main 2022 · Physics · Magnetic Effects of Current · medium

An electron enters a region of uniform magnetic field B (perpendicular to its velocity) which extends over a width d perpendicular to the initial…

Single correct option

An electron enters a region of uniform magnetic field BB (perpendicular to its velocity) which extends over a width dd perpendicular to the initial velocity. The electron's speed is such that its orbital radius is 2d2d. The time the electron spends inside the field region is
  1. A. πm3eB\dfrac{\pi m}{3eB}
  2. B. πm6eB\dfrac{\pi m}{6eB}
  3. C. πm12eB\dfrac{\pi m}{12eB}
  4. D. 2πm3eB\dfrac{2\pi m}{3eB}
Show answer and solution

Answer: (B) πm6eB\dfrac{\pi m}{6eB}

The electron moves on an arc of radius r=2dr=2d. It leaves the field when it has advanced a distance dd along the original direction: rsin⁡θ=dr\sin\theta=d, so sin⁡θ=12\sin\theta=\frac12 and the deflection angle is θ=30∘=π6\theta=30^\circ=\frac\pi6. Angular speed ω=eBm\omega=\dfrac{eB}{m}, independent of speed. Time =θω=π6⋅meB=πm6eB=\dfrac\theta\omega=\dfrac{\pi}{6}\cdot\dfrac{m}{eB}=\dfrac{\pi m}{6eB}. This is one-twelfth of the full period 2πmeB\dfrac{2\pi m}{eB}.
Report a mistake

Original question written in the JEE Main 2022 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake