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Physics · Moving Charges and Magnetic Field

JEE Main 2026 — 8 April, Evening Shift — Question 19

A 5mg5\mathrm{mg} particle carrying a charge of 5π×10−6C5\pi \times10^{-6}\mathrm{C} is moving with velocity of (3i^+2k^)×10−2m/s(3\hat{\mathrm{i}} + 2\hat{\mathrm{k}})\times10^{-2}\mathrm{m/s} in a region having magnetic field B⃗=0.1k^ Wb/m2\vec{\mathrm{B}} = 0.1\hat{\mathrm{k}}\,\mathrm{Wb/m}^2. It moves a distance of α\alpha meter along k^\hat{\mathbf{k}} when it completes 5 revolutions. The value of α\alpha is

Answer: 2

Numerical answer — enter this value.

Step-by-step solution

Pitch p=v∥T=v∥⋅2πmqBp = v_\parallel T = v_\parallel \cdot \frac{2\pi m}{qB}. Here v∥=2×10−2v_\parallel = 2\times10^{-2} m/s, m=5×10−6m=5\times10^{-6} kg, q=5π×10−6q=5\pi\times10^{-6} C, B=0.1B=0.1 T. Then p=0.02×2π×5×10−65π×10−6×0.1=0.02×20.1=0.4p = 0.02 \times \frac{2\pi\times5\times10^{-6}}{5\pi\times10^{-6}\times0.1} = 0.02 \times \frac{2}{0.1} = 0.4 m. For 5 revolutions, distance = 5p=25p = 2 m.

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Exam
JEE Main 2026
Paper
8 April, Evening Shift
Subject
Physics
Chapter
Moving Charges and Magnetic Field
Topic
Motion of Charged Particles in magnetic Fields

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