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JEE Main 2025 (4 Apr, Shift 2) · Moving Charge in a Magnetic Field

Moving Charge in a Magnetic Field question from JEE Main 2025 (4 Apr, Shift 2)

Q17JEE Main 2025 · 4 APR · SHIFT 2Easy

Moving Charge in a Magnetic Field

A particle of charge 1.6 μC1.6\ \mu\text{C} and mass 16 μg16\ \mu\text{g} is present in a strong magnetic field of 6.28 T. The particle is then fired perpendicular to magnetic field. The time required for the particle to return to original location for the first time is _________ s. (π=3.14)(\pi = 3.14)

Numerical answer type. Work it out, then open the answer.

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Answer: 0

  1. The particle returns to its starting point after one full revolution: T=2πmqBT = \dfrac{2\pi m}{qB}
  2. m=16 μg=16×10−9 kgm = 16\ \mu\text{g} = 16\times10^{-9}\text{ kg}, q=1.6×10−6 Cq = 1.6\times10^{-6}\text{ C}
  3. T=2×3.14×16×10−91.6×10−6×6.28=10−2 sT = \dfrac{2\times3.14\times16\times10^{-9}}{1.6\times10^{-6}\times6.28} = 10^{-2}\text{ s}
  4. So T=0.01 sT = 0.01\text{ s}, which rounds to 0 as an integer answer.
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