Angle between V of charge & B is 90∘ motion will be uniform circular motion time period is given by
T=qB2π m=1.6×10−6×6.282π×16×10−9 kg
T=0.01 seconds
NTA Answer is 10
Correct Answer is 0 (nearest integer)
JEE Main 2025 — Physics Electromagnetism
A particle of charge 1.6μC and mass 16μ 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)
Held on 4 Apr 2025 · Verified 6 Jul 2026.
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