Physics · Moving Charges and Magnetic Field

JEE Main 2025 — 4 April, Evening Shift — Question 62

A particle of charge $1.6 \mu \mathrm{C}$ and mass $16 \mu \mathrm{~g}$ 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 $\qquad$ s. $(\pi=3.14)$

Answer: 0.01

Numerical answer — enter this value.

Step-by-step solution

$w=\frac{2 B}{m}$ $T=\frac{2 \pi m}{q B}=\frac{2 \times 3.14 \times 16 \times 10^{-9}}{1.6 \times 10^{-6} \times 6.28}$ $T=0.01 \mathrm{sec}$

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Exam: JEE Main 2025
Paper: 4 April, Evening Shift
Subject: Physics
Chapter: Moving Charges and Magnetic Field
Topic: Motion of Charged Particles in magnetic Fields