JEE Main 2025 Apr 4 Shift 2, Physics Q21: Motion of a Charged Particle in a Magnetic Field

JEE Main2025Apr 4, Shift 2Physics

A particle of charge $1.6 μ C$ and mass $16 μg$ 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 ________ $\times 1 0^{- 3}$ s. ( $π = 3.14$ )

Your answer

Solution

Velocity perpendicular to $B$ produces uniform circular motion. The period is

$T = \frac{2 π m}{q B} = \frac{2 \times 3.14 \times 16 \times 1 0 ^{- 9}}{1.6 \times 1 0 ^{- 6} \times 6.28} = \frac{6.28 \times 16 \times 1 0 ^{- 9}}{1.6 \times 6.28 \times 1 0 ^{- 6}} = \frac{16 \times 1 0 ^{- 9}}{1.6 \times 1 0 ^{- 6}}$

$= 10 \times 1 0^{- 3}$ s.

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JEE Main 2025 Apr 4 Shift 2, Physics Q21: Motion of a Charged Particle in a Magnetic Field

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