Question 113 of 276

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 ________ S. $(\pi=3.14)$

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Answer: 0

Angle between $\overrightarrow{\mathrm{V}}$ of charge $\& \overrightarrow{\mathrm{~B}}$ is $90^{\circ}$ motion will be uniform circular motion time period is given by
$\mathrm{T}=\frac{2 \pi \mathrm{~m}}{\mathrm{qB}}=\frac{2 \pi \times 16 \times 10^{-9} \mathrm{~kg}}{1.6 \times 10^{-6} \times 6.28}$
$\mathrm{T}=0.01$ seconds
NTA Answer is 10
Correct Answer is 0 (nearest integer)