A particle of charge 1.6 \mu C and mass 16 \mu ~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 \underline\hspace2cm s. (\pi=3.14)

Circular Motion PYQ — JEE Main 2025 · 4 Apr · Shift 2 · Q74 with Solution

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Correct answer: 0.01

When a charged particle is fired perpendicular to a uniform magnetic field, it moves in a circular path.

The angular frequency is

\omega = \frac{qB}{m}

So the time period is

T = \frac{2\pi}{\omega} = \frac{2\pi m}{qB}

This is the time taken to return to the original location for the first time.

q = 1.6\,\mu C = 1.6 \times 10^{-6}\,C

m = 16\,\mu g = 16 \times 10^{-9}\,kg

because $1\,\mu g = 10^{-9}\,kg$ .

B = 6.28\,T

\pi = 3.14

T = \frac{2\pi m}{qB}

T = \frac{2 \times 3.14 \times 16 \times 10^{-9}}{(1.6 \times 10^{-6})(6.28)}

Now simplify numerator and denominator:

Numerator:

2 \times 3.14 \times 16 \times 10^{-9} = 100.48 \times 10^{-9}

Denominator:

1.6 \times 6.28 \times 10^{-6} = 10.048 \times 10^{-6}

Therefore,

T = \frac{100.48 \times 10^{-9}}{10.048 \times 10^{-6}}

T = 10 \times 10^{-3} = 10^{-2}\,s

T = 0.01\,s

The particle returns to its original location for the first time after

\boxed{0.01\,s}

If the answer has to be entered as an integer, this is not an integer-valued result. Numerically it is $0.01$ s, not $0$ s.