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. $(\pi=3.14)$

<p>To solve this problem, note that when a charged particle with mass $m$ and charge $q$ is fired perpendicular to a magnetic field of strength $B$, it undergoes uniform circular motion. The period (time to complete one full circle) is given by:</p> <p>$T = \frac{2\pi m}{qB}$</p> <p>Here's how to calculate it step by step:</p> <p>Convert the given quantities to SI units:</p> <p><p>Charge: $q = 1.6\,\mu\text{C} = 1.6 \times 10^{-6}\,\text{C}$</p></p> <p><p>Mass: $m = 16\,\mu\text{g} = 16 \times 10^{-9}\,\text{kg}$</p></p> <p><p>Magnetic field: $B = 6.28\,\text{T}$</p></p> <p><p>Write down the period formula:</p> <p>$T = \frac{2\pi m}{qB}$</p></p> <p><p>Substitute the values:</p> <p>$$ T = \frac{2\pi \times (16 \times 10^{-9}\,\text{kg})}{(1.6 \times 10^{-6}\,\text{C})(6.28\,\text{T})} $$</p></p> <p><p>Notice that $2\pi \approx 6.28$, which cancels with the given magnetic field value, simplifying the expression:</p> <p>$$ T = \frac{6.28 \times 16 \times 10^{-9}}{1.6 \times 10^{-6} \times 6.28} = \frac{16 \times 10^{-9}}{1.6 \times 10^{-6}} $$</p></p> <p><p>Simplify the fraction:</p> <p>$$ T = \frac{16}{1.6} \times \frac{10^{-9}}{10^{-6}} = 10 \times 10^{-3} = 0.01\,\text{s} $$</p></p> <p>Thus, the time required for the particle to return to its original location is:</p> <p>$\boxed{0.01\,\text{s}}$</p>