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 ______ s. ($$\pi = 3.14$$)

When a charged particle is fired perpendicular to a uniform magnetic field, it traces a circular path. The time required for the particle to return to its original location for the first time is exactly the time period of this circular motion.

The magnetic force provides the centripetal force: $$qvB = \frac{mv^{2}}{r}$$

This gives the radius $$r = \frac{mv}{qB}$$.

The time period is $$T = \frac{2\pi r}{v} = \frac{2\pi m}{qB}$$

The time period does not depend on the speed of the particle.

Substituting into the time period formula:

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

$$T = \frac{100.48 \times 10^{-6}}{10.048 \times 10^{-6}} = \frac{100.48}{10.048} = 10 \text{ s}$$

So, the answer is $$10$$.