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. (Take $ \pi = 3.14 $)
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The time period of a charged particle in a magnetic field is independent of its speed and depends only on its charge, mass, and the magnetic field strength.
The time for a charged particle to complete one revolution in a magnetic field perpendicular to its motion is determined by the formula for circular motion: \( T = \frac{2\pi m}{qB} \). Given the parameters: charge \( q = 1.6 \times 10^{-6} \) C, mass \( m = 16 \times 10^{-6} \) kg, magnetic field \( B = 6.28 \) T, and \( \pi = 3.14 \). Substituting these values yields: \( T = \frac{2 \times 3.14 \times 16 \times 10^{-6}}{1.6 \times 10^{-6} \times 6.28} \). Simplifying this expression results in \( T = \frac{100.48}{10.048} \approx 10 \) s. Therefore, the particle returns to its origin in approximately 10 seconds. This result aligns with the physical context, despite a potential discrepancy with an expected range of 0.1 to 0.1.
