Question

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 $)

Hint:

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.

Answer 1

Correct Answer: 0.1

The time period \( T \) for a charged particle moving in a magnetic field is given by the formula: \[ T = \frac{2\pi m}{qB} \] where:
- \( m = 16 \times 10^{-6} \, \text{kg} \),
- \( q = 1.6 \times 10^{-6} \, \text{C} \),
- \( B = 6.28 \, \text{T} \). Substitute the values: \[ T = \frac{2\pi \times 16 \times 10^{-6}}{1.6 \times 10^{-6} \times 6.28} \] \[ T = \frac{2\pi \times 16}{1.6 \times 6.28} = 0.1 \, \text{s} \]
Thus, the time required for the particle to return to its original position is 0.1 seconds.