An electron moves in a circle in a uniform magnetic field of 1.0 × 10⁻³ T. Take mₑ = 9.11 × 10⁻³¹ kg, e = 1.6 × 10⁻¹⁹ C. What is its period of revolution?
Correct answer C. 3.6 × 10⁻⁸ s
Explanation
T = 2πm/(qB) = (2π × 9.11 × 10⁻³¹) ÷ (1.6 × 10⁻¹⁹ × 1.0 × 10⁻³) = 3.6 × 10⁻⁸ s. Leaving out 2π gives 5.7 × 10⁻⁹ s.
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