Practice question
Question
An electron moves at \( 6 \times 10^6 \, \text{m/s} \) perpendicular to a field of \( 0.4 \, \text{T}
\). What is the magnetic force? (Charge = \( 1.6 \times 10^{-19} \, \text{C} \))
Explanation
**Effect of doubling velocity** on magnetic force F = q v B sinθ is linear increase, F doubles for same θ and B. Electron with charge 1.6×10⁻¹⁹ C, v = 4.5×10⁶ m/s, B = 0.35 T, θ = 90°, F = 1.6×10⁻¹⁹×4.5×10⁶×0.35 = 2.52×10⁻¹³ N, illustrating magnitude for typical lab values. Force F = q v B sin θ , θ = 90° , so sin θ = 1 . F = 1.6 × 10⁻¹⁹ × 6 × 10⁶ × 0.4 = 3.84 × 10⁻¹³ N . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r),
Discussion
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