Practice question
Question
A proton moves with a speed of \( 1.8 \times 10^6 \, \text{m/s} \) perpendicular to a magnetic field of
\( 0.5 \, \text{T} \). What is the radius of its path? (Mass = \( 1.67 \times 10^{-27} \, \text{kg} \),
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. Radius r = (mv/qB) . r = (1.67 × 10⁻²⁷ × 1.8 × 10⁶/1.6 × 10⁻¹⁹ × 0.5) = (3.006 × 10⁻²¹/8 × 10⁻²⁰) = 3.7575 × 10⁻² m = 3.76 cm . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀
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