Practice question
Question
A proton moves at \( 6 \times 10^7 \, \text{m/s} \) perpendicular to a field of \( 0.1 \, \text{T} \).
What is the magnetic force? (Charge = \( 1.6 \times 10^{-19} \, \text{C} \))
Explanation
**Lorentz force** on charge q moving with velocity v in magnetic field B is F = q v × B, magnitude F = q v B sinθ, θ angle between v and B (degrees), unit N. Direction perpendicular to both v and B via right-hand rule. When v ⊥ B, motion circular with radius r = m v/(q B), centripetal force provided by magnetic force. Force F = q v B sin θ , θ = 90° , so sin θ = 1 . F = 1.6 × 10⁻¹⁹ × 6 × 10⁷ × 0.1 = 9.6 × 10⁻¹³ N . Using F = q
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