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Question

A proton moves at \( 7.5 \times 10^7 \, \text{m/s} \) perpendicular to a field of \( 0.12 \, \text{T}
\). What is the magnetic force? (Charge = \( 1.6 \times 10^{-19} \, \text{C} \))

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Explanation

**Motion of charged particle perpendicular to uniform B** is circular because force F ⊥ v, no work done, speed constant, radius r = m v/(q B), period T = 2π m/(q B) independent of v. If v has component parallel to B, helical path results, pitch = v_parallel·T. Force F = q v B sin θ , θ = 90° , so sin θ = 1 . F = 1.6 × 10⁻¹⁹ × 7.5 × 10⁷ × 0.12 = 1.44 × 10⁻¹² N . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀

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