Skip to content

Question

A proton moves at \( 3.5 \times 10^7 \, \text{m/s} \) perpendicular to a field of \( 0.08 \, \text{T}
\). What is the radius of its path? (Mass = \( 1.67 \times 10^{-27} \, \text{kg} \), charge = \( 1.6
\times 10^{-19} \, \text{C} \))

Options

Choose one · Correct answer highlighted

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. r = (mv/qB) . r = (1.67 × 10⁻²⁷ × 3.5 × 10⁷/1.6 × 10⁻¹⁹ × 0.08) = (5.845 × 10⁻²⁰/1.28 × 10⁻²⁰) = 4.5664 ≈ 4.57 m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ = N I

Discussion

Comments

0 comments

No comments yet. Be the first to start the discussion.