Practice question
Question
A proton moves at \( 4 \times 10^7 \, \text{m/s} \) perpendicular to a field of \( 0.05 \, \text{T} \).
What is the radius of its path? (Mass = \( 1.67 \times 10^{-27} \, \text{kg} \), charge = \( 1.6 \times
10^{-19} \, \text{C} \))
Explanation
**Field outside long solenoid** considered negligible because magnetic lines are concentrated inside and return path spreads over large area outside, making B_out ≈ 0. This justifies using solenoid for uniform field experiments, with n = 950 m⁻¹, I = 1.4 A giving B = 4π×10⁻⁷×950×1.4 = 1.67×10⁻³ T. r = (mv/qB) . r = (1.67 × 10⁻²⁷ × 4 × 10⁷/1.6 × 10⁻¹⁹ × 0.05) = (6.68 × 10⁻²⁰/8 × 10⁻²¹) = 8.35 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 A B
Discussion
Comments
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.