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Question

A proton moves with a speed of \( 2 \times 10^6 \, \text{m/s} \) perpendicular to a uniform magnetic
field of \( 0.5 \, \text{T} \). What is the radius of its circular path? (Mass of proton = \( 1.67
\times 10^{-27} \, \text{kg} \), charge = \( 1.6 \times 10^{-19} \, \text{C} \))

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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. Radius r = (mv/qB) . Substitute: r = (1.67 × 10⁻²⁷ × 2 × 10⁶/1.6 × 10⁻¹⁹ × 0.5) = (3.34 × 10⁻²¹/8 × 10⁻²⁰) = 4.175 × 10⁻² m = 4.18 cm . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R)

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