Practice question
Question
A long wire carries \( 18 \, \text{A} \). At what distance is the magnetic field \( 6 \times 10^{-6} \,
\text{T} \)? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))
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. B = (μ₀ I/2 π r) , so r = (μ₀ I/2 π B) . r = (4 π × 10⁻⁷ × 18/2 π × 6 × 10⁻⁶) = (72 × 10⁻⁷/12 × 10⁻⁶) = 0.6 m . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B
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