Practice question
Question
A circular coil of radius \( 0.04 \, \text{m} \) with 50 turns carries \( 1.8 \, \text{A} \). What is
the magnetic field at the center? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))
Explanation
**Biot-Savart law** dB = μ₀/4π·I dl × r̂/r² underlies both straight wire and loop formulas. For square loop side a, area A = a², moment m = N I A, field pattern similar to dipole at large distances, with superposition for N turns. B = (μ₀ N I/2 R) . B = (4 π × 10⁻⁷ × 50 × 1.8/2 × 0.04) = (36 π × 10⁻⁶/0.08) = 4.5 π × 10⁻⁴ ≈ 1.41 × 10⁻³ T . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ = N
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