Practice question
Question
A circular coil of radius \( 0.11 \, \text{m} \) with 25 turns carries \( 3 \, \text{A} \). 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 × 25 × 3/2 × 0.11) = (30 π × 10⁻⁶/0.22) = 1.3636 π × 10⁻⁴ ≈ 4.28 × 10⁻⁴ 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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