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Question

A circular coil of 80 turns and radius \( 4 \, \text{cm} \) carries a current of \( 0.5 \, \text{A} \).
What is the magnetic field at its center? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

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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. Magnetic field B = (μ₀ N I/2 R) . B = (4 π × 10⁻⁷ × 80 × 0.5/2 × 0.04) = (16 π × 10⁻⁶/0.08) = 2 π × 10⁻⁴ ≈ 6.28 × 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

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