An insulated wire is wound into a flat spiral coil of N = 200 turns. The radius of the innermost turn is 3 cm and that of the outermost turn is 6 cm. The turns may be treated as uniformly distributed in radius. If the current is 20 mA, the magnetic dipole moment of the coil is α×10⁻² A·m². What is α?
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Question

An insulated wire is wound into a flat spiral coil of N = 200 turns. The radius of the innermost turn is 3 cm and that of the outermost turn is 6 cm. The turns may be treated as uniformly distributed in radius. If the current is 20 mA, the magnetic dipole moment of the coil is α×10⁻² A·m². What is α?

Options

G1 Physics · Accepted Answer

For a turn of radius r, the magnetic moment is dM = Iπr² dN. Because the turns are uniformly distributed in radius, dN = [N ÷ (r₂-r₁)]dr. Therefore, M = IπN ÷ (r₂-r₁) ∫ᵣ₁ʳ² r²dr. Thus, M = IπN(r₂³-r₁³) ÷ [3(r₂-r₁)]. Substitute: I = 20×10⁻³ A, N = 200, r₁ = 3×10⁻² m, r₂ = 6×10⁻² m. This gives M ≈ 2.64×10⁻² A·m². Hence, α = 2.64. Correct option: B.