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Field on axis of a long coil

Short answer

Use B(x)=μ₀NIR²/[2(R²+x²)^(3/2)] with R=0.6 m, I=10 A, N=50. (a) At x=0.9 m, B≈8.9×10⁻⁵ T. (b) Setting B = B₀/8 gives (R²+x²)^(3/2)=8R³ → x=√3 R≈1.04 m.

A long circuit coil consisting of 50 turns with diameter 1.2m carries a current of 10Amp. (a) Find the magnetic field at a point along the axis 90cm from the center. (b) At what distance from the center, along the axis, the field is 1/8 greater as at the center.

For a coil of N turns and radius R, the field on the axis at distance x from the centre is

B(x) = μ₀ N I R² / [2 (R² + x²)^(3/2)], and at the centre B₀ = μ₀NI/(2R).

Given: N = 50, I = 10 A, diameter 1.2 m so R = 0.6 m.

(a) B at x = 0.90 m

R² = 0.36 ;  x² = 0.81 ;  R² + x² = 1.17
(R² + x²)^(3/2) = 1.17 × √1.17 = 1.17 × 1.0817 = 1.2656
B = (4π×10⁻⁷ × 50 × 10 × 0.36) / (2 × 1.2656)
  = 2.262×10⁻⁴ / 2.531
  = 8.94×10⁻⁵ T

At the centre B₀ = (4π×10⁻⁷ × 500)/(2 × 0.6) = 5.24×10⁻⁴ T.

(b) Point where the field is 1/8 of the centre value

Reading of the question: on the axis the field is largest at the centre, so "1/8 greater as at the centre" is read as "1/8 of the centre value", B = B₀/8.

B/B₀ = R³ / (R² + x²)^(3/2) = 1/8
(R² + x²)^(1/2) = 2R   →   R² + x² = 4R²
x = √3 R = 1.732 × 0.6 = 1.04 m

Answer: (a) B ≈ 8.9×10⁻⁵ T (89 μT); (b) x ≈ 1.04 m from the centre.

✓ Answer checked by Claude Opus 5.5 (high effort)

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