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Field of finite wire on bisector

Short answer

Integrating the Biot–Savart law for a straight wire segment of length l along its perpendicular bisector gives B = (μ₀I/2πy) × l/√(l²+4y²). This reduces to μ₀I/(2πy) for an infinite wire.

A straight wire segment of length l carries current I. Show that the magnetic field B produced by that segment at a distance y from it along a perpendicular bisector is B = (μ0I/2πy)[l/√(l² + 4y²)].

The result to prove is B = (μ₀I/2πy) · l/√(l² + 4y²). It reduces to μ₀I/(2πy) for an infinite wire.

Derivation

Straight wire of length l, centre at O. Point P is at distance y from O along the perpendicular bisector.

        P
        |\
      y | \ r
        |  \
   -----+---o--->  x
   -l/2 O  x   +l/2

  1. Take an element dx at position x. Its distance from P is r = √(x² + y²). The angle φ between dx and r satisfies sin φ = y/r.

  2. Biot-Savart: dB = (μ₀I/4π) dx sin φ / r² = (μ₀I/4π) y dx/(x² + y²)^(3/2).

  3. All elements give B in the same direction (perpendicular to the plane), so integrate from −l/2 to +l/2. Using ∫dx/(x²+y²)^(3/2) = x/[y²√(x²+y²)]:

    B = (μ₀I y/4π) · [x/(y²√(x²+y²))] from −l/2 to l/2 = (μ₀I/4πy) · 2(l/2)/√(l²/4 + y²) = (μ₀I/4πy) · 2l/√(l² + 4y²).

  4. Simplify:

B = (μ₀ I / 2πy) · l/√(l² + 4y²)

Check: for l ≫ y, l/√(l²+4y²) → 1 and B → μ₀I/(2πy), the field of an infinite wire.

✓ Answer checked by Claude Opus 5.5 (high effort)

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