--- title: "Field of finite wire on bisector" subject: "Engineering Physics" chapter: 8 key: "semester_1/engineering_physics:wire-segment-perpendicular-bisector-field" source: https://www.bctengineer.com/q/semester_1/engineering_physics/wire-segment-perpendicular-bisector-field site: BCT Engineering --- # Field of finite wire on bisector Engineering Physics · Chapter 8: Electromagnetism - Asked in: 2069 Asar, 5 marks - Video: [Using Biot-Savart to Find the Magnetic Field from a Finite Wire](https://www.youtube.com/watch?v=BlIEavkDUF8) — Redmond Physics Tutoring - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/engineering_physics/wire-segment-perpendicular-bisector-field - Whole chapter: https://www.bctengineer.com/md/semester_1/engineering_physics/chapter-8 ## Question 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)]. ## 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. ## Answer 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. ```text 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The correct field at distance y on the perpendicular bisector of a segment of length l is: A. B = (μ₀I/2πy) · l/√(l² + 4y²) B. B = (μ₀/2πy) · l/(l² + 4y²), with no current C. B = (μ₀I/2πy) · l/(l² + 4y²) D. B = (μ₀I/4πy) · l/√(l² + y²) Answer: A. The question's printed form lacks I and the square root; the derivation gives the form with √. (Key point: Correct result) 2. For an element dx at distance r from P, dB is: A. (μ₀I/4π) dx/(x² + y²) B. (μ₀I/4π) y dx/(x² + y²)^(3/2) C. (μ₀I/4π) x dx/(x² + y²)^(3/2) D. (μ₀I/2π) y dx/(x² + y²)^(1/2) Answer: B. dB = (μ₀I/4π)dx sin φ/r² with sin φ = y/r. (Key point: Biot-Savart element) 3. Why can the dB contributions be added as magnitudes? A. They point along the wire B. They cancel in pairs C. All elements give B in the same direction, perpendicular to the plane D. They all point toward P Answer: C. Every dl × r̂ points the same way out of the plane. (Key point: Integration) 4. For l much greater than y, the result becomes: A. zero B. μ₀I/(4πy) C. μ₀I l/(2π y²) D. μ₀I/(2πy), the infinite-wire field Answer: D. l/√(l²+4y²) tends to 1. 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