Applications of Biot-Savart's Law > Magnetic Field Due to a Straight Current-carrying Conductor of Finite Size

Imagine a straight wire carrying current, like a long charging cable. At any point near the wire, a magnetic field is created — and this field weakens as you move away from the wire, similar to ripples spreading out from a stone dropped in water. Biot–Savart's Law lets us calculate the exact strength of this field at any point, based on the current, distance, and the geometry of the wire.

Setup: A straight conductor carries current I. Point P lies at perpendicular distance r from the wire. The two ends of the wire subtend angles ϕ₁​ and ϕ_2​ with the perpendicular drawn from P.

Derivation Steps

1. Consider a small current element Idl on the wire.
2. By Biot–Savart's Law, the field due to this element at P is dB = ​\[\frac{\mu_0}{4\pi}\frac{Idl\sin\theta}{r^2}\].
3. Integrate this contribution over the entire length of the wire, from one end (angle ϕ₁​) to the other (angle ϕ₂​).
4. All elemental fields at P point in the same direction (perpendicular to the plane containing the wire and P), so they add up directly.
5. The integration yields the net field at P.

Key Formula

Direction: Perpendicular to the plane containing the wire and point P, found using the right-hand palm rule.

Condition: Wire extends to infinity on both sides; P lies near the middle, so ϕ₁ = ϕ₂ = 90°.

Real-life example: This is the formula used to estimate the magnetic field around long transmission cables or lab wires that are much longer than the distance from the point of observation.

Condition: P lies near one end of the wire; the wire extends to infinity in only one direction, so ϕ₁ = 90°, ϕ₂ = 0°.

Note: This is exactly half the field of an infinite wire — a frequently tested conceptual point.

Condition: Wire has finite length l; point P lies on its perpendicular bisector, so ϕ₁ = ϕ₂.

Check: As l → ∞, this formula correctly reduces to the infinite wire case B = \[\frac {μ_0I}{2πr}\]​.

• Biot–Savart's Law gives the field due to a current element; integrating over a finite wire gives B = \[\frac {μ_0I}{4πr}\](sin ϕ₁ + sin ϕ₂).
• For an infinite wire, B = \[\frac {μ_0I}{2πr}\]​.
• For a semi-infinite wire, B = \[\frac {μ_0I}{4πr}\] (half of infinite wire).
• For a point on the perpendicular bisector of a finite wire of length l, B = \[\frac{\mu_0I}{2\pi r}\cdot\frac{l}{\sqrt{4r^2+l^2}}\].
• The field is always directly proportional to I and inversely proportional to r.
• Direction is found using the right-hand palm rule.