Biot-Savart law

Why it matters

The Biot–Savart law is the magnetic counterpart of Coulomb's law: it gives the magnetic field of any steady current path, symmetric or not. It is how you find the field of a finite PCB trace, a square search coil, a Helmholtz calibration coil or the loop in a current sensor — geometries where Ampère's law cannot be used directly.

Key ideas

The law. A steady current element I dl produces at a point P a field contribution dH = I dl × a_R / (4πR²), where R is the distance from the element to P and a_R is the unit vector from the element to P. In free space B = μ₀H.

Magnitude and direction. |dH| = I dl sin θ / (4πR²), with θ the angle between dl and a_R. The field is perpendicular to both dl and a_R, so it circles the current: right-hand grip rule — thumb along the current, fingers curl along H. Points on the line of a straight current (θ = 0) receive no contribution from it.

Assumptions. Steady (DC or slowly varying, quasi-static) currents, linear medium of uniform permeability, and currents forming closed circuits. For distributed currents, replace I dl by K dS (surface current, A/m) or J dv (volume current, A/m²).

Superposition. Fields from different parts of a circuit add vectorially. Break a circuit into straight segments and arcs, find each contribution with the standard results, and add with directions.

Standard results:

• Finite straight segment: depends on the angles the segment ends subtend at P. Letting the segment become infinite gives B = μ₀I/(2πρ).
• Semi-infinite wire, with P on the perpendicular through its end: half the infinite-wire value.
• Circular arc of angle α (radians) at its centre: B = μ₀Iα/(4πa); a full loop (α = 2π) gives μ₀I/(2a), a semicircle μ₀I/(4a).
• On the axis of a loop: falls off smoothly from the centre value, as 1/z³ far away (a magnetic dipole).

H and B. H (A/m) depends only on the currents; B (T = Wb/m²) = μH also depends on the medium. In air or free space μ = μ₀ = 4π × 10⁻⁷ H/m.

Link to other laws. Biot–Savart is equivalent to Ampère's circuital law for steady currents; it is also what gives ∇·B = 0, since the field of every element circles it and has no source.

Formulas

• dH = I dl × a_R / (4πR²) — Biot–Savart law; H in A/m, I in A, dl and R in m.
• dB = μ₀ I dl × a_R / (4πR²) — in free space; B in T; μ₀ = 4π × 10⁻⁷ H/m.
• H = I (sin α2 − sin α1) / (4πρ) — finite straight wire; ρ = perpendicular distance to P; α1, α2 = angles from the perpendicular foot to the two ends (signed).
• H = I / (2πρ) — infinite straight wire; B = μ₀I / (2πρ).
• H = I / (2a) — centre of a circular loop of radius a; H = Iα / (4πa) — arc of angle α.
• H = I a² / (2(a² + z²)^(3/2)) — on the axis of a loop, distance z from its centre.
• H = 2√2 I / (πL) — centre of a square loop of side L.
• H = N I / (2a) — N-turn flat coil at its centre.

Worked examples

Example 1 (standard). A long straight wire carries 5 A. Find B at 0.1 m.

1. B = μ₀I / (2πρ) = 4π × 10⁻⁷ × 5 / (2π × 0.1).
2. B = 2 × 10⁻⁷ × 5 / 0.1 = 10 μT (H = 7.96 A/m), circling the wire by the right-hand rule. (For scale, the Earth's field is about 50 μT.)

Example 2 (GATE level, square loop). A square loop of side L = 0.2 m carries I = 2 A. Find B at its centre.

1. Each side is a finite segment at perpendicular distance ρ = L/2 = 0.1 m, with ends at α = ±45°.
2. One side: B = μ₀I (sin 45° − sin(−45°)) / (4πρ) = 10⁻⁷ × 2 × (2 × 0.7071) / 0.1 = 2.828 × 10⁻⁶ T.
3. All four sides give fields in the same direction (into the plane for a clockwise current seen from the front), so B = 4 × 2.828 × 10⁻⁶ = 11.3 μT.
4. Check with B = 2√2 μ₀I / (πL) = 2√2 × 4π × 10⁻⁷ × 2 / (π × 0.2) = 11.3 μT ✓. A circular loop of the same "radius" 0.1 m would give μ₀I/(2a) = 12.6 μT — slightly more, because the circle is closer to the centre everywhere.

Example 3 (axis of a loop). A loop of radius a = 0.1 m carries 5 A. Find B at its centre and at z = 0.1 m on the axis.

1. Centre: B = μ₀I / (2a) = 4π × 10⁻⁷ × 5 / 0.2 = 31.4 μT.
2. Axis: B = μ₀Ia² / (2(a² + z²)^(3/2)) = 4π × 10⁻⁷ × 5 × 0.01 / (2 × (0.02)^1.5) = 11.1 μT.
3. Ratio = (a²/(a² + z²))^(3/2) = (0.5)^1.5 = 0.354 ✓.

Common mistakes

• Using the infinite-wire formula for a short wire when P is not far from it compared with its length.
• Reversing a_R (it points from the source element to the field point) and so getting the direction wrong.
• Writing B = μ₀I/(2r) for a straight wire (that is the loop-centre formula) or μ₀I/(2πr) for a loop.
• Adding contributions as magnitudes when some point in opposite directions (e.g. two straight leads of a hairpin, or arcs carrying current in opposite senses).
• Forgetting that a straight lead pointing directly at P contributes zero.
• Mixing H (A/m) and B (T): B = μ₀H in free space, a factor of about 1.26 × 10⁻⁶.

For GATE IN

Common items: field at the centre of composite loops (arcs plus straight leads, semicircles, square or rectangular loops), field of finite segments, on-axis field of a coil, two parallel wires with currents in the same or opposite directions (field at a point between or outside them), and conversion between H and B. Practise the finite-segment formula with correct angle signs.

Quick check

1. What is B at the centre of a 0.1 m radius loop carrying 2 A?
2. What is the field contribution of a straight wire at a point on its own line?
3. How does B at the centre of a loop change if the radius is doubled at the same current?
4. What is B at the centre of a semicircle of radius 5 cm carrying 4 A (straight leads radial)?
5. Far from a small loop on its axis, how does B fall off?

Answers: 1. 12.6 μT. 2. Zero. 3. It halves. 4. 25.1 μT. 5. As 1/z³.