Biot–Savart Law
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 4 practice questions.
The Biot–Savart law builds a field one small piece of current at a time: dB = (μ0/4π) I dl × r̂ / r². Add (integrate) the pieces, watching directions. Two results come up again and again: a long straight wire, μ0I/(2πr), and the center of a loop, μ0I/(2R).
1. Building a field piece by piece
2. How to set it up
Direction first: curl your right fingers along the current; your thumb gives B at the center of a loop. For a wire, the thumb follows I and fingers wrap with B.
Use symmetry: on a loop’s axis, the sideways parts of dB cancel in pairs and only the axial parts add.
Superpose: split odd shapes into straight pieces and arcs, find each one’s field, and add them as vectors.
A finite straight segment gives B = (μ0I/4πd)(sin θ2 − sin θ1), with the angles measured from the perpendicular.
3. Example, then try it
A circular loop of radius 5.0 cm carries 2.0 A. Find B at its center, and on its axis 12 cm from the center.
Center: B = μ0I/(2R) = (4π × 10⁻⁷)(2.0) ÷ (0.10) ≈ 2.5 × 10⁻⁵ T.
Axis: R² + z² = 0.0025 + 0.0144 = 0.0169 m², so B = (4π × 10⁻⁷)(2.0)(0.0025) ÷ [2(0.0169)3/2] ≈ 1.4 × 10⁻⁶ T.
The pie-slice loop in the figure below has radius R = 0.20 m, a 90° arc, and carries 5.0 A. Use Biot–Savart to find B at the center C.
Straight segments: along them dl is parallel to r̂, so dl × r̂ = 0. They add nothing.
Arc: every dl is perpendicular to r̂ and at distance R, so dB = μ0I dl/(4πR²), all out of the page.
Add them: ∫dl = Rθ, so B = μ0Iθ/(4πR).
Numbers: B = (10⁻⁷)(5.0)(π/2) ÷ 0.20 ≈ 3.9 × 10⁻⁶ T, out of the page.
Answer: about 3.9 × 10⁻⁶ T, out of the page (one quarter of a full loop’s μ0I/2R).
4. Common mistakes
Far from a current loop along its axis (z ≫ R), how does B depend on z?
For z ≫ R, (R² + z²)3/2 ≈ z³, so B ≈ μ0IR²/(2z³): a magnetic dipole field.
Ready to practice?
4 questions on Biot–Savart Law, each with a hint and an explanation for every wrong answer.