← Back to course

The Biot–Savart Law

You’ll be able to

Currents are the sources of magnetic fields

Just as dq was the point source of electric fields, the point source of magnetic fields is a current element I dl. The Biot–Savart law gives its contribution: dB = (μ₀/4π)·I dl × r̂ / r². The cross product means dB is perpendicular to both the current element and the line to the field point — magnetic field lines wrap around currents rather than radiating from them. The constant μ₀ = 4π × 10⁻⁷ T·m/A is the permeability of free space; the deliberate 4π makes later formulas clean.

Biot–Savart law
d→B = (μ₀/4π) · I d→l × r̂ / r², μ₀ = 4π × 10⁻⁷ T·m/A
Superpose by integrating over the whole current path. |dl × r̂| = dl·sinφ, where φ is the angle between the element and the line to the point.

The same integration playbook as Unit 1

Building B from dB follows the strategy you used for electric fields of continuous charge: slice the source, write the contribution of one slice, use symmetry to identify which component survives, and integrate the surviving scalar. The new wrinkle is the cross product — before integrating anything, establish the direction of dB with the right-hand rule and check whether all elements agree (straight wire: they do) or partially cancel (ring: off-axis parts cancel in pairs).

Worked example

A circular ring of radius R carries current I. Derive the magnetic field on the ring’s axis at distance x from its center.

  1. 1.Every element I dl sits a distance r = √(x² + R²) from the axial point, and dl is perpendicular to r̂, so |dl × r̂| = dl and dB = (μ₀/4π)·I dl/(x² + R²).
  2. 2.Each dB is perpendicular to r̂, tilted off the axis. By symmetry, the components perpendicular to the axis cancel between diametrically opposite elements; only the axial component survives: dB_x = dB·cosα with cosα = R/√(x² + R²).
  3. 3.So dB_x = (μ₀I/4π)·R·dl/(x² + R²)^(3/2). Everything except dl is constant around the ring.
  4. 4.Integrate: ∮ dl = 2πR, giving B = (μ₀I/4π)·R·(2πR)/(x² + R²)^(3/2) = μ₀IR²/[2(x² + R²)^(3/2)].
  5. 5.Check the limits: at the center (x = 0), B = μ₀I/(2R). Far away (x ≫ R), B ≈ μ₀IR²/(2x³) — a 1/x³ falloff, the signature of a magnetic dipole, since a current loop has no magnetic “monopole” to give 1/x².
Answer: B = μ₀IR² / [2(x² + R²)^(3/2)] along the axis; at the center, B = μ₀I/(2R)
Worked example

Derive the field a perpendicular distance d from an infinitely long straight wire carrying current I, then evaluate it for I = 10 A at d = 2.0 cm.

  1. 1.Put the wire on the x-axis with the field point P at distance d. An element I dx at position x is a distance r = √(x² + d²) from P, and the angle between the element and r̂ gives |dl × r̂| = dx·sinφ = dx·(d/r).
  2. 2.So dB = (μ₀I/4π)·d·dx/(x² + d²)^(3/2). Every element’s dB points the same way at P (wrapping around the wire), so the magnitudes simply add.
  3. 3.B = (μ₀I·d/4π) ∫₋∞⁺∞ dx/(x² + d²)^(3/2) — the same standard integral as the infinite line of charge in Unit 1: ∫ dx/(x² + d²)^(3/2) = x/[d²√(x² + d²)], which evaluates to 2/d².
  4. 4.Therefore B = (μ₀I·d/4π)(2/d²) = μ₀I/(2πd) — falling off as 1/d, just as the line charge’s E field did.
  5. 5.Numbers: B = (4π × 10⁻⁷)(10)/(2π × 0.020) = (2 × 10⁻⁷)(10)/(0.020) = 1.0 × 10⁻⁴ T.
Answer: B = μ₀I/(2πd); for 10 A at 2.0 cm, B = 1.0 × 10⁻⁴ T, circling the wire
Key Biot–Savart results
Straight wire: B = μ₀I/(2πd) Center of ring: B = μ₀I/(2R) On axis: B = μ₀IR²/[2(x² + R²)^(3/2)]
Wire field lines are circles around the wire: point the right thumb along I and the fingers curl in the direction of B.
Tip

Test every derived field in a limit you can check. The ring’s field must reduce to μ₀I/(2R) at the center and fall as 1/x³ far away; the wire’s must fall as 1/d. If your integral produces a 1/d² wire field, you dropped the sinφ factor from the cross product.

Checkpoint

At the center of a circular current loop of radius R, the magnetic field magnitude is:

Checkpoint

The field of a long straight wire is B at distance d. At distance 2d the field is:

Answer the 2 checkpoints as you read.

Sign in to save your progress