Biot-Savart law and the field on the axis of a circular loop

The Biot-Savart law

Think of a wire as a chain of tiny pieces. A current element is one such piece of length dl carrying current I, written as the vector I dl along the current. The Biot-Savart law gives the field dB it makes at a point P at distance r: dB = (μ₀/4π) I dl sin θ/r². Here θ is the angle between dl and the line r joining the element to P. The field dB is perpendicular to the plane containing dl and r. In vector form, dB = (μ₀/4π) I dl × r/r³; the cross product gives the direction by the right-hand screw rule. The field is zero along the line of the element (θ = 0°) and largest at right angles to it (θ = 90°). The constant μ₀ is the permeability of free space. It measures how readily a vacuum lets a magnetic field form: μ₀ = 4π × 10⁻⁷ T m A⁻¹.

Biot-Savart law vs Coulomb's law

Derivation: field on the axis of a circular loop

A circular loop of radius R and centre O carries a steady current I. P is a point on its axis at distance x from O. Find B at P.

1. 1
   Take a small element dl of the loop. Its distance from P is r, where r² = x² + R².
2. 2
   Every element of the loop is perpendicular to the line joining it to P, so θ = 90° and sin θ = 1.
3. 3
   Biot-Savart law: dB = (μ₀/4π) I dl/(x² + R²). dB is perpendicular to r, in the plane of r and the axis.
4. 4
   Resolve dB into dBx along the axis and dB⊥ perpendicular to the axis.
5. 5★
   The element diametrically opposite gives an equal dB⊥ in the opposite direction. So all the dB⊥ parts cancel in pairs.
6. 6
   Only the axial parts add: dBx = dB cos α, where α is the angle between dB and the axis, and cos α = R/(x² + R²)^(1/2).
7. 7
   So dBx = (μ₀ I dl/4π) × R/(x² + R²)^(3/2).
8. 8
   Add over the whole loop. The sum of all dl is the circumference: ∮ dl = 2πR.
9. 9★
   B = (μ₀ I/4π) × 2πR × R/(x² + R²)^(3/2) = μ₀IR²/[2(x² + R²)^(3/2)].
10. 10
    So, the field on the axis of a circular loop is μ₀IR²/[2(x² + R²)^(3/2)], directed along the axis. For a coil of N turns, multiply by N.

★ key step

Special cases and direction

• At the centre (x = 0): B = μ₀I/2R. For N turns, B = μ₀NI/2R.
• Far away on the axis (x much larger than R): B ≈ μ₀IR²/2x³. With m = IπR², this is B = (μ₀/4π)(2m/x³), the field of a magnetic dipole.
• Direction: curl the fingers of your right hand along the current; the thumb points along the field on the axis.
• Seen from one side, if the current flows anticlockwise, the field on the axis points towards you. That face acts as a north pole. Clockwise means the field points away (a south pole).
• B is largest at the centre and falls off on both sides along the axis.

How to draw the figure for the derivation

• Draw the loop as an ellipse (a circle seen at a slant). Mark the centre O and the radius R.
• Draw the axis as a horizontal line through O. Mark P on it at distance x from O.
• Mark a small element dl at the top of the loop, with an arrow for the current I. Join it to P and label the line r.
• At P, draw dB perpendicular to r. Resolve it into dBx along the axis and dB⊥ upwards.
• Mark the opposite element at the bottom, with its dB⊥ pointing downwards, to show the cancelling. Label the angle α between dB and the axis.

Worked example: field at the centre and on the axis