Field of a circular current loop

Adding up round a loop

Narrated video. Transcript:

Take a circular loop of radius R carrying a steady current I. Let us find the magnetic field at a point P on its axis, a distance x from the centre.

Pick a small element dl at the top of the loop. The line from dl to P has length r, and r² equals x² plus R².

dl is along the loop, at right angles to this line. So sin θ is 1, and Biot–Savart gives dB equals μ₀ over 4π, times I dl, over x² plus R².

dB is perpendicular to r. Split it into two parts: dB x along the axis, and dB perpendicular, at right angles to the axis.

Now take the element diametrically opposite, at the bottom. Its perpendicular part points the other way. The two cancel. Going round the whole loop, every perpendicular part cancels like this.

Only the parts along the axis add up. Each is dB cos θ, where cos θ equals R over √(x² + R²).

Adding all the dl round the loop just gives the circumference, 2πR. So B equals μ₀ I R², over 2, times x² plus R², to the power three-halves.

At the centre, x is zero, and this becomes B equals μ₀ I over 2R. For a coil of N closely wound turns, multiply by N: B equals μ₀ N I over 2R.

Far along the axis, where x is much bigger than R, B is about μ₀ I R² over 2x³. It falls off as 1 over x³, much faster than the field of a single element.

Direction and examples

Narrated video. Transcript:

Which way does the field point? Curl the fingers of your right hand along the current in the loop. Your extended thumb points along the field at the centre.

Here are the field lines of a loop. Near the wire they are small closed circles around it. Through the middle they run straight along the axis, then spread out and loop back outside.

Looking at the loop from one side, the current goes anticlockwise and the field lines come out at you. That face behaves like the north pole of a magnet. The other face is the south pole.

Example: a tightly wound coil of 100 turns, radius 10 centimetres, carries a current of 1 ampere. What is the field at its centre?

B equals μ₀ N I over 2R: 4π × 10⁻⁷ times 100 times 1, divided by 2 times 0.1. That is 1.26 × 10⁻⁴ over 0.2, or 6.28 × 10⁻⁴ tesla.

Second example: a straight wire carrying 12 amperes is bent into a semicircle of radius 2 centimetres. What is the field at the centre of the arc?

The straight parts point right at the centre. For them, dl and r are parallel, so they add nothing. The arc is half a loop, so it gives half the field of a full loop: μ₀ I over 4R.

4π × 10⁻⁷ times 12, over 4 times 0.02. That is 1.9 × 10⁻⁴ tesla, perpendicular to the plane of the arc.

How fast does the field fade along the axis? At x equals R it is already down to 0.35 of its centre value, and at x equals √3 R it is one-eighth.

Try it: the field of a loop

Interactive simulation. Things to try:

• Press “Side view” to see the field lines in a plane through the axis: they look like those of a bar magnet.
• With N = 100, I = 1 A and R = 10 cm, check that B at the centre is 6.28 × 10⁻⁴ T.
• Move P along the axis and watch B fall. The graph shows B(x) all along the axis.
• Double R at the centre: B halves. Double N or I: B doubles.

One-eighth of the centre value

At what distance x from the centre of a circular loop of radius R, on its axis, is the magnetic field one-eighth of its value at the centre?

1. x = R
2. x = √3 R
3. x = 2R
4. x = 8R

A coil of 100 turns

A tightly wound coil of 100 turns and radius 10 cm carries a current of 1 A. What is the magnetic field at its centre? (μ₀ = 4π × 10⁻⁷ T m A⁻¹)

1. 6.28 × 10⁻⁶ T
2. 2.0 × 10⁻⁴ T
3. 6.28 × 10⁻⁴ T
4. 1.26 × 10⁻³ T

Summary

• On the axis: B = μ₀ I R² / 2(x² + R²)^(3/2)
• At the centre (N turns): B = μ₀ N I / 2R
• Far along the axis (x ≫ R): B ≈ μ₀ I R² / 2x³
• Arc of a semicircle: B = μ₀ I / 4R

• The components of dB perpendicular to the axis cancel in pairs; only the axial parts add.
• Direction: right-hand thumb rule — fingers along the current, thumb along B.
• The field lines of a loop look like those of a bar magnet: one face acts as a north pole, the other as a south pole.
• Straight wire segments pointing at the centre contribute nothing there, because dl × r = 0.