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Helmholtz coils: spacing, field and uniformity

A Helmholtz coil is two identical coaxial coils carrying the same current, spaced for the most uniform field between them: one radius apart for circular coils (d = R) and about 0.5445 times the side for square coils. The circular pair’s centre field is B = (4/5)^(3/2)·μ₀NI/R ≈ 0.7155·μ₀NI/R.

Last updated: By: the AirInduct Sim engineering team

What is a Helmholtz coil?

A Helmholtz coil is a pair of identical coils on a common axis, carrying equal currents in the same direction. Each coil alone produces an axial field that peaks at its centre and falls away on both sides. Placed at the right distance, the fall-off of one coil is compensated by the rise of the other, leaving a region around the midpoint where the field is nearly constant in magnitude and direction.

That uniform region makes Helmholtz pairs standard laboratory equipment: calibrating magnetometers and Hall sensors, cancelling the Earth’s field, exposing samples or electronics to a known field for immunity testing, and characterising sensor coils. Three orthogonal pairs give control of the field in all three directions.

Why are circular Helmholtz coils spaced one radius apart?

The on-axis field of one circular loop of radius R, N turns and current I follows from Biot–Savart:

B(z) = μ₀ N I R² / (2 (R² + z²)^(3/2))

Put two such loops at z = ±d/2 and expand the total field about the midpoint. Odd derivatives cancel by symmetry. The second derivative is proportional to (d² − R²), so it vanishes exactly when d = R. With the first three derivatives zero, the leading deviation is fourth order:

B(z) / B(0) ≈ 1 − (144/125)·(z/R)⁴ (d = R)

Setting the correction to 1 % gives |z| ≈ 0.305·R. The exact Biot–Savart sum, computed below, gives 0.314·R, slightly wider because the sixth-order term has the opposite sign.

What is the field at the centre of a Helmholtz coil?

Substituting z = R/2 for each loop and adding the two contributions gives the circular-pair formula:

B₀ = (4/5)^(3/2) · μ₀ N I / R ≈ 0.7155 · μ₀ N I / R

For square loops of side s, each straight side is a finite wire whose field is also closed-form. On the axis of one square loop:

B(z) = μ₀ N I s² / (2π (s²/4 + z²) √(s²/2 + z²))

The Helmholtz condition for squares has no neat closed form; solving d²B/dz² = 0 numerically gives an optimal spacing of d = 0.54451·s, usually quoted as 0.5445·s (Firester analysed square Helmholtz systems in 1966). The centre field of the square pair is then twice B(z) at z = d/2.

The word Helmholtz is used in two senses in wireless power work, and it helps to keep them apart. The classic Helmholtz pair is a field-generating instrument: two coils, the same current, spaced for uniformity. In AirInduct Sim, a Helmholtz coil is a wound coil whose turns all sit at one radius or half-side, as opposed to a DTC (descending-turn) planar spiral whose turns step inward. A circular or square Helmholtz-to-Helmholtz topology in the simulator is therefore a transmitter and a receiver of that wound type, separated by whatever gap the application needs, and the solver reports their mutual inductance, self-inductances, coupling coefficient and peak field. The free Helmholtz coil calculator, by contrast, answers the instrument question: what field you get at the centre, what current you need, and how large the uniform region is. Both use the same Biot–Savart field of circular and square loops.

Worked example: a 100 mm, 50-turn pair at 2 A

Computed on this page with the code behind the Helmholtz coil calculator and the square Helmholtz coil calculator:

QuantityCircular, R = 100 mmSquare, s = 100 mm
Optimal spacing100 mm54.45 mm
Centre field at 2 A899.2 µT1.629 mT
Axial region within ±1 %±31.4 mm±17.1 mm
Axial region within ±0.1 %±17.3 mm±9.46 mm

The closed-form circular result (4/5)^(3/2)·μ₀NI/R gives 899.2 µT, identical to the summed loop fields. To produce 1 mT at the centre of the circular pair you need 2.224 A. Note that a 100 mm square is much smaller than a 100 mm-radius circle (whose diameter is 200 mm), which is why the square pair has a stronger centre field and a shorter uniform region here.

How sensitive is the uniformity to spacing errors?

The chart shows the normalised axial field B(z)/B(0) of the circular pair at the optimal spacing and at 1.4·R. Too wide a spacing produces a dip at the centre; too narrow a spacing produces a peak.

0.750.80.850.90.9511.051.11.15-50050Axial position from midpoint (mm)B(z) / B(0)
d = R (Helmholtz)d = 1.4·R
Circular pair, R = 100 mm, computed from the Biot–Savart field of each loop.

Moderate errors cost surprisingly little. At d = 1.2·R the ±1 % region is ±17.7 mm and at d = 0.8·R it is ±15.5 mm, against ±31.4 mm at d = R. Real windings have finite width and depth, so in practice the spacing is measured between the centroids of the two winding cross-sections. For a uniform region larger than about a third of the radius, use more coils (for example a Maxwell three-coil arrangement) or larger coils.

References

  1. A. H. Firester, “Design of square Helmholtz coil systems,” Review of Scientific Instruments 37(9), 1264–1265 (1966). doi.org/10.1063/1.1720478
  2. J. D. Jackson, Classical Electrodynamics, 3rd ed., Wiley (1999), ISBN 978-0-471-30932-1. www.wiley.com/en-us/Classical+Electrodynamics%2C+3rd+Edition-p-9780471309321
  3. F. W. Grover, Inductance Calculations: Working Formulas and Tables, Van Nostrand (1946); Dover reprint (2004), ISBN 978-0-486-49577-4. openlibrary.org/isbn/9780486495774

Frequently asked questions

Why is the Helmholtz spacing equal to the radius?

For two coaxial circular loops the second derivative of the on-axis field at the midpoint vanishes when the separation equals the radius. The first and third derivatives vanish by symmetry, so the field is flat to fourth order in z at the centre. That is the defining Helmholtz condition.

What is the optimal spacing for a square Helmholtz coil?

About 0.5445 times the side length. It is found the same way as for circular coils, by setting the second derivative of the axial field to zero at the centre, but for a square loop the result has to be solved numerically. Firester (1966) gives design data for square Helmholtz systems.

How uniform is the field of a Helmholtz coil?

For a circular pair the axial field stays within 1 % of its centre value for about ±0.31 R and within 0.1 % for about ±0.17 R. Off axis the uniform region is a little smaller. Larger uniform volumes need more coils, such as Maxwell or Lee–Whiting arrangements.

Is a Helmholtz coil used for wireless power transfer?

The classic Helmholtz pair produces a uniform test field rather than transferring power. In AirInduct Sim, “Helmholtz” names a wound coil whose turns sit at one radius, used as a transmitter or receiver; the uniform-field spacing rule applies to the field-generating pair, not to the power-transfer gap.