Ampere-Turns & Gauge Selection

A field coil working from a fixed voltage has a surprising property: the ampere-turns depend only on the wire gauge, not on how many turns you pile on. Set the supply and the bobbin, choose the NI your magnet needs, and the plate names the gauge.

Parameters

Supply voltage across coil

V

Target ampere-turns

(NI)_t

A·t

Selected gaugeRecommended: AWG 31 or heavier (exact 31.2)AWG

Inner diameter

D_i

Outer diameter

D_o

Winding length

L_w

Copper fill factor

k_{cu}

Bare copper area ÷ window area

Working temperature

T

°C

Fig. 2Meters read the selected AWG 31. The red arc marks shortfall below target; the power meter's red arc exceeds a comfortable 0.1 W/cm² of coil surface.

AWG 31 on this bobbin

Ampere-turns

NI

1.58 kA·t

Dissipation

P

14.1 W

Turns

N

12,257

Resistance

R

856 Ω

Current

I

129 mA

Current density

J

3.18 A/mm²

Required bare Ø0.221 mm

Exact gaugeAWG 31.2

☞ Meets target with 5 % margin. Surface loading ≈ 0.089 W/cm².

Ampere-turns vs. gauge (fixed voltage)

Power dissipated vs. gauge

How it works

Theory · assumptions · formulas

The gauge sets the ampere-turns

For a coil of $N$ turns of bare cross-section $A$ and mean turn length $\ell_{MLT}$ connected across a voltage $V$ :

R = \frac{\rho N \ell_{MLT}}{A}, \qquad I = \frac{V}{R} = \frac{V A}{\rho N \ell_{MLT}}

NI = \frac{V A}{\rho\,\ell_{MLT}}

The turns cancel. Adding turns raises resistance in exact proportion, so the current falls and NI stays put. Only the wire's cross-section, the mean turn length and the supply voltage matter. Solving for the area gives the required wire:

A_{req} = \frac{(NI)_t\,\rho\,\ell_{MLT}}{V}, \qquad d_{req} = \sqrt{4A_{req}/\pi}

So what do the turns do?

They set the power. Filling the window (fill factor $k_{cu}$ , window area $A_w$ ) gives $N = k_{cu} A_w / A$ , and

P = \frac{V^2}{R} = \frac{V^2 A^2}{\rho\,k_{cu} A_w\,\ell_{MLT}}

For a given gauge, winding more turns leaves NI untouched but cuts the power in proportion. Changing the gauge changes both: thicker wire raises NI linearly with $A$ but power with $A^2$ . Hence the classical procedure: choose the gauge for NI, then fill the bobbin completely to bring the dissipation down.

Temperature

Resistivity is taken at the working temperature, $\rho_T = \rho_{20}[1+\alpha(T-20)]$ . A coil that delivers its NI cold may fall 15 % short once it reaches 60 °C, so design at operating temperature.

Rule of thumbKeep current density below about 3–4 A/mm² and surface dissipation below about 0.1 W/cm² for a coil cooled by natural convection.