Picture of Electromagnet being wound

Here is the electromagnet coil in the process of being wound with #16 a.w.g wire.

Air Core Electromagnet Design

Electromagnet Design

For a physics experiment, I would like to produce a magnetic field in a region of space (air). To do this, here is a design for an air cored electromagnet to produce an axial B field in the space inside the bore. The volume of interest is 10 cm (4 inches) diameter and the length is 10 cm (4 inches). The flux density is 1000 Gauss (0.1 Tesla) at the center. This electromagnet is for DC, AC or pulsed operation would impose a different design. How reasonable is this? What would be the number of turns, layers, current, etc.?

For starts, it is noted that for a field of 1 Tesla in air, 8000 Amp-turns are required for each cm of length. So, for a path length of 10 cm and 1000 Gauss, 8000 Amp-turns are required. This does not include the Amp-turns required for the flux path outside the bore volume. This could be minimized by providing a ferromagnetic yoke for the return flux, but I will not do this here. Therefore, the total Amp-turns required will be some value greater than 8000 Amp-turns, but I guess less than double this amount.

The formula for the axial B field in the center of a single layer solenoid current-carrying coil is

(1)

Where B= flux density, Tesla, center of solenoid

μo = permeability of free space, 4π10-7 H/m,
n = number of turns,

I = current, Amps,

d = diameter, meters,

l = length, meters.

One can see that for the length l much larger than the diameter d, this formula reduces to

B =μo (n/l)I

Where (n/l) = turns/ unit length. The flux density is independent of the diameter. For No. 16 awg copper wire, the diameter is 1.35mm so (n/l) = 744 turns/meter and B is only 9.35 Gauss per Ampere! We choose an current of 6.3 Amps, a value for intermittent operation which would result in a flux density of 58.9 Gauss at the center. A larger diameter wire would result in less turns and voltage required, but a greater current for the same current density

For a single layer solenoid 10 cm long, the flux density is less than the 58.9 Gauss produced by a “long” solenoid, Formula (1) gives 41.5 Gauss. Additional layers are clearly required. However each layer has a larger diameter as it is wound over the previous layer. I use a sheet of paper between layers, so the increase of diameter equals 2x diameter of the wire + 2x thickness of the paper between layers. Formula (1) shows that the flux density produced by each layer is less than the layer underneath. In addition, each wire turn of the layer requires more wire, d and corresponding more resistance and power dissipation. These 2 factors result in diminishing returns. At some point it would be better to add more turns by increasing the length instead of adding an additional layer.

A program was written to calculate the flux density for multiple layers. The parameters are

No. 16 awg wire, 7.44 turns/cm, Current = 6.3 Amps.

Inner diameter, 10 cm

Length, 10cm.

The result are shown in this table:

No. layers	Flux Density, B	Ohms	Volts required for 6.3 Amps.	Watts required, V*I
1	41.5 Gauss	0.33 Ohms	2.08	13.1
2	82.4	0.66	4.16	26.2
4	162	1.35	8.5	54
8	316	2.85	18.0	113
10	389	3.66	23.1	146
12	461	4.50	28.4	178
16	598	6.30	39.7	250
20	727	8.25	52.0	327
25	880	10.9	68.6	430
30	1020	13.8	87	550
40	1280	20.2	127	800
50	1522	27.6	174	1100