A coil (encircling) shot drives a longitudinal field along the axis of the part, so it reveals discontinuities lying transverse to that axis. What sets the field inside the coil is the magnetomotive force, the ampere-turns NI – current multiplied by number of turns – not the current alone. A five-turn coil at 1 000 A and a one-turn wrap at 5 000 A produce the same field.

The part fights back. Its own end poles create a demagnetising field that opposes the applied field, and that opposing field grows as the part gets shorter and fatter. The empirical codes capture this with the length-to-diameter ratio L/D: required ampere-turns fall as L/D rises. Below L/D = 2 the demagnetising field dominates and the formula stops being meaningful; above L/D = 15 there is no further gain, so 15 is the value used.

Low fill factor means the coil is much larger than the part – the coil bore cross-section is at least ten times the part cross-section. The part sits far from the windings, out where the field has spread and weakened, so it needs more ampere-turns than a snug coil would. For the same reason the part is normally laid against the inside wall of the coil, where the flux is concentrated, and the wall-position formula NI = 45 000/(L/D) applies. Where the part must instead hang centred on the coil axis – on a fixture, or from a crane sling – the standards give a different relation, NI = 43 000 R / (6 L/D - 5) with R the coil radius in inches: out on the axis the field the part sees depends on the size of the coil itself, so the coil radius enters the calculation.

The result is a starting point, not a guarantee. A coil’s field is only usefully strong for roughly 230 mm (9 in.) either side of it, so long parts need several overlapping shots. Confirm the actual tangential field at the surface with a Hall-effect gaussmeter (30-60 G) or with artificial flaw shims before accepting the technique.

Worked example

Part length	300 mm
Part diameter	50 mm
Part position in the coil	wall
Coil inside diameter	300 mm
Coil turns	5
L/D used	6
Required ampere-turns	7500 A-turn
Ampere-turns, -10%	6750 A-turn
Ampere-turns, +10%	8250 A-turn
Coil current	1500 A
Current, lower tolerance	1350 A
Current, upper tolerance	1650 A
Coil positions needed	1

A 300 mm long, 50 mm diameter shaft laid against the wall of a large 5-turn coil. L/D = 300/50 = 6, inside the valid 2 to 15 band. NI = 45 000/6 = 7 500 ampere-turns; the ±10% band is 6 750 to 8 250. With 5 turns the machine is set to 7 500/5 = 1 500 A, tolerance 1 350 to 1 650 A. The part is shorter than 2 x 230 = 460 mm, so one coil position covers it. Suspended in the centre of a 300 mm bore coil instead, the requirement would rise to NI = 43 000 x (150/25.4)/(6 x 6 - 5) = 43 000 x 5.906/31 = 8 192 ampere-turns, about 1 638 A on the same 5 turns.

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