Fill factor is the fraction of the coil’s cross-sectional area actually occupied by the test object: eta = (d_coil / D_bore)^2 for a bobbin probe inside a tube, or (d_bar / D_coil)^2 for an encircling coil around a bar. Because it is an area ratio it goes as the square of the diameter ratio, and it is the single largest determinant of sensitivity in bobbin and encircling coil testing.

The physics is coupling. Flux crossing the annular gap between coil and part links no conductor, so it contributes to the empty-coil reactance but carries no flaw information. A probe at 0.6 fill factor is not 40 % less sensitive in some vague sense – 40 % of its field is doing nothing at all, and what remains is more easily swamped by wobble, because the same physical wobble is now a larger fraction of the gap.

Since fill factor is quadratic, small clearances matter more than intuition suggests. To first order the fill factor loses 4/D of its value for every millimetre of extra radial clearance, that is 400/D percent per millimetre. On a 16 mm bore that is about 25 % of the signal per millimetre, so a tenth of a millimetre of probe wear or debris under the probe is a measurable amplitude change.

Common practice is 85 to 90 % fill for bobbin probes. Going higher gives more signal but the probe stops passing dents, ovality, deposits and U-bends, and a stuck probe in a heat exchanger is far more expensive than a few percent of amplitude. Going lower loses the noise margin needed to see small OD-initiated defects. Whatever fill factor is chosen, the calibration standard must be examined with the same probe so the sensitivity is proven, not assumed.

Worked example

Coil (probe) diameter	15.5 mm
Bore diameter	16.57 mm
Additional radial clearance	0.1 mm
Fill factor	0.875 ratio
Fill factor	87.5 %
Radial clearance	0.535 mm
Coupling loss per mm of clearance	24.14 %/mm
Fill factor with added clearance	85.4 %
Coupling loss from added clearance	2.37 %

A 15.5 mm bobbin probe in a 19.05 mm OD x 1.24 mm wall tube (bore = 19.05 - 2.48 = 16.57 mm). Diameter ratio = 15.5/16.57 = 0.93543, so eta = 0.93543^2 = 0.8750 = 87.5 % fill - inside the usual 85 to 90 % window. Radial clearance = (16.57 - 15.5)/2 = 0.535 mm. First-order sensitivity = 400/16.57 = 24.14 % per mm. Adding 0.1 mm of clearance makes the effective bore 16.77 mm, so eta becomes (15.5/16.77)^2 = 0.85427 = 85.4 %, a relative loss of (87.502 - 85.427)/87.502 = 2.37 % - close to the 2.41 % the first-order rate predicts, the difference being the second-order term.

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