Coils and Adjacent Conductors

Outcomes

After this chapter, the learner should be able to:

• distinguish flexible-coil, adjacent-conductor and rigid-coil arrangements;
• define their field direction and effective area;
• apply the training calculation set correctly;
• calculate fill factor and L/D;
• plan overlap for long or curved components; and
• identify the assumptions that make a calculation invalid.

9.1 Flexible cable coil

An insulated cable wrapped tightly around a component forms a flexible coil and produces a longitudinal field. Discontinuities parallel to the windings - normally transverse to the component axis - are favoured. The effective area lies between and near the turns and depends on wall thickness, spacing and current.

The source-derived training formulae are:

For DC or rectified current:

I = 3H [T + (Y^2 / 4T)]

For AC:

I = 3H [10 + (Y^2 / 40)]

where I is amperes on the stated basis, H is kA/m, T is wall thickness or the radius of a solid circular bar in mm, and Y is spacing between adjacent windings in mm.

Example: AC flexible coil, H = 2 kA/m, Y = 10 mm:

I = 3 x 2 x [10 + (100/40)] = 75 A RMS

The formula and units shall be checked against the approved standard/procedure before field use.

9.2 Adjacent conductor

An insulated cable laid parallel to the surface creates a circular field around the cable. The effective strip extends approximately distance d on each side when the cable centreline is distance d from the surface in the source model. Move the cable by less than 2d between shots so strips overlap. Keep the return cable well outside the test region; the source describes a distance greater than 10d.

The source-derived relationship is:

I = 4 x pi x d x H

where I is amperes, d is cable-to-surface distance in mm and H is kA/m.

Example: H = 2 kA/m and d = 25 mm, using an AC RMS training basis:

I = 4 x 3.142 x 25 x 2 = 628 A RMS approximately.

A cable can also be arranged as a close-wrapped conductor around a branch or curved component. The source expresses the effective distance as:

d = NI / 4H

Confirm the definition of d and whether pi is included in the applicable formula set; diagrams and units in older training material are easy places for tiny algebraic gremlins to breed.

9.3 Rigid coil

A rigid coil creates a longitudinal field through its aperture. The component major axis is aligned with the coil axis, usually near the inside wall where the field is stronger. The method is suited to long, slender parts and favours transverse discontinuities.

Before using the low-fill-factor formula, confirm:

• component cross-sectional area is less than 10% of coil aperture area;
• the component is positioned as required near the coil wall;
• L/D is at least 5;
• L/D values above 20 are limited to 20 for the calculation;
• coil pitch and turn count meet the procedure; and
• the examination is repeated in overlapping coil-length sections.

For a non-circular section, the source uses D = perimeter / pi.

9.4 Fill factor

Fill factor (%) = component cross-sectional area / coil aperture area x 100

For circular areas, the pi terms cancel if diameters are used consistently:

Fill factor (%) = (component diameter / coil inside diameter)^2 x 100

Example: A 60 mm bar in a 200 mm coil:

(60/200)^2 x 100 = 9%

The result meets the less-than-10% condition in the supplied formula model.

9.5 Rigid-coil ampere-turn calculation

The source-derived low-fill-factor equation is:

NI = 0.4 H K / (L/D)

where N is effective turns, I is current in amperes, H is kA/m, and K is a waveform constant. The supplied source lists:

• K = 22,000 for AC, RMS basis;
• K = 22,000 for FWR, stated mean basis; and
• K = 11,000 for HWR, stated mean basis.

Because meter basis is critical, formal approval must reconcile these constants and indicated units with the current controlling document.

Worked example: 60 mm diameter, 350 mm long bar; 200 mm inside-diameter coil; 4.5 turns; AC; H = 2 kA/m.

1. Fill factor = 9%, so the low-fill-factor condition is met.
2. L/D = 350/60 = 5.83.
3. NI = 0.4 x 2 x 22,000 / 5.83 = 3,019 ampere-turns approximately.
4. I = 3,019 / 4.5 = 671 A RMS approximately.

Small differences result from rounding. Record the unrounded values used by the controlled calculation.

9.6 Pole extenders

If L/D is below the permitted minimum, ferromagnetic extenders of compatible cross-section may be attached to improve the effective magnetic length. They must contact properly and be included in the approved setup. Extenders do not change the physical inspection length; they modify the magnetic circuit.

9.7 Coverage and verification

Calculations estimate a starting current. Demonstrate field direction and system response at the least favourable locations. Long parts are shifted through the coil in controlled increments. Adjacent conductors are moved with overlap, and return cables are positioned so their field does not cancel or distort the intended field.

9.8 Common traps

• Using diameter instead of radius when calculating cross-sectional area.
• Forgetting to divide ampere-turns by N.
• Entering H in A/m when the formula expects kA/m.
• Applying the low-fill-factor equation above its permitted fill factor.
• Using actual L/D above 20 when the model caps it.
• Ignoring cable spacing and return-path field.

Chapter summary

Coil and adjacent-conductor techniques can cover components that portable poles cannot, but their calculation models have strict geometry and unit assumptions. A Level 2 technician verifies fill factor, L/D, waveform basis, effective area and overlap before treating a calculated current as usable.