2.3 Longitudinal Coil Magnetization and Calculations

2.3 Longitudinal Coil Magnetization and Calculations

Longitudinal magnetization—most commonly achieved using rigid multi-turn encircling coils or flexible cable wraps—is the foundational technique for detecting transverse discontinuities in cylindrical, tubular, and elongated ferromagnetic components. Unlike direct contact head shots that establish circular closed-loop magnetic fields without poles, longitudinal coil magnetization establishes an axial magnetic field running parallel to the long axis of the workpiece, terminating in external North and South magnetic poles.

Because the creation of external poles introduces a powerful self-demagnetizing field that directly opposes the applied magnetizing field, coil magnetization is governed by complex geometric relationships—specifically the component's Length-to-Diameter ($L/D$) ratio and the coil's fill factor. An ASNT NDT Level III must be thoroughly versed in calculating magnetizing ampere-turns ($NI$) across low, high, and intermediate fill-factor scenarios per ASTM E1444 and ASTM E709.

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Solenoid Electromagnetic Field Distribution

Field Structure of an Encircling Coil

An encircling coil (or solenoid) consists of multiple turns ($N$) of heavy copper conductor carrying an electric current ($I$). By the Biot-Savart Law and Ampere's Law, current circulating circumferentially around the coil turns induces a concentrated magnetic field along the longitudinal bore of the coil.

                      +-----------------------+
                      | (I) (I) (I) (I) (I)   | Coil Windings
             North    |=======================|    South
             Pole <---[===== Workpiece ======>]---> Pole
                      |=======================|
                      | (I) (I) (I) (I) (I)   |
                      +-----------------------+
                         Axial Magnetic Flux

Within the coil bore:

1. The magnetic flux lines travel parallel to the central axis of the coil.
2. When an elongated ferromagnetic component (such as an axle, bolt, tube, or tie rod) is placed within the bore, the high-permeability steel concentrates the magnetic flux lines internally.
3. The flux lines exit one end of the workpiece, creating a North magnetic pole, curve through the surrounding air, and re-enter the opposite end, creating a South magnetic pole.

Discontinuity Detectability and Directionality

Because the magnetic lines of force travel longitudinally through the component:

• Transverse Discontinuities: Flaws oriented perpendicular to the long axis of the component (such as circumferential fatigue cracks, transverse weld fissures, radial heat-treat cracks, and circumferential grinding checks) cut across the longitudinal lines of force at an angle near $90^\circ$. These flaws force magnetic flux to leak out into the air, creating sharp, prominent indications.
• Longitudinal Discontinuities: Flaws oriented parallel to the long axis (such as longitudinal seams, forging laps, and rolling stringers) lie parallel to the longitudinal flux lines. They create zero flux leakage and are completely undetectable with a coil. A direct contact circular head shot or central conductor shot must be used instead.

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End-Pole Effects and the Self-Demagnetizing Field

The fundamental physical challenge in longitudinal magnetization is the phenomenon of self-demagnetization.

               North Pole                         South Pole
                 +----+                             +----+
                 | +N | ===== Applied Field (H) ==> | -S |
                 |    |                             |    |
                 |    | <== Demagnetizing (Hd) ==== |    |
                 +----+                             +----+

The Demagnetizing Field ($H_d$)

When magnetic flux exits the North pole and enters the South pole at the ends of the workpiece, the free magnetic poles establish an internal magnetic field ($H_d$) that acts within the steel in the exact opposite direction to the applied field ($H$).

The net magnetizing field ($H_{\text{net}}$) acting inside the steel is therefore: $H_{\text{net}} = H_{\text{applied}} - H_d$

The magnitude of this opposing demagnetizing field depends entirely on the proximity of the two poles to each other, which is governed by the component's Length-to-Diameter ($L/D$) ratio:

• Low $L/D$ Ratio (Short, Stubby Parts): The North and South poles are physically close together. The demagnetizing field $H_d$ is massive, canceling out the majority of the applied field. Even several thousand amperes of current may fail to produce adequate net magnetic flux in the steel.
• High $L/D$ Ratio (Long, Slender Parts): The poles are far apart. The opposing demagnetizing field is weak, allowing the applied field to magnetize the steel efficiently with modest current.

Geometric Constraints on $L/D$

Under ASTM E1444 and ASTM E709:

1. Minimum $L/D$ Ratio of 2: Standard longitudinal coil magnetization formulas are completely invalid for parts with $L/D < 2$. The self-demagnetizing field in such parts is so overwhelming that saturation cannot be attained without burning the coil.
2. Ideal $L/D$ Bracket: The standard empirical coil formulas operate with high accuracy for $L/D$ ratios between 4 and 15.
3. Capping $L/D$ at 15: For long parts with $L/D > 15$, the demagnetizing effect becomes negligible. In calculations, an $L/D$ value of 15 is substituted into the formulas to prevent under-estimating required ampere-turns.

The Pole Extender Solution for Low $L/D$ Components

When a short component has an $L/D < 2$ (or an $L/D < 4$ where high test sensitivity is required):

• Ferromagnetic Pole Extenders: Soft iron or low-carbon steel extension blocks of matching cross-sectional area are placed in direct mechanical contact with both ends of the workpiece during the coil shot.
• Physical Effect: The magnetic poles are displaced to the outer ends of the extension blocks, effectively increasing the overall length ($L_{\text{effective}} = L_{\text{part}} + L_{\text{extenders}}$). This raises the effective $L/D$ ratio above 4, eliminating self-demagnetization across the actual test piece.

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Coil Fill-Factor Classification and Calculation Formulas

The mathematical formula used to determine required magnetizing current depends directly on the fill factor of the coil-part combination.

$\text{Fill Factor Ratio } (\tau) = \frac{\text{Cross-Sectional Area of Coil Bore } (A_{\text{coil}})}{\text{Cross-Sectional Area of Part } (A_{\text{part}})}$

For circular coils and solid cylindrical parts: $\tau = \frac{\pi R_{\text{coil}}^2}{\pi r_{\text{part}}^2} = \frac{D_{\text{coil}}^2}{D_{\text{part}}^2}$

                                  Fill Factor Ratio (Tau)
                                            |
               +----------------------------+----------------------------+
               |                                                         |
         Tau > 10 (Low Fill-Factor)                                Tau <= 2 (High Fill-Factor)
               |                                                         |
     * Off-Center (Coil Wall):                                 * Form-Fitting / Cable Wrap:
       NI = 45,000 / (L/D)                                       NI = 35,000 / [(L/D) + 2]
     * Centered on Axis:
       NI = 43,000 * R / [6(L/D) - 5]

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1. Low Fill-Factor Coils ($A_{\text{coil}} / A_{\text{part}} > 10$)

A low fill-factor condition exists when a small-diameter component is tested in a large-diameter stationary coil—a standard scenario on wet horizontal benches (e.g., a 2-inch diameter shaft inside a 16-inch diameter 5-turn bench coil, where $\tau = 16^2 / 2^2 = 64 > 10$).

Under ASTM E1444 and ASTM E709, two distinct formulas apply depending on part positioning:

Case A: Part Positioned Off-Center (Resting on Inner Coil Wall)

In standard shop practice, heavy shafts rest on the bottom of the coil bore against the inner diameter wall. Because magnetic flux density within a solenoid is highest near the inner conductor windings, less current is required:

$NI = \frac{45,000}{L/D} \quad (\pm 10\%)$

Where:

• $N$ = Number of turns in the coil (typically 5 turns on bench units)
• $I$ = Magnetizing current in Amperes
• $NI$ = Total magnetizing ampere-turns
• $L/D$ = Length-to-diameter ratio of the part (if $L/D > 15$, use $15$; if $L/D < 2$, pole extenders required)

Case B: Part Positioned on Coil Central Axis (Centered)

When specialized non-magnetic fixturing holds the component centered along the longitudinal axis of the coil bore, the component sits in the region of lowest solenoid flux density. A higher current is mandatory:

$NI = \frac{43,000 \cdot R_{\text{coil}}}{6(L/D) - 5} \quad (\pm 10\%)$

Where:

• $R_{\text{coil}}$ = Inside radius of the coil, in inches
• $L/D$ = Length-to-diameter ratio

Critical Level III Comparison: The centered formula requires approximately $30%$ to $50%$ more current than the off-center formula for identical parts! Mistakenly applying the off-center formula to a centered part results in severe under-magnetization.

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2. High Fill-Factor Coils ($A_{\text{coil}} / A_{\text{part}} \le 2$)

A high fill-factor condition occurs when the coil closely hugs the component contours—such as form-fitting rigid coils or flexible $4/0$ welding cable wrapped directly around the workpiece (typically 3 to 5 turns). High fill-factor configurations provide superior magnetic coupling and minimal flux leakage:

$NI = \frac{35,000}{(L/D) + 2} \quad (\pm 10\%)$

Where:

• $L/D$ = Length-to-diameter ratio
• $N$ = Number of cable wraps or coil turns

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3. Intermediate Fill-Factor Coils ($2 < A_{\text{coil}} / A_{\text{part}} \le 10$)

When the coil cross-sectional area is between 2 and 10 times the part cross-sectional area, neither the high nor low fill-factor formulas apply directly. ASTM E1444 and ASTM E709 prescribe a linear interpolation formula:

$(NI)_{\text{intermediate}} = (NI)_h \cdot \left(\frac{10 - \tau}{8}\right) + (NI)_l \cdot \left(\frac{\tau - 2}{8}\right)$

Where:

• $\tau = A_{\text{coil}} / A_{\text{part}}$
• $(NI)_h$ = Ampere-turns calculated using the High Fill-Factor formula: $\frac{35,000}{(L/D) + 2}$
• $(NI)_l$ = Ampere-turns calculated using the Low Fill-Factor formula: $\frac{45,000}{L/D}$

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Step-by-Step Worked Mathematical Calculations

Problem 1: Off-Center Low Fill-Factor Shaft

Problem: A solid AISI 4140 steel drive axle has a length of $24.0\text{ inches}$ ($610\text{ mm}$) and an outer diameter of $2.0\text{ inches}$ ($50.8\text{ mm}$). It is inspected on a wet horizontal bench using a 5-turn stationary coil with an inside diameter of $16.0\text{ inches}$. The axle rests against the bottom of the coil bore. Calculate the required bench current setting.

Step 1: Calculate the Length-to-Diameter ($L/D$) Ratio: $\frac{L}{D} = \frac{24.0\text{ in}}{2.0\text{ in}} = 12.0$ Check: $2 \le 12.0 \le 15$ (Valid without adjustment).

Step 2: Check the Fill Factor Ratio ($\tau$): $A_{\text{coil}} = \pi \cdot R^2 = \pi \cdot (8.0\text{ in})^2 = 201.06\text{ sq in}$ $A_{\text{part}} = \pi \cdot r^2 = \pi \cdot (1.0\text{ in})^2 = 3.14\text{ sq in}$ $\tau = \frac{201.06}{3.14} = 64.0$ Check: $\tau = 64.0 > 10 \rightarrow$ Low Fill-Factor applies.

Step 3: Calculate Ampere-Turns ($NI$) for Off-Center Position: $NI = \frac{45,000}{L/D} = \frac{45,000}{12.0} = 3750\text{ Ampere-Turns}$

Step 4: Calculate Machine Amperage ($I$) for a 5-Turn Coil ($N = 5$): $I = \frac{NI}{N} = \frac{3750\text{ A-turns}}{5\text{ turns}} = 750\text{ Amperes}$ Conclusion: Set machine to $750\text{ Amperes}$ (acceptable range $\pm 10% = 675\text{ to }825\text{ A}$). Verify field with a QQI shim.

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Problem 2: Centered Low Fill-Factor Shaft

Problem: Take the identical shaft from Problem 1 ($L = 24.0\text{ in}$, $D = 2.0\text{ in}$, $L/D = 12.0$), but mounted in non-magnetic v-blocks along the central axis of the 16-inch diameter 5-turn coil ($R_{\text{coil}} = 8.0\text{ inches}$). Calculate the required bench current.

Step 1: Apply the Centered Low Fill-Factor Formula: $NI = \frac{43,000 \cdot R_{\text{coil}}}{6(L/D) - 5}$ $NI = \frac{43,000 \cdot 8.0}{6(12.0) - 5} = \frac{344,000}{72.0 - 5} = \frac{344,000}{67.0} \approx 5134.3\text{ Ampere-Turns}$

Step 2: Calculate Machine Amperage ($I$): $I = \frac{NI}{N} = \frac{5134.3\text{ A-turns}}{5\text{ turns}} \approx 1027\text{ Amperes}$ Technical Comparison: Centering the shaft increases required amperage from $750\text{ A}$ to $1027\text{ A}$—a $37%$ increase in current!

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Problem 3: High Fill-Factor Cable Wrap

Problem: A heavy forging tie rod with a length of $32.0\text{ inches}$ and a diameter of $4.0\text{ inches}$ ($L/D = 8.0$) is inspected in the field using 4 turns of flexible $4/0$ cable wrapped tightly around the bar ($A_{\text{coil}} \approx A_{\text{part}} \rightarrow \tau \approx 1$). Calculate required current.

Step 1: Apply High Fill-Factor Formula: $NI = \frac{35,000}{(L/D) + 2} = \frac{35,000}{8.0 + 2} = \frac{35,000}{10.0} = 3500\text{ Ampere-Turns}$

Step 2: Calculate Cable Current ($I$ for $N = 4$): $I = \frac{NI}{N} = \frac{3500\text{ A-turns}}{4\text{ turns}} = 875\text{ Amperes}$ Conclusion: Set portable power pack to deliver $875\text{ Amperes}$.

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Effective Inspection Length and Multi-Shot Overlap Requirements

A critical operational rule governing longitudinal coil testing is that the effective magnetic field does NOT extend indefinitely along the part.

       [ 6 to 9 in ]        [ Coil Width ]        [ 6 to 9 in ]
      <-------------> <-------------------------> <------------->
      |=============| | (I)  (I)  (I)  (I)  (I) | |=============|
      |  Effective  | |=========================| |  Effective  |
      |  Field Zone | |  Peak Internal Field    | |  Field Zone |
      +-------------+ +-------------------------+ +-------------+
      <---------------- Total Effective Length ----------------->

The 6 to 9 Inch Effective Field Rule

Under ASTM E1444 and ASTM E709:

• The effective longitudinal magnetic field extends approximately 6 to 9 inches ($150\text{ to }230\text{ mm}$) beyond either face of the coil.
• Beyond this 6 to 9 inch boundary, the axial flux diverges sharply into the air, dropping the tangential field strength below the mandatory 30 Gauss threshold.
• Total Effective Zone Per Shot: $\text{Effective Inspection Length} = \text{Coil Width} + (12 \text{ to } 18\text{ inches})$

Multi-Shot Overlap Requirements for Long Parts

When testing parts whose overall length exceeds the effective inspection zone (e.g., a 48-inch shaft tested in a 6-inch wide coil):

1. Multiple Coil Positions: The coil must be repositioned sequentially along the length of the part.
2. Mandatory Overlap: Adjacent coil positions must overlap by $10%$ to $20%$ of the effective field length (a minimum of 2 to 3 inches).
3. Zero Dead Zones: Inspecting a long shaft with discrete coil shots without overlap leaves intermediate "dead zones" where transverse fatigue cracks remain completely undetected.

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Summary of Level III Coil Magnetization Formulas