 Home - General Resources -Eddy Current Inspection Formula- Standard Depth of Penetration
When Permeability in Henries/meter and Conductivity in Siemens/meter Are Known

When electrical conductivity
(siemens/meter) and absolute permeability are known. Where:

 d = Standard Depth of Penetration (m) p = 3.14 f = Test Frequency (Hz) m = Magnetic Permeability (H/m) s = Electrical Conductivity (siemens/meter)

Eddy currents are more concentrated at the surface and decrease in intensity with distance below the surface of the metal. This effect is known as the "skin effect." The depth at which eddy current density has decreased to 1/e, or about 37% of the surface density, is called the standard depth of penetration (d). Although eddy currents penetrate deeper than one standard depth of penetration, they decrease rapidly with depth. At two standard depths of penetration (2d), the eddy current density has decreased to 1/e squared or 13.5% of the surface density. At three depths (3d), the eddy current density is down to only 5% of the surface density.

The depth of penetration is dependent of test drive frequency, the test material's conductivity and magnetic permeability. The depth of penetration decreases with increasing frequency, conductivity and permeability. It is important to know the standard depth of penetration because for some testing (like flaw detection) the inspection should be conducted at a frequency that places the depth of any likely flaws at 1d or less where eddy currents are strong. When making conductivity tests, the sample should be at least 3d thick so that changes in the thickness of the sample do not affect the measurements. When electrical conductivity in % IACS and permeability in H/mm are known, the standard depth of penetration can be calculated using the equation to the right.

Working with this Equation
In many places, magnetic permeability is presented as relative permeability. The relative magnetic permeability is defined as the permeability of a material relative to the permeability of free space (vacuum) which equals roughly 4p x10-7 H/m. Therefore, it may be necessary to multiply the given relative permeability value by 4p x10-7 H/m or 1.257 x10-6 H/m to get the absolute permeability value which is needed in this equation. Since the magnetic permeability of nonmagnetic materials is close to that of free space, the free space value is commonly used.

Also, electrical conductivity values are often expressed as a percent of the conductive of pure, annealed copper measured at 25oC. At this temperature, the conductivity of pure copper is 5.8 x107 siemens/meter or 100% IACS. IACS is an acronym for International Annealed Copper Standard.   A material that has a conductivity of 3.2x107 siemens/meter can be also be expressed as 57% IACS and visa versa.

Example 1

What is the standard depth of penetration when performing an eddy current inspection on a piece of 304 Corrosion Resistant Steel? The material has a relative permeability of 1.05 and a conductivity of 2.5% IACS. The frequency used to drive the eddy current probe is 50 kHz.

First, since relative permeability is given, it must be converted to an absolute permeability value.
Given the equation and the permeability of free space (u0) of 1.257x10-6 H/m
Rearranging this equation to solve for absolute permeability results in:  Plugging the given values into the equation produces an absolute permeability value. Since conductivity is expressed in %IACS, it must be converted to siemens/meter. Then plug-in the values and solve the equation.   Tip: You may have noticed that there is a term with 10-6 and a term with 106 in the denominator of the equation and they can be canceled to make the equation easier to solve.

Example 2

Determine the frequency needed to achieve a depth of penetration of 2mm in 7075-T76 aluminum plate.

Aluminum is nonmagnetic so its relative permeability is one and its permeability is the same as that of free space (1.257x10-6 H/mm). With a measurement or from the Electrical Conductivity and Resistivity Property Tables, determine the conductivity value for the material. The value given is 38.5%IACS is given in the table so it must be converted to Siemens/m. The depth of penetration equation must be rearranged to solve for frequency (f). First square both sides to eliminate the square root. Multiply both sides by pfms. Then divide both sides by to isolate d2pms. Plug in the known values and solve. Notice that the depth of penetration (2mm) had to be converted to meters (0.002m) because d must be in meters for this equation to work.   