Why Does Copper at 1 GHz Carry Current in Only a 2 µm Layer?
Because an alternating field drives eddy currents inside the conductor that oppose it, so the field and the current density decay exponentially from the surface inwards, with a characteristic depth δ = 1/√(πfμσ) that shrinks as the frequency rises. For copper (σ = 5.8 × 10⁷ S/m, non-magnetic) δ is 2.09 µm at 1 GHz, compared with 66.1 µm at 1 MHz and 9.35 mm at 50 Hz.
What you'll learn
- What the skin depth measures and the assumptions behind its formula
- How δ = 1/√(πfμσ) falls out of the wave equation in a good conductor
- A worked calculation for copper at 1 GHz, plus a table from 50 Hz to 10 GHz
- How fast the current density falls below the surface
- What skin depth does to the resistance of a real wire
Conventions and assumptions
A good conductor, meaning the conduction current far exceeds the displacement current: σ ≫ ωε. For copper at 1 GHz σ/(ωε0) is about 10⁹, so this holds easily. δ is defined as the depth at which the field amplitude has fallen to 1/e (36.8%) of its surface value. Copper is taken as σ = 5.8 × 10⁷ S/m, the standard textbook value for annealed copper at room temperature, with μ = μ0 = 4π × 10⁻⁷ H/m.
Where does δ = 1/√(πfμσ) come from?
In a conductor Maxwell's equations reduce, for a sinusoidal field with e^(jωt) time dependence, to a wave equation with propagation constant γ = √(jωμ(σ + jωε)). When σ ≫ ωε the jωε term can be dropped, leaving γ ≈ √(jωμσ) = (1 + j)√(ωμσ/2). The real part is the attenuation constant α, so the field varies as e^(−αz) with depth z, and the depth where it falls to 1/e is δ = 1/α = √(2/(ωμσ)). Substituting ω = 2πf gives the more common form δ = 1/√(πfμσ). The two forms give the same number; mixing ω and f between them is the classic slip.
Worked example: copper at 1 GHz
- Multiply out the denominator: πfμσ = π × 10⁹ × 4π × 10⁻⁷ × 5.8 × 10⁷ = 2.290 × 10¹¹ m⁻².
- Take the square root: 4.785 × 10⁵ m⁻¹.
- Invert it: δ = 2.09 × 10⁻⁶ m, about 2.09 µm.
- Scale to other frequencies with 1/√f: at 1 MHz, a thousand times lower in frequency, δ is √1000 ≈ 31.6 times larger, 66.1 µm.
| Frequency | Skin depth δ |
|---|---|
| 50 Hz | 9.35 mm |
| 1 kHz | 2.09 mm |
| 1 MHz | 66.1 µm |
| 100 MHz | 6.61 µm |
| 1 GHz | 2.09 µm |
| 10 GHz | 0.661 µm |
Does the good-conductor assumption hold across the table?
For copper, yes, by a huge margin: σ/(ωε0) is about 2 × 10¹⁶ at 50 Hz and still about 10⁸ at 10 GHz, so the displacement current is negligible at every frequency in the table. It is not automatic for every material. Taking illustrative textbook values for seawater, σ = 4 S/m and a relative permittivity of 81, conduction and displacement currents become equal at f = σ/(2πε) ≈ 888 MHz. Around and above that frequency seawater is not a good conductor and the simple formula for δ no longer applies; the full propagation constant has to be used instead.
How fast does the current fall below the surface?
The current density follows the field: J(z) = J0 e^(−z/δ) in magnitude, with its phase also lagging by z/δ radians. At one skin depth it is 36.8% of the surface value, at three skin depths 5.0%, and at five skin depths 0.67%. Power density goes as the square, so at one skin depth it is already down to 13.5%. Current does flow below δ; the skin depth is simply the distance over which it shrinks by a factor of e. For resistance calculations, though, it is exactly equivalent to imagine a uniform current flowing in a surface layer one δ thick.
What does this do to a wire's resistance?
Take a copper wire 1 mm in diameter (radius 0.5 mm). Its DC resistance is 1/(σπa²) = 21.95 mΩ per metre. At 1 GHz the current effectively flows only in a layer 2.09 µm thick around the circumference, so the resistance is roughly 1/(σ × 2πa × δ) = 2.63 Ω per metre: about 120 times the DC value, matching the rule of thumb Rac/Rdc ≈ a/(2δ) for a radius much larger than δ. The same idea gives the surface resistance Rs = 1/(σδ), which for copper at 1 GHz is 8.25 mΩ per square. This is why high-frequency conductors are often silver-plated (only the surface carries current), why hollow tubes work as well as solid rods, and why stranded litz wire, made of many thin insulated strands, helps at lower frequencies where the strands can be thinner than δ.
Common mistakes
- Using ω in 1/√(πfμσ) or f in √(2/(ωμσ)). Each form needs its own variable.
- Dropping μ0, or using a value of μ for a magnetic material without saying so: a large μr makes δ much smaller.
- Saying no current flows below one skin depth. It falls to 36.8%, not to zero.
- Applying the good-conductor formula to a lossy dielectric or seawater at high frequency, where σ ≫ ωε may not hold.
- Mixing units: δ in metres comes out of SI inputs; at microwave frequencies it is micrometres, not millimetres.
Check your understanding
- Q1. Copper's skin depth at 1 MHz is 66.1 µm. What is it at 4 MHz? Answer: four times the frequency halves δ, so 33.0 µm.
- Q2. Taking aluminium as σ = 3.5 × 10⁷ S/m, what is its skin depth at 1 GHz? Answer: δ = 1/√(π × 10⁹ × 4π × 10⁻⁷ × 3.5 × 10⁷) = 2.69 µm.
- Q3. At what frequency is copper's skin depth 10 µm? Answer: f = 1/(πμ0σδ²) = 43.7 MHz.
Related
Official source
Last verified
The GATE 2027 EC syllabus lists propagation through various media and skin depth under plane waves in Section 8, Electromagnetics (checked against the official PDF from IIT Madras on 10 October 2026). Every skin depth, resistance and decay figure was computed in code; the conductivities are standard textbook values, not measurements.