---
title: "Skin depth and power in dissipative medium"
subject: "Electromagnetism"
chapter: 4
key: "semester_3/electromagnetism:skin-depth-time-average-power-density"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/skin-depth-time-average-power-density
site: BCT Engineering
---

# Skin depth and power in dissipative medium

Electromagnetism · Chapter 4: Wave equation and Wave propagation

- Asked in: 2068 Baisakh, 7 marks
- Video: [skin depth in conductors detailed concept and derivation Lec 07 btech physics](https://www.youtube.com/watch?v=2Cvkc4viJ8s) — RGSClassesEngineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/skin-depth-time-average-power-density
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-4

## Question

Explain the term skin depth. Using Poynting vector, deduce the time average power density for a dissipative medium.

## Short answer

Skin depth δ is the distance over which the wave amplitude decays to 1/e of its surface value, given by δ = 1/α (≈ 1/√(πfμσ) for a good conductor). The time-average Poynting vector in a dissipative medium is S_av = (E₀²/(2|η|)) e^(−2αz) cos θ_η a_z W/m², obtained by averaging the instantaneous Poynting vector over one period.

## Answer

**Skin depth (δ)** is the distance a wave travels in a conducting (dissipative) medium over which its amplitude falls to 1/e (about 36.8 %) of its value at the surface. It equals the reciprocal of the attenuation constant: δ = 1/α.

For a good conductor (σ ≫ ωε), α = β = √(πfμσ), so

δ = 1/√(πfμσ)

Skin depth decreases as frequency, permeability or conductivity increases. Example: copper at 1 MHz has δ ≈ 66 μm, so high-frequency current flows only in a thin surface layer.

#### Time-average power density in a dissipative medium

Let the wave travel in +z with E along a_x, in a medium with γ = α + jβ and intrinsic impedance η = |η|∠θ_η (θ_η is 0 for a lossless medium, up to 45° for a good conductor).

```text
E = E₀ e^(−αz) cos(ωt − βz) a_x
H = (E₀/|η|) e^(−αz) cos(ωt − βz − θ_η) a_y
```

Instantaneous Poynting vector:

```text
S = E × H
  = (E₀²/|η|) e^(−2αz) cos(ωt − βz) cos(ωt − βz − θ_η) a_z
```

Use cos A cos B = ½[cos(A − B) + cos(A + B)]:

```text
S = (E₀²/2|η|) e^(−2αz) [cos θ_η + cos(2ωt − 2βz − θ_η)] a_z
```

The second term averages to zero over one period, so

**S_av = (E₀² / 2|η|) e^(−2αz) cos θ_η  a_z  W/m²**

The same result follows from the phasor form S_av = ½ Re(E_s × H_s*).

Remarks:
- Power density falls as e^(−2αz), twice as fast as the field amplitude.
- At z = δ the power density is e⁻² ≈ 13.5 % of its surface value.
- The factor cos θ_η is the "power factor" of the medium; for a lossless medium θ_η = 0 and S_av = E₀²/2η.
- The power lost per unit volume appears as ohmic heat (σE²).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is skin depth defined?
   A. delta = 1/alpha, where the amplitude falls to 1/e of the surface value
   B. delta = 1/beta where the power falls to zero
   C. delta = 2 pi/alpha
   D. delta = alpha, the amplitude at the surface
   Answer: A. Skin depth is the reciprocal of the attenuation constant. (Key point: Skin depth)

2. How does the time-average term appear in S = (E0^2/|eta|) e^(-2az) cos(wt - bz) cos(wt - bz - theta)?
   A. The product of cosines averages to zero
   B. The cos(A+B) term is the constant and cos(A-B) averages out
   C. Averaging doubles the 2w term
   D. cos A cos B = (1/2)[cos(A-B) + cos(A+B)] gives a constant term plus a 2w term that averages to zero
   Answer: D. A - B = theta is constant, so cos(theta) survives; cos(2wt - 2bz - theta) averages to zero. (Key point: Instantaneous Poynting vector)

3. What is the time-average power density in a dissipative medium?
   A. (E0^2/2|eta|) e^(-alpha z)
   B. (E0^2/|eta|) e^(-2 alpha z) sin(theta_eta)
   C. (E0^2/2|eta|) e^(-2 alpha z) cos(theta_eta)
   D. (E0^2/2|eta|) cos(theta_eta), constant with z
   Answer: C. Power decays as e^(-2az) and is reduced by the power factor cos(theta_eta). (Key point: Result)

4. What fraction of the surface power density remains at z = delta?
   A. e^-1, about 37 %
   B. e^-2, about 13.5 %
   C. 50 %
   D. e^-4, about 1.8 %
   Answer: B. Power goes as e^(-2 alpha z); with alpha*delta = 1 this is e^-2. (Key point: Power at one skin depth)

## Related in Electromagnetism

- [Short note: Skin depth](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Askin-depth-short-note) — Electromagnetism, chapter 4
- [Poynting numerical, wave in free space](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Apoynting-numerical-em-wave-free-space) — Electromagnetism, chapter 4
- [Average J over surface and ∂ρv/∂t](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Aaverage-j-charge-rate-change) — Electromagnetism, chapter 2
- [Displacement current density in transformer](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Adisplacement-current-density-transformer) — Electromagnetism, chapter 3

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