---
title: "Derive α and β for lossy dielectric"
subject: "Electromagnetism"
chapter: 4
key: "semester_3/electromagnetism:lossy-dielectric-alpha-beta-derive"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/lossy-dielectric-alpha-beta-derive
site: BCT Engineering
---

# Derive α and β for lossy dielectric

Electromagnetism · Chapter 4: Wave equation and Wave propagation

- Asked in: 2067 Mangsir, 8 marks
- Video: [4.4 WAVE PROPAGATION IN LOSSY DIELECTRICS for I.E.S./G.A.T.E.](https://www.youtube.com/watch?v=1gZuY_xxbQs) — EPOV CHANNEL
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/lossy-dielectric-alpha-beta-derive
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-4

## Question

Consider a wave propagating in a lossy dielectric with propagation constant γ = α + jβ. Derive expressions for α and β if the medium is characterized by permittivity ε, permeability μ and conductivity σ.

## Short answer

From γ² = jωμ(σ+jωε) and γ = α+jβ, equate real and imaginary parts to obtain α = ω√(με/2)[√(1+(σ/ωε)²)−1]^(1/2) Np/m and β = ω√(με/2)[√(1+(σ/ωε)²)+1]^(1/2) rad/m.

## Answer

For a medium with ε, μ and σ, Maxwell's equations in phasor form give the wave equation ∇²E = γ²E, where

γ² = jωμ(σ + jωε)

(σ + jωε is the total conduction plus displacement current term.) Put γ = α + jβ.

#### Step 1: Expand γ²

```text
γ² = (α + jβ)² = α² − β² + j2αβ
γ² = jωμσ − ω²με
```

Equate real and imaginary parts:

```text
(1)  α² − β² = −ω²με
(2)  2αβ     = ωμσ
```

#### Step 2: Use the magnitude of γ²

```text
|γ²| = α² + β² = ωμ √(σ² + ω²ε²)       (3)
```

Taking (3) and (1) together:

```text
Adding (3) and (1):    2α² = ωμ√(σ² + ω²ε²) − ω²με
Subtracting (1) from (3): 2β² = ωμ√(σ² + ω²ε²) + ω²με
```

#### Step 3: Final expressions

```text
α = ω √(με/2) · [ √(1 + (σ/ωε)²) − 1 ]^(1/2)   Np/m

β = ω √(με/2) · [ √(1 + (σ/ωε)²) + 1 ]^(1/2)   rad/m
```

Here σ/ωε = tan δ is the loss tangent.

#### Special cases

- Lossless dielectric (σ = 0): α = 0, β = ω√(με).
- Low-loss (σ ≪ ωε): α ≈ (σ/2)√(μ/ε), β ≈ ω√(με).
- Good conductor (σ ≫ ωε): α = β ≈ √(πfμσ).

## Check yourself

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

1. What is gamma^2 for a medium with sigma, eps, mu?
   A. jw mu (sigma + jw eps)
   B. jw mu (sigma - jw eps)
   C. w^2 mu eps
   D. jw eps (sigma + jw mu)
   Answer: A. From Maxwell's equations gamma^2 = jw mu (sigma + jw eps) = jw mu sigma - w^2 mu eps. (Key point: Starting equation)

2. Equating real and imaginary parts of (alpha + j beta)^2 gives which pair?
   A. alpha^2 + beta^2 = -w^2 mu eps and alpha beta = w mu sigma
   B. alpha^2 - beta^2 = w mu sigma and 2 alpha beta = -w^2 mu eps
   C. alpha^2 - beta^2 = w^2 mu eps and 2 alpha beta = w mu sigma
   D. alpha^2 - beta^2 = -w^2 mu eps and 2 alpha beta = w mu sigma
   Answer: D. Real part: alpha^2 - beta^2 = -w^2 mu eps. Imaginary part: 2 alpha beta = w mu sigma. (Key point: Real and imaginary parts)

3. Why is the magnitude |gamma^2| = alpha^2 + beta^2 = w mu sqrt(sigma^2 + w^2 eps^2) used?
   A. It gives alpha and beta directly
   B. It replaces the imaginary-part equation
   C. Adding it to (1) isolates 2 alpha^2 and subtracting (1) isolates 2 beta^2
   D. It shows alpha = beta for every medium
   Answer: C. Adding gives 2 alpha^2 = w mu sqrt(...) - w^2 mu eps; subtracting gives 2 beta^2 = w mu sqrt(...) + w^2 mu eps. (Key point: Using the magnitude)

4. In a good conductor (sigma >> w eps), what do the results reduce to?
   A. alpha = 0, beta = w sqrt(mu eps)
   B. alpha = beta = sqrt(pi f mu sigma)
   C. alpha = (sigma/2) sqrt(mu/eps), beta = w sqrt(mu eps)
   D. alpha = w sqrt(mu eps), beta = 0
   Answer: B. For sigma >> w eps, alpha = beta = sqrt(pi f mu sigma). The second option is the low-loss case. (Key point: Special cases)

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