---
title: "Loss tangent, α, β, η at 5 MHz"
subject: "Electromagnetism"
chapter: 4
key: "semester_3/electromagnetism:non-magnetic-material-loss-tangent-5mhz"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/non-magnetic-material-loss-tangent-5mhz
site: BCT Engineering
---

# Loss tangent, α, β, η at 5 MHz

Electromagnetism · Chapter 4: Wave equation and Wave propagation

- Asked in: 2068 Baisakh, 8 marks
- Video: [Electromagnetic Waves Explained: Free Space and Medium Analysis](https://www.youtube.com/watch?v=nEVlWCEIDIA) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/non-magnetic-material-loss-tangent-5mhz
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-4

## Question

For a non-magnetic material having ε_r = 2.25 and σ = 10⁻⁴ mho/m, find the numeric values at 5 MHz for: (a) the loss tangent (b) the attenuation constant (c) the phase constant (d) the intrinsic impedance.

## Short answer

(a) loss tangent ≈ 0.16. (b) α ≈ 0.0125 Np/m. (c) β ≈ 0.158 rad/m. (d) η ≈ 249.6∠4.54° Ω.

## Answer

Given: ε_r = 2.25, σ = 10⁻⁴ S/m, μ = μ₀ (non-magnetic), f = 5 MHz.

```text
ω  = 2π × 5×10⁶ = 3.1416×10⁷ rad/s
ε  = 2.25 × 8.854×10⁻¹² = 1.9922×10⁻¹¹ F/m
μ₀ = 4π×10⁻⁷ H/m
```

#### (a) Loss tangent

```text
tan δ = σ/(ωε) = 10⁻⁴ / (3.1416×10⁷ × 1.9922×10⁻¹¹)
      = 10⁻⁴ / 6.2587×10⁻⁴
      = 0.160
```

**Answer (a): tan δ ≈ 0.16.** Since 0.16 ≪ 1, the medium is a low-loss dielectric.

Common factor: ω√(με/2) = 3.1416×10⁷ × √(4π×10⁻⁷ × 1.9922×10⁻¹¹ / 2) = 0.1111 (used below).

#### (b) Attenuation constant

```text
α = ω√(με/2) · √(√(1 + tan²δ) − 1)
√(1 + 0.02553) = 1.01269 → 1.01269 − 1 = 0.01269 → √ = 0.11265
α = 0.1111 × 0.11265 = 0.0125 Np/m
```

Check with the low-loss formula α ≈ (σ/2)√(μ/ε) = 0.5×10⁻⁴×√(4π×10⁻⁷/1.9922×10⁻¹¹) = 0.0126 Np/m.

**Answer (b): α ≈ 0.0125 Np/m.**

#### (c) Phase constant

```text
β = ω√(με/2) · √(√(1 + tan²δ) + 1)
  = 0.1111 × √(2.01269) = 0.1111 × 1.41869
  = 0.158 rad/m
```

**Answer (c): β ≈ 0.158 rad/m.**

#### (d) Intrinsic impedance

```text
|η| = √(μ/ε) / (1 + tan²δ)^(1/4)
√(μ/ε) = √(4π×10⁻⁷ / 1.9922×10⁻¹¹) = 251.2 Ω
(1.02553)^(1/4) = 1.00634
|η| = 251.2 / 1.00634 = 249.6 Ω
θ_η = ½ tan⁻¹(tan δ) = ½ × 9.08° = 4.54°
```

**Answer (d): η ≈ 249.6∠4.54° Ω** (about 250 Ω).

## Check yourself

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

1. For eps_r = 2.25, sigma = 1e-4 S/m at 5 MHz, how is tan delta computed and what is it?
   A. sigma/(w eps) = 0.16, so a low-loss dielectric
   B. w eps/sigma = 6.26, so a good conductor
   C. sigma/(w eps) = 1.6, so a quasi-conductor
   D. sigma/(w eps) = 0.016, so a lossless medium
   Answer: A. w = 3.14x10^7, eps = 1.99x10^-11, w*eps = 6.26x10^-4, so tan delta = 0.16 which is much less than 1. (Key point: Loss tangent)

2. What is alpha?
   A. About 0.125 Np/m
   B. About 0.158 Np/m
   C. About 0.0063 Np/m
   D. About 0.0125 Np/m, close to the low-loss value (sigma/2) sqrt(mu/eps)
   Answer: D. alpha = w sqrt(mu eps/2) sqrt(sqrt(1 + tan^2 d) - 1) = 0.1111 x 0.11265 = 0.0125 Np/m. (Key point: Attenuation constant)

3. Which expression gives beta?
   A. w sqrt(mu eps/2) sqrt(sqrt(1 + tan^2 d) - 1) = 0.0125 rad/m
   B. w sqrt(mu eps) / (1 + tan d) = 0.14 rad/m
   C. w sqrt(mu eps/2) sqrt(sqrt(1 + tan^2 d) + 1) = 0.158 rad/m
   D. sigma/2 sqrt(mu/eps) = 0.0126 rad/m
   Answer: C. beta uses the plus sign inside the bracket; here it is nearly w sqrt(mu eps). (Key point: Phase constant)

4. What is the phase angle of the intrinsic impedance?
   A. The full loss angle, 9.08 degrees
   B. Half of the loss angle: (1/2) arctan(tan d) = 4.54 degrees
   C. 0 degrees, since mu is real
   D. 45 degrees
   Answer: B. theta_eta = (1/2) arctan(sigma/w eps) = 4.54 degrees, giving eta of about 249.6 angle 4.54 ohm. (Key point: Intrinsic impedance)

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