---
title: "∇×H in coaxial cable regions"
subject: "Electromagnetism"
chapter: 3
key: "semester_3/electromagnetism:curl-h-in-coax-regions"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/curl-h-in-coax-regions
site: BCT Engineering
---

# ∇×H in coaxial cable regions

Electromagnetism · Chapter 3: Magnetic field

- Asked in: 2067 Mangsir, 10 marks
- Video: [EMF48 Magnetic Field Intensity due to Long Co axial Transmission Line](https://www.youtube.com/watch?v=YhhZzu9XRIY) — Lectures in Electrical Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/curl-h-in-coax-regions
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-3

## Question

For a given co-axial cable with inner conductor of radius 'a', outer conductor of inner radius 'b' and outer radius 'c', with current 'I' in the inner conductor and current '−I' in the outer conductor, determine ∇×H for 0 ≤ r ≤ a, a ≤ r ≤ b, b ≤ r ≤ c.

## Short answer

In a coaxial cable carrying ±I, ∇×H equals the current density: for 0≤r≤a, ∇×H = I/(πa²) a_z; for a≤r≤b, ∇×H = 0; for b≤r≤c, ∇×H = −I/[π(c²−b²)] a_z.

## Answer

By Maxwell's (Ampere's) law for steady fields, ∇×H = J, so we find H from Ampere's circuital law and then take the curl (or directly the current density).

**Set-up:** the inner conductor (radius a) carries +I in a_z, uniformly distributed. The outer conductor (inner radius b, outer radius c) carries −I (return current). By symmetry H = H_φ(r) a_φ, and Ampere's law gives H_φ · 2πr = I_enc.

#### Region 0 ≤ r ≤ a
J = I/(πa²) a_z, so I_enc = I r²/a².

```text
H_φ = I r / (2π a²)
∇×H = (1/r) d(r H_φ)/dr a_z = (1/r) d/dr [ I r² / (2π a²) ] a_z
    = I / (π a²) a_z
```

**∇×H = I/(πa²) a_z**, which equals J, as it must.

#### Region a ≤ r ≤ b
I_enc = I (the whole inner conductor current).

```text
H_φ = I / (2π r)
r H_φ = I / 2π = constant
∇×H = (1/r) d(I/2π)/dr a_z = 0
```

**∇×H = 0** (no current in the dielectric).

#### Region b ≤ r ≤ c
The outer conductor carries −I spread over the area π(c² − b²). The part of it enclosed within radius r is −I (r² − b²)/(c² − b²).

```text
I_enc = I − I (r² − b²)/(c² − b²) = I (c² − r²)/(c² − b²)
H_φ   = I (c² − r²) / [2π r (c² − b²)]
r H_φ = I (c² − r²) / [2π (c² − b²)]
∇×H   = (1/r) d/dr [ I (c² − r²) / (2π (c² − b²)) ] a_z
      = (1/r) × [ −2 r I / (2π (c² − b²)) ] a_z
      = −I / [π (c² − b²)] a_z
```

**∇×H = −I/[π(c² − b²)] a_z**, which equals the current density J of the outer conductor (negative because the current returns in −a_z).

#### Summary

| Region | H_φ | ∇×H |
|---|---|---|
| 0 ≤ r ≤ a | I r / (2πa²) | I/(πa²) a_z |
| a ≤ r ≤ b | I / (2πr) | 0 |
| b ≤ r ≤ c | I (c² − r²) / [2πr(c² − b²)] | −I/[π(c² − b²)] a_z |

For r > c, I_enc = 0, so H = 0 and ∇×H = 0 (the cable has no external field).

## Check yourself

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

1. How is curl H found for the coaxial cable?
   A. curl H = J, with H from Ampere's circuital law (H_phi 2 pi r = I_enc)
   B. curl H = dD/dt
   C. curl H = 0 everywhere
   D. curl H = B / mu0
   Answer: A. For steady fields curl H = J, and symmetry plus Ampere's law gives H_phi. (Key point: Method)

2. For 0 <= r <= a, the enclosed current and curl H are:
   A. I_enc = I, curl H = 0
   B. I_enc = I r^2/a^2, curl H = I/(pi a^2) a_z
   C. I_enc = I r/a, curl H = I/(2 pi a) a_z
   D. I_enc = 0, curl H = 0
   Answer: B. Uniform current gives I_enc = I r^2/a^2, H_phi = Ir/(2 pi a^2), and curl H = J = I/(pi a^2) a_z. (Key point: Inside the inner conductor)

3. For a <= r <= b, curl H is zero because:
   A. H is zero there
   B. r H_phi = I/2 pi is constant, and there is no current density
   C. The outer conductor shields it
   D. I_enc = 0
   Answer: B. H_phi = I/(2 pi r), so r H_phi is constant and its derivative vanishes, matching J = 0 in the dielectric. (Key point: In the dielectric)

4. For b <= r <= c, curl H is:
   A. +I/[pi (c^2 - b^2)] a_z
   B. 0
   C. -I/[pi (c^2 - b^2)] a_z
   D. -I/(pi a^2) a_z
   Answer: C. I_enc = I(c^2 - r^2)/(c^2 - b^2), and the return current density is -I/[pi(c^2-b^2)] a_z. (Key point: In the outer conductor)

## Related in Electromagnetism

- [H in coaxial cable regions](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Acoax-cable-h-different-regions) — Electromagnetism, chapter 3, asked 3×
- [Coaxial cable charge density, D and E](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Acoax-cable-charge-density-d-e) — Electromagnetism, chapter 2
- [Field between coaxial conductors](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Acoax-cylinders-field-and-line-charge) — Electromagnetism, chapter 2
- [Capacitance of coaxial capacitor](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Acoax-capacitance-laplace-cylindrical) — Electromagnetism, chapter 2, asked 2×

## Related in other subjects

- [Optimum conductor radius for minimum cable stress](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Acable-optimum-conductor-radius) — Power System Analysis I, chapter 2
- [Actual grade on inner and outer lanes](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-64p-grade-inner-outer) — Transportation Engineering I, chapter 3
- [VBT for [Fe(CN)₆]³⁻ inner/outer orbital](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Avbt-fecn6) — Engineering Chemistry, chapter 6
- [Inner/outer orbital complexes; VBT of [Fe(CN)₆]³⁻](https://www.bctengineer.com/question.md?key=semester_2%2Fengineering_chemistry%3Ainner-outer-vbt-fecn6) — Engineering Chemistry, chapter 6