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

# H in coaxial cable regions

Electromagnetism · Chapter 3: Magnetic field

- Asked in: 2075 Chaitra, 6 marks; 2068 Chaitra, 7 marks; 2067 Shrawan, 6 marks (asked 3 times)
- 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/coax-cable-h-different-regions
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-3

## Question

Derive the expression for magnetic field intensity in different regions due to a co-axial cable carrying uniformly distributed dc current I in the inner conductor and −I in the outer conductor.

## Short answer

Inside inner conductor (ρ<a): H = Iρ/(2πa²) a_φ. Between conductors (a<ρ<b): H = I/(2πρ) a_φ. Inside outer conductor (b<ρ<c): H = I(c²−ρ²)/[2πρ(c²−b²)] a_φ. Outside (ρ>c): H = 0.

## Answer

**Setup.** Let the inner conductor have radius a, carrying a uniformly distributed current I along +z. The outer conductor has inner radius b and outer radius c, and carries −I (uniformly distributed). By symmetry H has only an a_φ component that depends on ρ. Apply Ampere's circuital law ∮H·dl = I_enc on a circle of radius ρ, where ∮H·dl = H_φ(2πρ).

```text
 cross-section
        ___________
      /  .-------.  \      c = outer radius of outer conductor
     |  /  .---.  \  |     b = inner radius of outer conductor
     | |  ( I ⊙ )  | |     a = radius of inner conductor
     |  \  '---'  /  |     inner: current I out of page
      \  '-------'  /      outer: current −I (into page)
        '-----------'
```

#### Region 1: 0 ≤ ρ < a (inside the inner conductor)
The current enclosed is the fraction of the uniform current within radius ρ: I_enc = I (πρ²)/(πa²) = Iρ²/a².

```text
H_φ (2πρ) = Iρ²/a²
H = Iρ/(2πa²)  a_φ
```

#### Region 2: a < ρ < b (the space between conductors)
I_enc = I.

```text
H = I/(2πρ)  a_φ
```

#### Region 3: b < ρ < c (inside the outer conductor)
The enclosed current is I plus the portion of −I within radius ρ:

```text
I_enc = I − I (ρ² − b²)/(c² − b²) = I (c² − ρ²)/(c² − b²)

H = I/(2πρ) · (c² − ρ²)/(c² − b²)  a_φ
```

#### Region 4: ρ > c (outside the cable)
I_enc = I − I = 0, so

```text
H = 0
```

#### Summary
| Region | H (a_φ component) |
|---|---|
| ρ < a | Iρ/(2πa²) |
| a < ρ < b | I/(2πρ) |
| b < ρ < c | I(c² − ρ²)/[2πρ(c² − b²)] |
| ρ > c | 0 |

H rises linearly from zero at the axis to I/(2πa) at ρ = a, falls as 1/ρ in the gap, decreases to zero across the outer conductor, and is zero outside, so the coaxial cable has no external magnetic field (shielding).

## Check yourself

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

1. Which law and path are used to find H in each region?
   A. Gauss's law with a cylinder
   B. Ampere's circuital law with a circular path of radius ρ
   C. Biot-Savart with a sphere
   D. Faraday's law
   Answer: B. By symmetry H is a_φ and depends on ρ, so ∮H·dl = H_φ 2πρ = I_enc. (Key point: Method)

2. For ρ < a, the enclosed current is:
   A. I
   B. Iρ²/a²
   C. Iρ/a
   D. 0
   Answer: B. Uniform current: I_enc = I(πρ²)/(πa²), so H = Iρ/(2πa²). (Key point: Inside inner conductor)

3. For b < ρ < c, I_enc is:
   A. I
   B. I(c² − ρ²)/(c² − b²)
   C. I(ρ² − b²)/(c² − b²)
   D. 0
   Answer: B. I minus the portion of −I within ρ: I − I(ρ²−b²)/(c²−b²) = I(c²−ρ²)/(c²−b²). (Key point: Outer conductor region)

4. Why is H = 0 for ρ > c?
   A. Distance is too large
   B. Enclosed current is I − I = 0, so no external field (shielding)
   C. H falls as 1/ρ² to zero
   D. The outer conductor absorbs H
   Answer: B. Net enclosed current is zero, so the coax has no external magnetic field. (Key point: Outside the cable)

## Related in Electromagnetism

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- [Field between coaxial conductors](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Acoax-cylinders-field-and-line-charge) — Electromagnetism, chapter 2
- [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
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