---
title: "Ampere's law and boundary H components"
subject: "Electromagnetism"
chapter: 3
key: "semester_3/electromagnetism:amperes-law-surface-current-boundary"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/amperes-law-surface-current-boundary
site: BCT Engineering
---

# Ampere's law and boundary H components

Electromagnetism · Chapter 3: Magnetic field

- Asked in: 2076 Asoj, 10 marks
- Video: [Boundary Conditions for Magnetic Field | Magnetism | Electromagnetics Theory](https://www.youtube.com/watch?v=5u-xDWpRULA) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/amperes-law-surface-current-boundary
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-3

## Question

State Ampere's circuital law. Let the permeability be 5 μH/m in region A where x < 0 and 20 μH/m in region B where x > 0. If there is a surface current K = 150a_y − 200a_z A/m at x = 0, and H_A = 300a_x − 400a_y + 500a_z A/m, find: (a) |H_LA| (b) |H_NA| (c) |H_LB| (d) |H_NB|.

## Short answer

Ampere's law: ∮H·dl = I_enc. Boundary conditions at x=0: (a) |H_LA| = 640.3 A/m, (b) |H_NA| = 300 A/m, (c) |H_LB| = 694.6 A/m, (d) |H_NB| = 75 A/m. (Normal B continuous; tangential H jump equals K: a_x × (H_B − H_A) = K.)

## Answer

**Ampere's circuital law:** the line integral of H around any closed path equals the current enclosed by the path, ∮H·dl = I_enc (point form ∇×H = J).

#### Boundary conditions used
The boundary is the plane x = 0, so the normal is a_x and tangential components are the a_y and a_z components. Region A (x < 0): μ_A = 5 μH/m. Region B (x > 0): μ_B = 20 μH/m. K = 150 a_y − 200 a_z A/m. H_A = 300 a_x − 400 a_y + 500 a_z A/m.

- Normal B is continuous: μ_A H_NA = μ_B H_NB.
- Tangential H jumps by the surface current: a_N12 × (H_B − H_A) = K, with a_N12 = a_x pointing from A to B.

#### (a) |H_LA| (tangential component in A)
```text
H_LA = −400 a_y + 500 a_z
|H_LA| = √(400² + 500²) = √410000 = 640.3 A/m
```

#### (b) |H_NA| (normal component in A)
```text
H_NA = 300 a_x ,  |H_NA| = 300 A/m
```

#### (d) |H_NB| (normal component in B)
```text
μ_B H_NB = μ_A H_NA
H_NB = (5/20)(300) = 75 a_x ,  |H_NB| = 75 A/m
```

#### (c) |H_LB| (tangential component in B)
Let H_LB − H_LA = ΔH_y a_y + ΔH_z a_z. Then

```text
a_x × (ΔH_y a_y + ΔH_z a_z) = ΔH_y a_z − ΔH_z a_y = 150 a_y − 200 a_z
→ ΔH_z = −150 ,  ΔH_y = −200

H_LB = (−400 − 200) a_y + (500 − 150) a_z = −600 a_y + 350 a_z
|H_LB| = √(600² + 350²) = √482500 = 694.6 A/m
```

#### Result
| Quantity | Value |
|---|---|
| (a) \|H_LA\| | 640.3 A/m |
| (b) \|H_NA\| | 300 A/m |
| (c) \|H_LB\| | 694.6 A/m |
| (d) \|H_NB\| | 75 A/m |

Hence H_B = 75 a_x − 600 a_y + 350 a_z A/m.

## Check yourself

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

1. Which relation gives |H_NB|?
   A. H_NB = H_NA
   B. μ_A H_NA = μ_B H_NB
   C. H_NB = K
   D. μ_B H_NA = μ_A H_NB
   Answer: B. Normal B is continuous: H_NB = (5/20)(300) = 75 A/m. (Key point: Normal component)

2. |H_LA| for H_A = 300 a_x − 400 a_y + 500 a_z (normal a_x) is:
   A. 640.3 A/m
   B. 300 A/m
   C. 707 A/m
   D. 900 A/m
   Answer: A. Tangential part is −400 a_y + 500 a_z, magnitude √410000 = 640.3 A/m. (Key point: Tangential in A)

3. The tangential jump condition used is:
   A. H_LB = H_LA
   B. a_N12 × (H_B − H_A) = K
   C. B_LB = B_LA
   D. a_N × K = H
   Answer: B. Tangential H changes by the surface current K. (Key point: Surface current condition)

4. |H_LB| is:
   A. 640.3 A/m
   B. 694.6 A/m
   C. 350 A/m
   D. 600 A/m
   Answer: B. Solving gives H_LB = −600 a_y + 350 a_z, magnitude √482500 = 694.6 A/m. (Key point: Tangential in B)

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