---
title: "Boundary μr = 10, H given in region 1"
subject: "Electromagnetism"
chapter: 3
key: "semester_3/electromagnetism:boundary-mur-10-region-1-given"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/boundary-mur-10-region-1-given
site: BCT Engineering
---

# Boundary μr = 10, H given in region 1

Electromagnetism · Chapter 3: Magnetic field

- Asked in: 2073 Shrawan, 3+3 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/boundary-mur-10-region-1-given
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-3

## Question

The region y < 0 (region 1) is air and y > 0 (region 2) has μ_r = 10. If there is a uniform magnetic field H = 5a_x + 6a_y + 7a_z A/m in region 1, find B and H in region 2.

## Short answer

Tangential H continuous: H2x=5, H2z=7 A/m. Normal B continuous: μ0*6 = 10μ0*H2y, so H2y=0.6 A/m. H2 = 5a_x + 0.6a_y + 7a_z A/m. B2 = μ0*10*H2 = 62.8a_x + 7.54a_y + 88.0a_z μT.

## Answer

Region 1 (y < 0) is air (μ_r1 = 1) and region 2 (y > 0) has μ_r2 = 10. The interface is the plane y = 0, so the normal is a_y and the tangential components are x and z. There is no surface current stated (K = 0).

H₁ = 5 a_x + 6 a_y + 7 a_z A/m.

**Step 1: B in region 1 (for reference).**

```text
B₁ = μ₀ H₁ = 4π×10⁻⁷ (5, 6, 7) = 6.283 a_x + 7.540 a_y + 8.796 a_z  μT
```

**Step 2: boundary conditions.**
- Tangential H is continuous (K = 0): H₂x = 5, H₂z = 7 A/m.
- Normal B is continuous: μ₁H₁y = μ₂H₂y

```text
H₂y = (μ₀ × 6) / (10 μ₀) = 0.6 A/m
```

**Step 3: H in region 2.**

```text
H₂ = 5 a_x + 0.6 a_y + 7 a_z   A/m
```

**Step 4: B in region 2.**

```text
B₂ = μ₀ μ_r2 H₂ = 4π×10⁻⁷ × 10 × (5, 0.6, 7)
   = 62.83 a_x + 7.54 a_y + 87.96 a_z   μT
```

**Answer: H₂ = 5 a_x + 0.6 a_y + 7 a_z A/m; B₂ = 62.8 a_x + 7.54 a_y + 88.0 a_z μT** (note B_y is unchanged at 7.54 μT).

## Check yourself

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

1. At the interface y = 0 with no surface current, which components of H are continuous?
   A. The y component (normal)
   B. The x and z components (tangential)
   C. All three components
   D. None of them
   Answer: B. With K = 0, tangential H is continuous, so H2x = 5 and H2z = 7 A/m. (Key point: Tangential H)

2. How is H2y found from H1y = 6 A/m?
   A. H2y = 6 because H is continuous
   B. H2y = 60 from mu_r2 times H1y
   C. H2y = 0.6 from mu1 H1y = mu2 H2y
   D. H2y = 0 because region 2 is magnetic
   Answer: C. Normal B is continuous, so mu0 (6) = 10 mu0 H2y, giving H2y = 0.6 A/m. (Key point: Normal B continuity)

3. Which is B2 in region 2?
   A. 62.8 a_x + 7.54 a_y + 88.0 uT
   B. 6.28 a_x + 7.54 a_y + 8.80 uT
   C. 62.8 a_x + 75.4 a_y + 88.0 uT
   D. 6.28 a_x + 0.754 a_y + 8.80 uT
   Answer: A. B2 = mu0 mu_r2 H2 = 4pi x 10^-7 x 10 x (5, 0.6, 7); B_y stays 7.54 uT because normal B is continuous. (Key point: B in region 2)

## Close variants set in other exams

- [Boundary condition with μr = 10, find B and H](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Aboundary-mur-10-region-2) — Electromagnetism, chapter 3, asked 2×

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