---
title: "Boundary condition with μr = 10, find B and H"
subject: "Electromagnetism"
chapter: 3
key: "semester_3/electromagnetism:boundary-mur-10-region-2"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/boundary-mur-10-region-2
site: BCT Engineering
---

# Boundary condition with μr = 10, find B and H

Electromagnetism · Chapter 3: Magnetic field

- Asked in: 2076 Chaitra, 6 marks
- Video: [Boundary Conditions for Magnetic Field - EMF - EMT](https://www.youtube.com/watch?v=ZLOQOL8oKx0) — Padmasri Naban
- 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-2
- 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 2, find B and H in region 2.

## Short answer

H₂ = 5 a_x + 6 a_y + 7 a_z A/m (given). B₂ = μ₀ μ_r2 H₂ = (4π×10⁻⁷)(10)H₂ = 62.8 a_x + 75.4 a_y + 88.0 a_z μT (≈ 6.28×10⁻⁵ a_x + 7.54×10⁻⁵ a_y + 8.80×10⁻⁵ a_z T).

## Answer

Region 1 (y < 0) is air (μ_r1 = 1). Region 2 (y > 0) has μ_r2 = 10. The boundary is the plane y = 0 with normal a_y. The field given is in region 2: H₂ = 5 a_x + 6 a_y + 7 a_z A/m.

#### H in region 2
The field H₂ is given directly: **H₂ = 5 a_x + 6 a_y + 7 a_z A/m**, with |H₂| = √(25+36+49) = 10.49 A/m.

#### B in region 2
```text
B₂ = μ₀ μ_r2 H₂ = (4π×10⁻⁷)(10)(5 a_x + 6 a_y + 7 a_z)
   = 1.2566×10⁻⁵ (5 a_x + 6 a_y + 7 a_z)
   = 62.8 a_x + 75.4 a_y + 88.0 a_z  μT   (μWb/m²)
```
In tesla: B₂ = 6.28×10⁻⁵ a_x + 7.54×10⁻⁵ a_y + 8.80×10⁻⁵ a_z T.

#### Corresponding fields in region 1 (using boundary conditions)
No surface current is stated, so tangential H is continuous and normal B is continuous:

```text
H_t1 = H_t2 = 5 a_x + 7 a_z
B_N1 = B_N2 :  μ₀ H_y1 = 10 μ₀ (6)  →  H_y1 = 60
H₁ = 5 a_x + 60 a_y + 7 a_z  A/m
B₁ = μ₀H₁ = 6.28 a_x + 75.4 a_y + 8.80 a_z  μT
```

**Answer:** H₂ = 5 a_x + 6 a_y + 7 a_z A/m; B₂ = 62.8 a_x + 75.4 a_y + 88.0 a_z μT. (Region 1: H₁ = 5 a_x + 60 a_y + 7 a_z A/m.)

## Check yourself

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

1. B₂ is obtained from H₂ by:
   A. B = μ0 H
   B. B = μ0 μ_r H
   C. B = H/μ
   D. B = μ_r/μ0 H
   Answer: B. B₂ = μ0(10)H₂ = 1.2566×10^−5 (5,6,7) T. (Key point: B in medium)

2. B₂ is approximately:
   A. 6.28 a_x + 7.54 a_y + 8.80 a_z μT
   B. 62.8 a_x + 75.4 a_y + 88.0 a_z μT
   C. 5 a_x + 6 a_y + 7 a_z μT
   D. 628 a_x + 754 a_y + 880 a_z μT
   Answer: B. 1.2566×10^−5 × (5,6,7) = 62.8, 75.4, 88.0 μT. (Key point: Value of B2)

3. Which boundary conditions give H₁?
   A. Tangential H continuous, normal B continuous
   B. Normal H continuous, tangential B continuous
   C. Both H continuous
   D. Both B continuous
   Answer: A. With no surface current, H_t is continuous and B_N is continuous, so H_y1 = 10 × 6 = 60 A/m. (Key point: Boundary condition used for region 1)

## Close variants set in other exams

- [Boundary μr = 10, H given in region 1](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Aboundary-mur-10-region-1-given) — Electromagnetism, chapter 3, asked 2×

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