---
title: "Boundary at z=0, find B2"
subject: "Electromagnetism"
chapter: 3
key: "semester_3/electromagnetism:magnetic-boundary-z0-find-b2"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/magnetic-boundary-z0-find-b2
site: BCT Engineering
---

# Boundary at z=0, find B2

Electromagnetism · Chapter 3: Magnetic field

- Asked in: 2074 Asoj, 8 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/magnetic-boundary-z0-find-b2
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-3

## Question

Consider a boundary at z = 0 for which B1 = 2a_x − 3a_y + a_z mT, μ1 = 4 μH/m (z > 0), μ2 = 7 μH/m (z < 0) and K = 80a_x A/m at z = 0. Find B2.

## Short answer

Normal B continuous: B2z = 1 mT. Tangential H1t = (2a_x - 3a_y)/4e-6 = 500a_x - 750a_y A/m. Using a_z × (H1t - H2t) = 80a_x gives H2t = 500a_x - 670a_y A/m. Then B2t = μ2 H2t = 3.5a_x - 4.69a_y mT. So B2 = 3.5a_x - 4.69a_y + 1a_z mT.

## Answer

The boundary is the plane z = 0, so the normal is a_z and the tangential components are the x and y components. Region 1 (z > 0, μ₁ = 4 μH/m) is above, region 2 (z < 0, μ₂ = 7 μH/m) is below. A surface current K = 80 a_x A/m flows on the boundary. The normal pointing from region 2 to region 1 is a_N12 = a_z.

B₁ = 2 a_x − 3 a_y + 1 a_z mT.

**Step 1: normal component (B continuous).**

```text
B₂z = B₁z = 1 mT
```

**Step 2: tangential H in region 1.**

```text
H₁t = B₁t / μ₁
    = (2 a_x − 3 a_y)×10⁻³ / (4×10⁻⁶)
    = 500 a_x − 750 a_y   A/m
```

**Step 3: apply the boundary condition** a_N12 × (H₁ − H₂) = K. Let H₁t − H₂t = h_x a_x + h_y a_y.

```text
a_z × (h_x a_x + h_y a_y) = h_x a_y − h_y a_x = 80 a_x
⇒ h_x = 0,  h_y = −80
```

So H₂t = H₁t − (0 a_x − 80 a_y):

```text
H₂t = 500 a_x + (−750 + 80) a_y = 500 a_x − 670 a_y   A/m
```

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

```text
B₂t = μ₂ H₂t = 7×10⁻⁶ (500 a_x − 670 a_y)
    = 3.5×10⁻³ a_x − 4.69×10⁻³ a_y   T
```

**Step 5: combine.**

**Answer: B₂ = 3.5 a_x − 4.69 a_y + 1 a_z mT**

## Check yourself

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

1. B₂z equals:
   A. 1 mT, since normal B is continuous
   B. 1.75 mT
   C. 0.57 mT
   D. 0
   Answer: A. Normal B is continuous across the boundary. (Key point: Normal component)

2. H₁t = B₁t/μ₁ =
   A. 500 a_x − 750 a_y A/m
   B. 2 a_x − 3 a_y A/m
   C. 8 a_x − 12 a_y A/m
   D. 500 a_x − 750 a_y mA/m
   Answer: A. (2,−3)×10^−3/(4×10^−6) = (500, −750) A/m. (Key point: Tangential H in region 1)

3. a_z × (H₁t − H₂t) = 80 a_x gives:
   A. H₁t − H₂t = −80 a_y
   B. H₁t − H₂t = 80 a_x
   C. H₁t − H₂t = 80 a_y
   D. No change
   Answer: A. a_z × (h_y a_y) = −h_y a_x, so h_y = −80; hence H₂t = 500 a_x − 670 a_y. (Key point: Surface current jump)

4. B₂ =
   A. 3.5 a_x − 4.69 a_y + 1 a_z mT
   B. 2 a_x − 3 a_y + 1 a_z mT
   C. 3.5 a_x − 5.25 a_y + 1 a_z mT
   D. 3.5 a_x − 4.69 a_y + 1.75 a_z mT
   Answer: A. B₂t = 7×10^−6 (500, −670) = (3.5, −4.69) mT with B₂z = 1 mT. (Key point: Final answer)

## Related in Electromagnetism

- [Boundary condition, find E2](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Aboundary-condition-find-e2-medium) — Electromagnetism, chapter 2
- [Dielectric boundary at y=0, find fields](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Adielectric-boundary-y-d1-find-region2) — Electromagnetism, chapter 2

## Related in other subjects

- [Two-mass system X2/F with F-V analogy (B1, K3 link)](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-fv-m1-m2-b1k3-k1) — Control Systems, chapter 2
- [Two-mass system between walls with F-I analogy](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Atf-fi-m1-m2-k1k2k3-b1b2b3) — Control Systems, chapter 2