---
title: "Derive magnetic boundary conditions"
subject: "Electromagnetism"
chapter: 3
key: "semester_3/electromagnetism:magnetic-boundary-conditions-derive"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/magnetic-boundary-conditions-derive
site: BCT Engineering
---

# Derive magnetic boundary conditions

Electromagnetism · Chapter 3: Magnetic field

- Asked in: 2075 Asoj, 5 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-conditions-derive
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-3

## Question

Derive the expression for magnetic boundary conditions.

## Short answer

Normal B is continuous: B1n = B2n. Tangential H satisfies a_N12 × (H1 - H2) = K, where K is surface current density. If K=0, H1t = H2t. These give refraction law tan θ1/tan θ2 = μ1/μ2.

## Answer

Boundary conditions give the relations between the field components just on either side of the interface between two media. Take region 1 (μ₁) and region 2 (μ₂), with a_N pointing from region 2 into region 1. We resolve every field into a normal and a tangential part and use the two integral laws.

#### Normal component of B
Use a small cylindrical pillbox across the boundary, with top and bottom faces of area ΔS in each medium and height Δh → 0. The magnetic Gauss law is ∮ B·dS = 0:

```text
B₁n ΔS − B₂n ΔS + (side contribution → 0) = 0
⇒ B₁n = B₂n
⇒ μ₁H₁n = μ₂H₂n
```

The normal component of **B is continuous**; the normal component of H is not (H₁n/H₂n = μ₂/μ₁).

#### Tangential component of H
Take a small rectangular loop abcda of width Δw straddling the boundary, with height Δh → 0. Let the surface current density at the boundary be K (A/m), perpendicular to the loop plane. Ampere's law ∮ H·dl = I_enc:

```text
H₁t Δw − H₂t Δw + (sides → 0) = K Δw
⇒ H₁t − H₂t = K
```

In vector form: **a_N12 × (H₁ − H₂) = K**, where a_N12 is the unit normal from region 2 to region 1.

Corresponding statement for B: B₁t/μ₁ − B₂t/μ₂ = K.

#### Special case
If there is no surface current (K = 0, true for most real interfaces, since conductivities are finite):

```text
H₁t = H₂t         (tangential H continuous)
B₁t / μ₁ = B₂t / μ₂
```

#### Summary

| Component | Condition |
|---|---|
| Normal B | B₁n = B₂n |
| Tangential H | H₁t − H₂t = K (zero if K = 0) |

These conditions also give the refraction law of magnetic field lines: tan θ₁ / tan θ₂ = μ₁/μ₂.

## Check yourself

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

1. A pillbox and ∮B·dS = 0 give:
   A. H₁n = H₂n
   B. B₁n = B₂n
   C. B₁t = B₂t
   D. H₁n − H₂n = K
   Answer: B. Normal B is continuous, so μ₁H₁n = μ₂H₂n. (Key point: Normal B)

2. A rectangular loop with Ampere's law gives:
   A. H₁t − H₂t = K
   B. B₁t = B₂t
   C. H₁t + H₂t = K
   D. H₁t = μH₂t
   Answer: A. Enclosed current is KΔw, so H₁t − H₂t = K. (Key point: Tangential H)

3. If K = 0:
   A. H_t is continuous and B₁t/μ₁ = B₂t/μ₂
   B. B_t is continuous
   C. H_n is continuous
   D. Nothing is continuous
   Answer: A. Tangential H is continuous, hence B_t/μ is too. (Key point: No surface current)

4. The refraction law of field lines is:
   A. tanθ₁/tanθ₂ = μ₁/μ₂
   B. sinθ₁/sinθ₂ = μ₁/μ₂
   C. tanθ₁ = tanθ₂
   D. cosθ₁/cosθ₂ = μ₂/μ₁
   Answer: A. Combining the two conditions gives tanθ₁/tanθ₂ = μ₁/μ₂. (Key point: Refraction)

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