---
title: "Stokes' theorem and line integral of H"
subject: "Electromagnetism"
chapter: 1
key: "semester_3/electromagnetism:stokes-theorem-closed-line-integral"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/stokes-theorem-closed-line-integral
site: BCT Engineering
---

# Stokes' theorem and line integral of H

Electromagnetism · Chapter 1: Introduction

- Asked in: 2072 Chaitra, 1+3+4 marks
- Video: [Stokes Theorem : Basics, Proof, Physical Significance and Applications](https://www.youtube.com/watch?v=D2eHgZ4kMHU) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/stokes-theorem-closed-line-integral
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-1

## Question

State and derive the expression for Stokes' theorem. Evaluate the closed line integral of H from P1(5, 4, 1) to P2(5, 6, 1) to P3(0, 6, 1) to P4(0, 4, 1) to P1 using straight line segments, if H = 0.1y³ a_x + 0.4x a_z A/m.

## Short answer

Stokes' theorem equates a closed line integral to the surface integral of the curl. For the given H and rectangular path, the closed line integral evaluates to −76 A.

## Answer

#### Stokes' theorem (statement)

The line integral of **H** around a closed path equals the surface integral of ∇ × **H** over any surface bounded by the path: ∮ H · dl = ∫ (∇ × H) · dS.

#### Derivation

1. Divide the surface into small patches ΔS_k. For a small rectangle Δx × Δy in the xy-plane, adding the four sides gives
   ```text
   ∮ H · dl ≈ H_x(y)Δx + H_y(x+Δx)Δy − H_x(y+Δy)Δx − H_y(x)Δy
            = (∂H_y/∂x − ∂H_x/∂y) Δx Δy = (∇ × H)_z ΔS
   ```
   So for any small patch, ∮ H · dl = (∇ × H) · ΔS (the same holds for the other two components).
2. Sum over all patches. The common edge between two adjacent patches is traversed in opposite directions, so interior contributions cancel, leaving only the outer boundary.
3. In the limit of ΔS → 0: ∮_L H · dl = ∫_S (∇ × H) · dS.

#### Numerical: H = 0.1y³ a_x + 0.4x a_z A/m

Path: P1(5, 4, 1) → P2(5, 6, 1) → P3(0, 6, 1) → P4(0, 4, 1) → P1. All points lie in the plane z = 1, so dz = 0 and the a_z term of H does not contribute.

```text
Segment 1: P1→P2, x = 5, z = 1: dl = dy a_y.  H · dl = H_y dy = 0   →  0
Segment 2: P2→P3, y = 6, z = 1: dl = dx a_x.  H_x = 0.1(6)³ = 21.6
           ∫ H_x dx from x = 5 to 0 = 21.6 (0 − 5) = −108 A
Segment 3: P3→P4, x = 0: dl = dy a_y.  H · dl = 0   →  0
Segment 4: P4→P1, y = 4, z = 1: dl = dx a_x.  H_x = 0.1(4)³ = 6.4
           ∫ H_x dx from x = 0 to 5 = 6.4 (5) = +32 A
```

```text
∮ H · dl = 0 − 108 + 0 + 32 = −76 A
```

**Check using Stokes' theorem:**

```text
∇ × H = (∂H_z/∂y − ∂H_y/∂z) a_x + (∂H_x/∂z − ∂H_z/∂x) a_y + (∂H_y/∂x − ∂H_x/∂y) a_z
      = −0.4 a_y − 0.3y² a_z
```

The path goes counter-clockwise in the xy-plane, so dS = dx dy a_z:

```text
∫ (∇ × H) · dS = ∫₀⁵ ∫₄⁶ (−0.3 y²) dy dx = −0.3 (5)(6³ − 4³)/3 = −0.3(5)(50.67) = −76 A
```

**Answer: ∮ H · dl = −76 A**

## Check yourself

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

1. Along the path in the plane z = 1, why does the a_z term 0.4x a_z not contribute?
   A. Because x = 0
   B. Because dz = 0 along the whole path
   C. Because it is constant
   D. Because it is cancelled by H_x
   Answer: B. All points are at z = 1, so dz = 0 and H_z dz contributes nothing. (Key point: Which term contributes)

2. Which segments contribute to the line integral, and how much?
   A. P1→P2 and P3→P4, which are along y
   B. P2→P3 gives -108 A and P4→P1 gives +32 A
   C. All four give equal values
   D. P2→P3 gives +108 A and P4→P1 gives -32 A
   Answer: B. Only segments along x see H_x: 21.6(0-5) = -108 for P2→P3 and 6.4(5) = +32 for P4→P1; segments along y have H_y = 0. (Key point: Segment contributions)

3. What is the closed line integral, and how is it checked?
   A. -76 A; the surface integral of curl H gives the same -76 A
   B. 76 A; the curl is zero
   C. -140 A; the surface integral gives 140
   D. -108 A; the curl is -0.4 a_y only
   Answer: A. 0 - 108 + 0 + 32 = -76 A. The curl is -0.4 a_y - 0.3y^2 a_z, and integrating -0.3y^2 over the rectangle gives -76 A. (Key point: Result and check)

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