---
title: "H due to side of triangular loop"
subject: "Electromagnetism"
chapter: 3
key: "semester_3/electromagnetism:h-due-to-triangle-side-1"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/h-due-to-triangle-side-1
site: BCT Engineering
---

# H due to side of triangular loop

Electromagnetism · Chapter 3: Magnetic field

- Asked in: 2068 Baisakh, 8 marks
- Video: [Find H at the center C of an equilateral triangular loop of side 4 m carrying 5 A of current  - EM](https://www.youtube.com/watch?v=TKlrBW-UxQw) — Engineer Thileban Explains
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/h-due-to-triangle-side-1
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-3

## Question

The triangular loop shown in the figure carries a current of 10 A. Find H at (0, 0, 5) due to side 1 of the loop. [Figure: a right-triangle loop in the z = 0 plane with corners at (0, 1, 0), (2, 0, 0) and the origin, carrying 10 A; side 1 is the side lying along the x-axis between the origin and (2, 0, 0).]

## Short answer

Apply the Biot-Savart law to the segment on the x-axis with dl = dx a_x and R = −x a_x + 5 a_z. Integration yields dl×R = −5 dx a_y and H₁ = −59.1 a_y mA/m at (0,0,5).

## Answer

**Assumptions (the figure is described in words):** the loop lies in the z = 0 plane with corners at O(0,0,0), (2,0,0) and (0,1,0). Side 1 is the segment on the x-axis from O to (2,0,0). The current of 10 A is taken to flow counter-clockwise when seen from +z, so in side 1 it flows from O toward (2,0,0), i.e. along +a_x. (If the figure shows the opposite sense, the answer just reverses sign.)

**Method:** Biot-Savart law, H = ∫ I dl × R / (4π R³), where R is from the source element to the field point P(0, 0, 5).

**Step 1: set up for side 1.** Source point (x, 0, 0), 0 ≤ x ≤ 2.

```text
dl = dx a_x
R  = (0 − x) a_x + 0 a_y + (5 − 0) a_z = −x a_x + 5 a_z
|R| = √(x² + 25)
```

**Step 2: cross product.**

```text
dl × R = dx a_x × (−x a_x + 5 a_z) = 5 dx (a_x × a_z) = −5 dx a_y
```

**Step 3: integrate.**

```text
H₁ = (I / 4π) ∫₀² (−5 dx) / (x² + 25)^(3/2) a_y
```

Use ∫ dx/(x² + a²)^(3/2) = x / (a² √(x² + a²)), with a = 5:

```text
∫₀² 5 dx / (x² + 25)^(3/2) = 5 × [ x / (25 √(x² + 25)) ]₀²
                           = 5 × 2 / (25 × √29)
                           = 10 / (25 × 5.385)
                           = 0.07428
```

**Step 4: result.**

```text
H₁ = −(10 / 4π) × 0.07428 a_y
   = −0.7958 × 0.07428 a_y
   = −0.0591 a_y   A/m
```

**Answer: H₁ ≈ −59.1 a_y mA/m** at (0, 0, 5) due to side 1.

## Check yourself

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

1. Which law is used to find H at (0,0,5) from one side of the loop?
   A. Gauss's law
   B. Biot-Savart law
   C. Faraday's law
   D. Coulomb's law
   Answer: B. H = integral of I dl x R /(4 pi R^3) is used for a finite straight segment. (Key point: Method)

2. For a source point (x,0,0) and field point (0,0,5), R is:
   A. x a_x + 5 a_z
   B. -x a_x + 5 a_z
   C. -x a_x - 5 a_z
   D. 5 a_z
   Answer: B. R points from the source to the field point: (0 - x) a_x + (5 - 0) a_z, with magnitude sqrt(x^2+25). (Key point: Vector R)

3. What is dl x R with dl = dx a_x?
   A. 5 dx a_y
   B. -x dx a_z
   C. -5 dx a_y
   D. 0
   Answer: C. a_x x (-x a_x + 5 a_z) = 5 (a_x x a_z) = -5 a_y. (Key point: Cross product)

4. What is H due to side 1?
   A. +0.0591 a_y A/m
   B. -0.0591 a_y A/m
   C. -0.7958 a_y A/m
   D. -0.0591 a_x A/m
   Answer: B. H = -(10/4 pi)(0.07428) a_y = -0.0591 a_y A/m, about -59.1 mA/m. (Key point: Result)

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