---
title: "H at centre of square loop"
subject: "Electromagnetism"
chapter: 3
key: "semester_3/electromagnetism:h-at-centre-square-loop"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/h-at-centre-square-loop
site: BCT Engineering
---

# H at centre of square loop

Electromagnetism · Chapter 3: Magnetic field

- Asked in: 2068 Chaitra, 6 marks
- Video: [EC 8451 ELECTROMAGNETIC FIELDS-MAGNETIC FIELD AT THE CENTRE OF THE SQUARE LOOP](https://www.youtube.com/watch?v=WHNqNPROWSo) — Shaik Majeeth
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/h-at-centre-square-loop
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-3

## Question

A current carrying square loop with vertices A(0, −2, 2), B(0, 2, 2), C(0, 2, −2), D(0, −2, −2) is carrying a dc current of 20 A in the direction along A-B-C-D-A. Find magnetic field intensity H at the centre of the current carrying loop.

## Short answer

Each side subtends ±45° at the centre. Using H_side = (I/4πd)(sin45°+sin45°) with d=2 m gives 1.1254 A/m per side. All four contributions add in the −a_x direction, yielding total H = −4.50 a_x A/m.

## Answer

The loop lies in the plane x = 0 (the yz-plane), centred at the origin. Vertices: A(0, −2, 2), B(0, 2, 2), C(0, 2, −2), D(0, −2, −2). Each side is 4 m long, so the centre O is d = 2 m from each side. I = 20 A, direction A→B→C→D→A.

**Formula for a finite straight segment** at perpendicular distance d:

H = (I / 4πd)(sin α₂ − sin α₁)

For a symmetric segment, the ends subtend ±45° from the perpendicular at the centre (half-length 2 m, d = 2 m), so sin α₂ = sin 45°, sin α₁ = −sin 45°.

**Step 1: magnitude due to one side.**

```text
H_side = 20 / (4π × 2) × (0.7071 + 0.7071)
       = 0.7958 × 1.4142
       = 1.1254 A/m
```

**Step 2: direction of each side** (use dl × a_R, with a_R from the element to O).

```text
Side AB: current +a_y at z = 2, a_R = −a_z     → a_y × (−a_z) = −a_x
Side BC: current −a_z at y = 2, a_R = −a_y     → (−a_z) × (−a_y) = a_z × a_y = −a_x
Side CD: current −a_y at z = −2, a_R = +a_z    → (−a_y) × a_z = −a_x
Side DA: current +a_z at y = −2, a_R = +a_y    → a_z × a_y = −a_x
```

All four add up in the −a_x direction (current is clockwise when viewed from the +x side).

**Step 3: total.**

```text
H = 4 × 1.1254 (−a_x) = −4.50 a_x  A/m
```

Exact form: H = −(10√2/π) a_x.

**Answer: H = −4.50 a_x A/m** at the centre of the loop.

## Check yourself

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

1. For the square loop with vertices at (0, +-2, +-2), what are the side length and distance d from the centre to a side?
   A. Side 2 m, d = 1 m
   B. Side 4 m, d = 2 m
   C. Side 4 m, d = 4 m
   D. Side 2.83 m, d = 2 m
   Answer: B. Adjacent vertices are 4 m apart and the centre is 2 m from each side. (Key point: Geometry)

2. For a symmetric side, the ends subtend 45 degrees from the perpendicular. The field of one side is:
   A. I/(2 pi d) = 1.59 A/m
   B. (I/4 pi d)(sin45 + sin45) = 1.125 A/m
   C. I/(4 pi d) = 0.796 A/m
   D. (I/4 pi d)(sin45 - sin45) = 0
   Answer: B. H_side = (20/8 pi)(0.7071 + 0.7071) = 1.125 A/m. (Key point: Finite segment formula)

3. How do the contributions of the four sides combine at the centre?
   A. They cancel in pairs
   B. They add in the same direction, -a_x
   C. Two add and two cancel
   D. They add in different axes
   Answer: B. Each side gives a field along -a_x (checked by dl x a_R), so all four add. (Key point: Adding the four sides)

4. What is H at the centre?
   A. -1.13 a_x A/m
   B. -9.0 a_x A/m
   C. -4.50 a_x A/m
   D. +4.50 a_x A/m
   Answer: C. H = 4 x 1.1254 = 4.50 A/m, directed along -a_x, equal to -(10 sqrt(2)/pi) a_x. (Key point: Total H)

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