---
title: "H due to finite current filament"
subject: "Electromagnetism"
chapter: 3
key: "semester_3/electromagnetism:h-due-to-finite-filament-12a"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/h-due-to-finite-filament-12a
site: BCT Engineering
---

# H due to finite current filament

Electromagnetism · Chapter 3: Magnetic field

- Asked in: 2075 Chaitra, 6 marks
- Video: [Magnetic Field Intensity - Finite and Infinite Conductor - EMF](https://www.youtube.com/watch?v=jCZH0cYez-0) — Padmasri Naban
- 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-finite-filament-12a
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-3

## Question

Find the vector magnetic field intensity H in Cartesian coordinates at P(−1.5, −4, 3) caused by a current filament of 12 A in the a_z direction on the z-axis and extending from z = −3 to z = 3.

## Short answer

Biot-Savart for finite filament: H = (I/(4πρ))(sin α2 - sin α1) a_φ. ρ = 4.272 m, sin α1 = -0.8146, sin α2 = 0, Hφ = 0.1821 A/m, direction a_φ = 0.9363 a_x - 0.3511 a_y, so H = 0.17 a_x - 0.064 a_y A/m.

## Answer

The filament lies on the z-axis from z = −3 to z = 3 and carries I = 12 A in a_z. By the Biot-Savart law, the field of a finite straight filament is

H = (I / 4πρ) (sin α₂ − sin α₁) a_φ

where ρ is the perpendicular distance from the filament to the point, and α₁, α₂ are the angles (measured from the perpendicular through P) to the lower and upper ends.

**Step 1: geometry.** P(−1.5, −4, 3), so the point is at height z = 3.

```text
ρ  = √((−1.5)² + (−4)²) = √18.25 = 4.272 m
lower end:  z₁ = −3 − 3 = −6 m (below P)
upper end:  z₂ =  3 − 3 =  0 m (level with P)
sin α₁ = −6 / √(4.272² + 6²) = −6 / 7.366 = −0.8146
sin α₂ =  0 / 4.272          =  0
```

**Step 2: magnitude.**

```text
Hφ = 12 / (4π × 4.272) × (0 − (−0.8146))
   = 0.2236 × 0.8146
   = 0.1821 A/m
```

**Step 3: direction.** a_φ = a_z × a_ρ, with a_ρ = (−1.5 a_x − 4 a_y)/4.272 = −0.3511 a_x − 0.9363 a_y.

```text
a_φ = a_z × a_ρ = 0.9363 a_x − 0.3511 a_y
```

**Step 4: vector form.**

```text
H = 0.1821 (0.9363 a_x − 0.3511 a_y)
  = 0.1705 a_x − 0.0639 a_y   A/m
```

**Answer: H ≈ 0.17 a_x − 0.064 a_y A/m** (magnitude 0.182 A/m).

## Check yourself

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

1. Field of a finite straight filament is:
   A. (I/4πρ)(sinα₂ − sinα₁) a_φ
   B. (I/2πρ) a_φ
   C. (I/4πρ²) a_φ
   D. Iρ/(4π)
   Answer: A. Angles are measured from the perpendicular through P to the ends. (Key point: Formula)

2. The perpendicular distance ρ from the z-axis to P(−1.5,−4,3) is:
   A. 4.272 m
   B. 5.5 m
   C. 3 m
   D. 4.27 m with sign negative
   Answer: A. ρ = √(1.5² + 4²) = √18.25 = 4.272 m. (Key point: Geometry)

3. Why is sin α₂ = 0?
   A. The upper end z = 3 is level with P
   B. The filament is infinite
   C. The current is zero there
   D. P lies on the filament
   Answer: A. The upper end is at the same height as P, so the angle from the perpendicular is 0. (Key point: Why sin α₂ = 0)

4. a_φ at P is found as:
   A. a_z × a_ρ = 0.936 a_x − 0.351 a_y
   B. a_ρ × a_z
   C. −a_ρ
   D. a_z
   Answer: A. Right-hand rule gives H = 0.1705 a_x − 0.0639 a_y A/m. (Key point: Direction)

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