---
title: "H of infinite filament via vector potential"
subject: "Electromagnetism"
chapter: 3
key: "semester_3/electromagnetism:h-infinite-filament-vector-potential"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/h-infinite-filament-vector-potential
site: BCT Engineering
---

# H of infinite filament via vector potential

Electromagnetism · Chapter 3: Magnetic field

- Asked in: 2069 Chaitra, 5 marks
- Video: [Magnetic Field Intensity for infinite length filament](https://www.youtube.com/watch?v=P36GgI6rhKI) — WIT Solapur - Professional Learning Community
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/h-infinite-filament-vector-potential
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-3

## Question

Derive the expression for the magnetic field intensity at a point due to an infinite filament carrying a dc current I, placed on the z-axis, using the concept of vector magnetic potential.

## Short answer

Vector magnetic potential is A = -(μ₀I/2π) ln ρ a_z (plus a constant). Taking the curl, B = ∇×A = (μ₀I/2πρ) a_φ, thus the magnetic field intensity is H = (I/2πρ) a_φ A/m.

## Answer

The vector magnetic potential A is defined by B = ∇×A, with the integral form for a line current:

A = ∫ μ₀ I dl / (4π R)

#### Step 1: A of a filament of length 2L (then let L → ∞)
Place the filament along the z-axis from z = −L to z = L, carrying I in the +a_z direction. The field point is P(ρ, φ, 0), and R = √(ρ² + z²). The current element is dl = dz a_z, so A has only a z-component:

```text
A_z = (μ₀ I / 4π) ∫₋L^L dz / √(ρ² + z²)
    = (μ₀ I / 4π) [ln(z + √(ρ² + z²))]₋L^L
    = (μ₀ I / 4π) ln[ (√(ρ²+L²) + L) / (√(ρ²+L²) − L) ]
```

For L ≫ ρ, √(ρ²+L²) ≈ L, so the argument is ≈ 2L·2L/ρ² = 4L²/ρ²:

```text
A_z ≈ (μ₀ I / 4π) ln(4L²/ρ²) = (μ₀ I / 2π) ln(2L/ρ)
```

As L → ∞ this grows without limit, but only the ρ-dependence matters for B. Writing A_z = −(μ₀ I / 2π) ln ρ + C, where C is a constant (reference level):

**A = −(μ₀ I / 2π) ln ρ a_z + C a_z**

#### Step 2: find B and H
A depends only on ρ, so in cylindrical coordinates:

```text
B = ∇×A = −(∂A_z/∂ρ) a_φ
  = (μ₀ I / 2πρ) a_φ
```

#### Step 3: H
```text
H = B/μ₀ = (I / 2πρ) a_φ   A/m
```

This is the same result as from the Biot-Savart law and Ampere's circuital law. The constant C disappears on differentiation, confirming that A is defined only up to a constant.

## Check yourself

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

1. For a line current, the vector magnetic potential is:
   A. A = integral of mu0 I dl / (4 pi R)
   B. A = integral of mu0 I dl / (4 pi R^2)
   C. A = grad of V
   D. A = integral of I dl x R
   Answer: A. A = integral of mu0 I dl/(4 pi R), with B = curl A. (Key point: Definition of A)

2. For a filament along the z-axis, what is the direction of A?
   A. a_phi
   B. a_rho
   C. a_z, parallel to the current
   D. Both a_rho and a_z
   Answer: C. A is parallel to the current element dl = dz a_z, so it has only a z component. (Key point: Direction of A)

3. As L tends to infinity, the rho-dependence of A_z is:
   A. +(mu0 I/2 pi) ln(rho)
   B. -(mu0 I/2 pi) ln(rho) + constant
   C. mu0 I/(2 pi rho)
   D. independent of rho
   Answer: B. A_z = (mu0 I/2 pi) ln(2L/rho), whose rho-dependence is -(mu0 I/2 pi) ln rho plus a constant. (Key point: Form of A)

4. How is H obtained from A_z(rho)?
   A. H = A / mu0
   B. H = grad A
   C. H = dA/dt
   D. B = curl A = -(dA_z/d rho) a_phi, then H = B/mu0 = I/(2 pi rho) a_phi
   Answer: D. The curl gives B = (mu0 I/2 pi rho) a_phi; dividing by mu0 gives H. The constant C vanishes on differentiation. (Key point: Getting H)

## Related in Electromagnetism

- [H due to finite current filament](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Ah-due-to-finite-filament-12a) — Electromagnetism, chapter 3
- [H of infinite filament by Biot-Savart](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Abiot-savart-infinite-filament-derive) — Electromagnetism, chapter 3
- [Scalar vs vector magnetic potential](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Ascalar-vs-vector-magnetic-potential) — Electromagnetism, chapter 3, asked 4×
- [Magnetic flux from vector potential](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Amagnetic-flux-from-vector-potential-a) — Electromagnetism, chapter 3
- [Vector field to cylindrical at point](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Avector-field-transform-cylindrical) — Electromagnetism, chapter 1
- [H due to bent filament at P](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Amagnetic-field-bent-filament-5ma) — Electromagnetism, chapter 3

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