---
title: "H of infinite filament by Biot-Savart"
subject: "Electromagnetism"
chapter: 3
key: "semester_3/electromagnetism:biot-savart-infinite-filament-derive"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/biot-savart-infinite-filament-derive
site: BCT Engineering
---

# H of infinite filament by Biot-Savart

Electromagnetism · Chapter 3: Magnetic field

- Asked in: 2070 Chaitra, 5 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/biot-savart-infinite-filament-derive
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-3

## Question

Derive the expression for magnetic field intensity (H) due to an infinite current carrying filament using Biot-Savart's law.

## Short answer

Using Biot-Savart law for an infinite filament along z, the field at distance ρ is H = (I/(2πρ)) a_φ. Integration of dH from -∞ to ∞ yields this result, showing field is circular and falls as 1/ρ.

## Answer

**Biot-Savart law:** the magnetic field intensity at a point due to a current element I dl is

dH = I dl × a_R / (4π R²) = I dl × R / (4π R³)

where R is the vector from the element to the field point.

#### Set-up
Let the infinite filament lie along the z-axis carrying current I in the +a_z direction. Find H at P(ρ, φ, 0) (by symmetry, any z). 

```text
        z
        ^
        |  element I dz at (0,0,z)
        *
        |\
        | \ R
        |  \
   -----+---*--> ρ   (P at distance ρ)
```

- Element position: (0, 0, z); field point: (ρ, φ, 0).
- R = ρ a_ρ − z a_z, so R = √(ρ² + z²).
- dl = dz a_z.

#### Derivation
```text
dl × R = dz a_z × (ρ a_ρ − z a_z) = ρ dz a_φ
dH = I ρ dz a_φ / (4π (ρ² + z²)^(3/2))
```

Integrate from z = −∞ to ∞:

```text
H = (I ρ / 4π) a_φ ∫₋∞^∞ dz / (ρ² + z²)^(3/2)
```

Use the substitution z = ρ tan α, dz = ρ sec²α dα, (ρ² + z²)^(3/2) = ρ³ sec³α:

```text
∫ dz/(ρ²+z²)^(3/2) = (1/ρ²) ∫₋π/2^π/2 cos α dα = (1/ρ²)[sin α]₋π/2^π/2 = 2/ρ²
```

Hence

```text
H = (I ρ / 4π) × (2/ρ²) a_φ
```

**H = (I / 2πρ) a_φ  A/m**

The field is circular around the wire (right-hand rule), and its magnitude falls as 1/ρ. Check with Ampere's law: ∮H·dl = H(2πρ) = I.

## Check yourself

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

1. Which expression is the Biot-Savart law for the field of a current element?
   A. dH = I dl x a_R / (4 pi R^2)
   B. dH = I dl . a_R / (4 pi R)
   C. dH = I dl x a_R / (2 pi R)
   D. dH = I dl / (4 pi R^3)
   Answer: A. dH = I dl x a_R /(4 pi R^2), where R is measured from the element to the field point. (Key point: Biot-Savart law)

2. For a filament along z with R = rho a_rho - z a_z, what is dl x R?
   A. -z dz a_rho
   B. rho dz a_z
   C. rho dz a_phi
   D. 0
   Answer: C. dz a_z x (rho a_rho - z a_z) = rho dz a_phi, which gives the circular direction of H. (Key point: Cross product)

3. With z = rho tan(alpha), what does the integral of dz/(rho^2+z^2)^(3/2) from -inf to inf equal?
   A. 1/rho
   B. 2/rho^2
   C. pi/rho
   D. 2/rho
   Answer: B. It becomes (1/rho^2) times integral of cos(alpha) d alpha from -pi/2 to pi/2 = 2/rho^2. (Key point: The integral)

4. What is H of an infinite filament?
   A. I/(4 pi rho) a_phi
   B. I/(2 pi rho^2) a_phi
   C. I rho/(2 pi) a_phi
   D. I/(2 pi rho) a_phi
   Answer: D. H = (I rho/4pi)(2/rho^2) a_phi = I/(2 pi rho) a_phi, falling as 1/rho and circling the wire. (Key point: Final result)

## Related in Electromagnetism

- [State Biot-Savart's law](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Abiot-savart-law-state-explain) — Electromagnetism, chapter 3, asked 2×
- [H of infinite filament via vector potential](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Ah-infinite-filament-vector-potential) — Electromagnetism, chapter 3
- [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
- [E due to line charge using Gauss's law](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Aline-charge-field-gauss-law) — Electromagnetism, chapter 2, asked 3×
- [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
- [Derive magnetic boundary conditions](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Amagnetic-boundary-conditions-derive) — Electromagnetism, chapter 3

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