---
title: "H due to bent filament at P"
subject: "Electromagnetism"
chapter: 3
key: "semester_3/electromagnetism:magnetic-field-bent-filament-5ma"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/magnetic-field-bent-filament-5ma
site: BCT Engineering
---

# H due to bent filament at P

Electromagnetism · Chapter 3: Magnetic field

- Asked in: 2076 Asoj, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/magnetic-field-bent-filament-5ma
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-3

## Question

Let a filamentary current of 5 mA be directed from infinity to the origin on the positive z-axis and then back out to infinity on the positive x-axis. Find H at P(0, 1, 0).

## Short answer

For each semi-infinite straight wire, H = I/(4πρ). For the z-axis segment (current −a_z, ρ=1 m), H₁ = 0.398 a_x mA/m. For the x-axis segment (current +a_x), H₂ = 0.398 a_z mA/m. Total at P(0,1,0): H = 0.398 a_x + 0.398 a_z mA/m.

## Answer

**Setup.** A 5 mA current comes in from infinity along the +z-axis to the origin (direction −a_z), then goes out along the +x-axis to infinity (direction +a_x). Find H at P(0, 1, 0) using the Biot-Savart law for a straight filament:

```text
H = (I/4πρ)(cosα₁ − cosα₂) a_φ         or  H = (I/4πρ)(sinβ₂ − sinβ₁) a_φ
```
For a semi-infinite wire whose end lies at the foot of the perpendicular from P, the factor is 1, so H = I/(4πρ).

```text
   z
   ↑ ↓ I (from ∞)
   |
   |
   O ---------→ x   I (to ∞)
    \
     P(0,1,0)  at y = 1
```

#### Segment 1: z-axis from z = ∞ to 0 (current −a_z)
P lies at a perpendicular distance ρ = 1 m from the z-axis, and the foot of the perpendicular is the origin (the end of the segment).

```text
Magnitude: H₁ = I/(4πρ) = 5×10⁻³/(4π × 1) = 0.398×10⁻³ A/m
Direction: dl × a_R : (−a_z) × (a_y) = +a_x

H₁ = 0.398 a_x  mA/m
```
(From Biot-Savart: with dl = −dz a_z and R = a_y − z a_z, dl × R = dz a_x; H₁ = (I/4π)∫₀^∞ dz/(1+z²)^(3/2) a_x = (I/4π) a_x.)

#### Segment 2: x-axis from 0 to ∞ (current +a_x)
P is a perpendicular distance 1 m from the x-axis, again with the foot at the origin.

```text
Magnitude: H₂ = I/(4π × 1) = 0.398 mA/m
Direction: dl × a_R : a_x × a_y = +a_z

H₂ = 0.398 a_z  mA/m
```

#### Total
```text
H = H₁ + H₂ = 0.398 a_x + 0.398 a_z  mA/m
|H| = 0.563 mA/m
```

**Answer: H(0,1,0) = 0.398 a_x + 0.398 a_z mA/m** (= 0.4 (a_x + a_z) mA/m).

## Check yourself

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

1. For a semi-infinite filament with P at the foot of the perpendicular from its end, H is:
   A. I/(4πρ)
   B. I/(2πρ)
   C. I/(8πρ)
   D. I/(4πρ²)
   Answer: A. (I/4πρ)(cosα₁ − cosα₂) with factor 1 gives I/(4πρ) = 0.398 mA/m for ρ = 1 m. (Key point: Field of semi-infinite wire)

2. For current −a_z and P at (0,1,0), a_R = a_y; the direction of H₁ is:
   A. +a_x
   B. −a_x
   C. +a_z
   D. −a_y
   Answer: A. (−a_z) × a_y = +a_x. (Key point: Direction for z-axis segment)

3. For current +a_x, H₂ direction at P is:
   A. −a_z
   B. +a_z
   C. +a_x
   D. +a_y
   Answer: B. a_x × a_y = a_z. (Key point: Direction for x-axis segment)

4. Total H at P is:
   A. 0.398 a_x + 0.398 a_z mA/m
   B. 0.796 a_z mA/m
   C. 0.563 a_y mA/m
   D. 0
   Answer: A. Superposition of the two perpendicular contributions: 0.398(a_x + a_z) mA/m. (Key point: Total)

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