---
title: "Internal inductance independent of geometry and current"
subject: "Power System Analysis I"
chapter: 4
key: "semester_4/power_system_analysis_i:internal-inductance-geometry-current"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/internal-inductance-geometry-current
site: BCT Engineering
---

# Internal inductance independent of geometry and current

Power System Analysis I · Chapter 4: Line Parameter Calculations

- Asked in: 2071 Magh, 5 marks
- Video: [Internal Flux Linkage with in conductor, full derivation | electrical power system](https://www.youtube.com/watch?v=yGergiJFLFk) — ELECTRICAL HUB
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/internal-inductance-geometry-current
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-4

## Question

Show that the inductance due to internal flux linkages of a conductor is independent of its geometry and the current through it.

## Short answer

Internal flux integration yields L_int = μ0/(8π) = 0.5×10⁻⁷ H/m = 0.05 mH/km. Radius cancels in the integration and current cancels in the definition, so internal inductance depends only on conductor permeability, not on geometry or current.

## Answer

The internal inductance of a round conductor works out to $\mu_0/8\pi$ H/m, which contains neither the radius nor the current. So it does not depend on the conductor's size or the current.

#### Derivation

Consider a long round conductor of radius $r$ carrying current $I$, uniformly distributed (DC or low frequency).

```text
        _______
      /   ...   \      x  = radius of tube
     |  ( x )dx  |     dx = thickness
      \_________/      r  = conductor radius
```

**1. Field inside.** At radius $x < r$, the enclosed current is $I_x = I\dfrac{x^2}{r^2}$. By Ampere's law, $H_x \cdot 2\pi x = I_x$:

$$
H_x = \frac{Ix}{2\pi r^2}, \qquad B_x = \mu_0 H_x = \frac{\mu_0 I x}{2\pi r^2}
$$

**2. Flux in a tube.** Flux per metre length in a tube of thickness $dx$:

$$
d\phi = B_x\,dx = \frac{\mu_0 I x}{2\pi r^2}dx
$$

**3. Partial linkage.** This flux links only the current inside radius $x$, i.e. a fraction $x^2/r^2$ of the total:

$$
d\lambda = \frac{x^2}{r^2}d\phi = \frac{\mu_0 I x^3}{2\pi r^4}dx
$$

**4. Integrate** from $0$ to $r$:

$$
\lambda_{int} = \int_0^r \frac{\mu_0 I x^3}{2\pi r^4}dx = \frac{\mu_0 I}{2\pi r^4}\cdot\frac{r^4}{4} = \frac{\mu_0 I}{8\pi}\ \text{Wb-T/m}
$$

**5. Internal inductance.**

$$
L_{int} = \frac{\lambda_{int}}{I} = \frac{\mu_0}{8\pi} = \frac{4\pi\times10^{-7}}{8\pi} = \frac12\times10^{-7}\ \text{H/m}
$$

#### Conclusion

- The radius $r$ cancels in step 4, so $L_{int}$ does **not depend on the conductor size (geometry)**.
- The current $I$ cancels in step 5, so it does **not depend on the current**.
- It depends only on the permeability of the conductor material ($\mu_r = 1$ for copper and aluminium), giving $0.05$ mH/km.

This is why the internal flux can be included simply by replacing $r$ with $r' = re^{-1/4} = 0.7788r$.

## Check yourself

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

1. At radius x inside a round conductor of radius r, the enclosed current (uniform density) is:
   A. I x^2 / r^2
   B. I x / r
   C. I
   D. I x^3 / r^3
   Answer: A. Current is proportional to enclosed area, pi x^2 over pi r^2. (Key point: Enclosed current)

2. Why is the flux in a tube at radius x multiplied by x^2/r^2 to get linkage?
   A. It links the full current I
   B. It links only the fraction of current inside radius x
   C. Because the field is zero beyond x
   D. To convert from H to B
   Answer: B. A tube's flux encircles only the current inside it, so linkage is a fraction x^2/r^2 of the total. (Key point: Partial linkage)

3. What is the internal inductance of a round conductor?
   A. mu0/(2*pi) H/m
   B. mu0/(4*pi) H/m
   C. mu0/(8*pi) = 0.5e-7 H/m
   D. mu0*r/(8*pi) H/m
   Answer: C. Integrating gives lambda = mu0 I/(8 pi), so L_int = mu0/8pi, which has no r or I. (Key point: Result)

4. Why is L_int independent of conductor size and current?
   A. Because the conductor is non-magnetic and so the flux is zero
   B. Because skin effect confines all current to the surface
   C. Because the flux is assumed to be outside the conductor
   D. r cancels on integration and I cancels in lambda/I
   Answer: D. The r^4 from the integral cancels with 1/r^4, and dividing by I removes the current. (Key point: Why independent)

## Related in Power System Analysis I

- [Internal inductance independent of geometry](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Ainternal-inductance-independent-geometry) — Power System Analysis I, chapter 4, asked 2×
- [Internal inductance independent of current](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Ainternal-inductance-independent-current) — Power System Analysis I, chapter 4, asked 3×
- [Inductance of composite-conductor single-phase line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Asingle-phase-composite-conductors) — Power System Analysis I, chapter 4

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