---
title: "Current for flux in cast steel series circuit with air gap"
subject: "Electrical Machines I"
chapter: 1
key: "semester_4/electrical_machines_i:cast-steel-series-circuit-airgap-200turns"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/cast-steel-series-circuit-airgap-200turns
site: BCT Engineering
---

# Current for flux in cast steel series circuit with air gap

Electrical Machines I · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2075 Bhadra, 8 marks
- Video: [54 - Solved Problems on Magnetic Circuits](https://www.youtube.com/watch?v=iOhQn_EtvIU) — SkanCity Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/cast-steel-series-circuit-airgap-200turns
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-1

## Question

Find the value of I required to establish a magnetic flux of Φ = 0.75×10⁻⁴ Wb in the series magnetic circuit as shown in figure below. Given, that the relative permeability for the steel is μr = 1424. [Figure: all cast steel rectangular core, area (throughout) = 1.5×10⁻⁴ m²; coil of N = 200 turns carrying current I on the left limb; an air gap in the right limb; mean steel path length l(cdefab) = 100×10⁻³ m and air gap length l(bc) = 2×10⁻³ m.]

## Short answer

For B=0.5 T, steel mmf = 27.9 AT, air-gap mmf = 795.8 AT; total mmf = 823.7 AT from 200 turns. Required current = 4.12 A.

## Answer

The steel path and the air gap are in series and carry the same flux; the area is the same throughout (fringing neglected).

**Data:** $\phi = 0.75\times10^{-4}$ Wb, $A = 1.5\times10^{-4}$ m², $l_s = 100\times10^{-3}$ m, $l_g = 2\times10^{-3}$ m, $\mu_r = 1424$, $N = 200$.

#### Flux density

$$
B = \frac{\phi}{A} = \frac{0.75\times10^{-4}}{1.5\times10^{-4}} = 0.5\ \text{T}
$$

#### Steel portion

$$
\begin{aligned}
H_s &= \frac{B}{\mu_0\mu_r} = \frac{0.5}{4\pi\times10^{-7}\times1424} = 279.4\ \text{AT/m} \\
H_s l_s &= 279.4\times0.1 = 27.9\ \text{AT}
\end{aligned}
$$

#### Air gap

$$
\begin{aligned}
H_g &= \frac{B}{\mu_0} = \frac{0.5}{4\pi\times10^{-7}} = 3.979\times10^{5}\ \text{AT/m} \\
H_g l_g &= 3.979\times10^{5}\times2\times10^{-3} = 795.8\ \text{AT}
\end{aligned}
$$

#### Current

$$
\begin{aligned}
NI &= 27.9 + 795.8 = 823.7\ \text{AT} \\
I &= \frac{823.7}{200} = 4.12\ \text{A}
\end{aligned}
$$

**Answer: I ≈ 4.12 A.** About 97% of the mmf is used by the 2 mm air gap.

## Check yourself

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

1. With phi = 0.75e-4 Wb and A = 1.5e-4 m^2, B is:
   A. 0.5 T
   B. 5 T
   C. 0.05 T
   D. 1.125 T
   Answer: A. B = phi/A = 0.5 T, same in steel and gap. (Key point: Flux density)

2. The gap ampere-turns are about:
   A. 27.9 AT
   B. 795.8 AT
   C. 3979 AT
   D. 79.6 AT
   Answer: B. H_g = B/mu0 = 3.979e5 AT/m, times 2 mm = 795.8 AT. (Key point: Gap ampere-turns)

3. The steel needs far fewer ampere-turns than the gap because:
   A. Its length is less
   B. Its H = B/(mu0 mu_r) is 1424 times smaller than in air for the same B
   C. It has more turns
   D. Flux is lower in steel
   Answer: B. Same B but mu_r = 1424 reduces H, giving only 27.9 AT for 0.1 m. (Key point: Steel ampere-turns)

4. The current required is:
   A. 4.12 A
   B. 0.14 A
   C. 8.24 A
   D. 823.7 A
   Answer: A. I = (27.9 + 795.8)/200 = 4.12 A. (Key point: Current)

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