---
title: "Coil current for flux in three-limb cast steel core"
subject: "Electrical Machines I"
chapter: 1
key: "semester_4/electrical_machines_i:cast-steel-three-limb-current"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/cast-steel-three-limb-current
site: BCT Engineering
---

# Coil current for flux in three-limb cast steel core

Electrical Machines I · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2082 Kartik (new course), 6 marks
- Video: [ANALYSIS OF PARALLEL MAGNETIC CIRCUITS   1](https://www.youtube.com/watch?v=lCyiphcg-20) — Archana Manuel
- 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-three-limb-current
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-1

## Question

A cast steel magnetic structure made of a bar of section 2 cm × 2 cm is shown in figure. Determine the current that the 500 turns magnetizing coil on the left limb should carry so that a flux of 2 mWb is produced in the right limb. Take μr = 600 and neglect leakage. [Figure: three-limb rectangular core; the 500-turn coil carrying current I is on the left limb A; the centre limb B carries flux Φ1 and the right limb C carries flux Φ2; the left and right windows are each 25 cm wide (left limb to centre limb 25 cm, centre limb to right limb 25 cm) and the limbs are 15 cm high; flux Φ in the left limb divides into Φ1 and Φ2.]

## Short answer

Using magnetic circuit analysis, the parallel limbs have equal MMF drop. With flux 2 mWb in the right limb, total flux is 10.667 mWb, total MMF is 27299 AT, so the required current is 54.6 A.

## Answer

The centre limb and the right limb are two parallel paths between the same two junctions, so they have the same mmf drop. The left limb (with the coil) carries the total flux.

**Assumed mean path lengths** (from the figure): left path A = 25 + 15 + 25 = 65 cm, centre limb B = 15 cm, right path C = 25 + 15 + 25 = 65 cm. Area $a = 2 \times 2 = 4$ cm² $= 4\times10^{-4}$ m² throughout.

```text
   +-----25-----+-----25-----+
   |            |            |
 [500T]  phi ^  | phi1  v    | phi2 v
   | A          | B (15)     | C
   +------------+------------+
```

#### Reluctances

$$
\begin{aligned}
S &= \frac{l}{\mu_0\mu_r a}, \quad \mu_0\mu_r a = 4\pi\times10^{-7}\times600\times4\times10^{-4} \\
S_A = S_C &= \frac{0.65}{3.016\times10^{-7}} = 2.155\times10^{6}\ \text{AT/Wb} \\
S_B &= \frac{0.15}{3.016\times10^{-7}} = 4.974\times10^{5}\ \text{AT/Wb}
\end{aligned}
$$

#### Fluxes

MMF across the parallel limbs (B and C):

$$
\begin{aligned}
F_{BC} &= \phi_2 S_C = 2\times10^{-3}\times2.155\times10^6 = 4310.4\ \text{AT} \\
\phi_1 &= \frac{F_{BC}}{S_B} = \frac{4310.4}{4.974\times10^5} = 8.667\ \text{mWb} \\
\phi &= \phi_1 + \phi_2 = 8.667 + 2 = 10.667\ \text{mWb}
\end{aligned}
$$

(Equivalently $\phi_1 = \phi_2 \times 65/15$ because the limbs differ only in length.)

#### Total mmf and current

$$
\begin{aligned}
NI &= \phi S_A + F_{BC} \\
&= 10.667\times10^{-3}\times2.155\times10^{6} + 4310.4 \\
&= 22989 + 4310 = 27299\ \text{AT} \\
I &= \frac{27299}{500} = 54.6\ \text{A}
\end{aligned}
$$

**Answer: I ≈ 54.6 A.**

(Note: with μr = 600 held constant the flux densities come out very high, e.g. 5 T in the right limb; a real cast-steel core would saturate, so the result is valid only for the linear data given.)

## Check yourself

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

1. Why do the centre limb B and the right limb C have the same mmf drop?
   A. They carry equal flux
   B. They are two parallel paths between the same two junctions
   C. They have equal lengths
   D. They have the coil on them
   Answer: B. Parallel magnetic paths between the same two points share the same mmf drop, like parallel resistors share voltage. (Key point: Parallel limbs)

2. With phi2 = 2 mWb in the right limb (65 cm) and the centre limb 15 cm of same area, phi1 is about:
   A. 0.46 mWb
   B. 2 mWb
   C. 8.67 mWb
   D. 10.67 mWb
   Answer: C. phi1 = phi2 x 65/15 = 8.67 mWb because the flux is inversely proportional to path length for equal area and mu. (Key point: Flux in centre limb)

3. Which flux must the left limb (with the coil) carry?
   A. phi2 only
   B. phi1 only
   C. phi1 + phi2 = 10.67 mWb
   D. phi1 - phi2
   Answer: C. The left limb supplies both parallel paths, so it carries the sum of the fluxes. (Key point: Flux in coil limb)

4. How is the total NI found?
   A. phi S_A + F_BC (left limb drop plus parallel drop)
   B. phi2 S_C only
   C. phi S_A x F_BC
   D. Sum of all three limb drops
   Answer: A. The left limb is in series with the parallel pair, so NI = phi S_A + F_BC = 27299 AT, giving I = 54.6 A. (Key point: Total mmf)

## Related in Electrical Machines I

- [Coil current for 2 mWb in central limb of core](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Athree-limb-two-coils-central-core) — Electrical Machines I, chapter 1
- [Current for flux in cast steel series circuit with air gap](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Acast-steel-series-circuit-airgap-200turns) — Electrical Machines I, chapter 1
- [Air gap flux density in core with three windings](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Athree-winding-core-airgap-flux-density) — Electrical Machines I, chapter 1
- [Fluxes in three-limb core with two air gaps](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Athree-limb-core-two-airgaps-300turns) — Electrical Machines I, chapter 1
- [Magnetizing current for ring with 5 mm air gap](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aring-airgap-magnetizing-current-1mwb) — Electrical Machines I, chapter 1
- [No-load current components, iron loss and core flux](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Ano-load-current-10a-0p3pf-500turns) — Electrical Machines I, chapter 2

## Related in other subjects

- [Three-limb core, ampere-turns for flux](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Athree-limb-core-amp-turn) — Electrical Machines, chapter 1
- [Three-coil core: air-gap flux](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Athree-coil-core-air-gap-flux) — Electrical Machine, chapter 1
- [Three-coil core, flux in air gap](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Athree-coil-core-airgap-flux) — Electrical Machines, chapter 1
- [Magnetic circuit: current in coil B](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Amagnetic-circuit-coil-b-current) — Electrical Machine, chapter 1