---
title: "Two-coil magnetic circuit, current for 1.2 T"
subject: "Electrical Machines"
chapter: 1
key: "semester_4/electrical_machines:two-coil-core-1-2t-current"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/two-coil-core-1-2t-current
site: BCT Engineering
---

# Two-coil magnetic circuit, current for 1.2 T

Electrical Machines · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2068 Bhadra, 8 marks
- Video: [(English)EM ||  Question 1.6 || Magnetic Circuits || Core with Two Air Gaps](https://www.youtube.com/watch?v=P32UoFLVtXM) — Electrical Engineering Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/two-coil-core-1-2t-current
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-1

## Question

For the magnetic circuit shown below, calculate the value of current I required to produce a magnetic flux density of 1.2 Tesla. Given: cross-sectional area of core = 16 sq.cm; air gap length $l_g$ = 0.06 cm; mean length of core $l_c$ = 40 cm; relative permeability $\mu_r$ = 6000.
[Figure: rectangular iron core with a coil of $N_1 = 6000$ turns carrying current I on the left limb, a second coil of $N_2 = 500$ turns carrying current $I = 2$ A on the right limb, and an air gap $l_g$ in the right limb; the current directions are drawn in the figure]

## Short answer

At 1.2 T, required mmf: iron path (40 cm) = 63.66 AT, gap (0.06 cm) = 572.96 AT, total 636.6 AT. Coil 2 provides 500*2 = 1000 AT, so coil 1 must oppose. Net mmf: 6000*I - 1000 = 636.6 => I = 0.2728 A ≈ 0.273 A.

## Answer

The mmf of the two coils acts together on one core. The net mmf must equal the mmf needed for the iron and the gap at 1.2 T.

**Given:** $A = 16$ cm$^2$, $l_g = 0.06$ cm, $l_c = 40$ cm, $\mu_r = 6000$, $B = 1.2$ T, $N_1 = 6000$ (current $I$, unknown), $N_2 = 500$ (current 2 A).

**Mmf needed**

$$
\begin{aligned}
F_c &= \frac{B l_c}{\mu_0\mu_r} = \frac{1.2\times0.40}{4\pi\times10^{-7}\times6000} = 63.66\ \text{AT} \\
F_g &= \frac{B l_g}{\mu_0} = \frac{1.2\times6\times10^{-4}}{4\pi\times10^{-7}} = 572.96\ \text{AT} \\
F &= 636.6\ \text{AT}
\end{aligned}
$$

The flux is $\Phi = BA = 1.2\times16\times10^{-4} = 1.92$ mWb.

**Net mmf of the coils**

The current directions are drawn in the figure, which is not available. Coil 2 supplies $N_2I_2 = 500\times2 = 1000$ AT, which already exceeds the 636.6 AT needed. A positive $I$ in coil 1 must therefore oppose coil 2. Taking the coils as opposing:

$$
\begin{aligned}
N_1I - N_2I_2 &= 636.6 \\
6000\,I &= 636.6 + 1000 = 1636.6 \\
I &= 0.2728\ \text{A}
\end{aligned}
$$

(If the coils aided each other, the required current would be negative with magnitude 0.0606 A, i.e. coil 1 would have to be reversed.)

**Answer: $I \approx 0.273$ A**

## Check yourself

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

1. Which part of the core needs the larger mmf to produce 1.2 T?
   A. The air gap (about 573 AT) rather than the iron (about 64 AT)
   B. The iron core (about 573 AT) rather than the gap (about 64 AT)
   C. Both need equal mmf
   D. The gap needs no mmf because it is short
   Answer: A. Although the gap is only 0.06 cm, its permeability is mu0 versus 6000 mu0 for iron, so F_g = B*l_g/mu0 = 572.96 AT against F_c = 63.66 AT. (Key point: Mmf for the gap)

2. What is the total mmf needed for the iron plus the gap at 1.2 T?
   A. 572.96 AT
   B. 636.6 AT
   C. 63.66 AT
   D. 1636.6 AT
   Answer: B. F = F_c + F_g = 63.66 + 572.96 = 636.6 AT. (Key point: Total mmf)

3. What current I results?
   A. 0.106 A
   B. 0.0606 A
   C. 1.64 A
   D. 0.273 A
   Answer: D. I = 1636.6/6000 = 0.2728 A. (Key point: Final current)

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