---
title: "Magnetic circuit: current in coil B"
subject: "Electrical Machine"
chapter: 1
key: "semester_4/electrical_machine:magnetic-circuit-coil-b-current"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/magnetic-circuit-coil-b-current
site: BCT Engineering
---

# Magnetic circuit: current in coil B

Electrical Machine · Chapter 1: Magnetic Circuits and Induction

- Asked in: 2073 Bhadra, 8 marks
- Video: [Introduction to Magnetic Circuits: MMF, Flux, Reluctance | What is a Magnetic Circuit?](https://www.youtube.com/watch?v=0egh0XBItFc) — Electrical and Electronics Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machine/magnetic-circuit-coil-b-current
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-1

## Question

For the magnetic circuit shown in the figure below, find out the current to be passed through coil B so that magnetic flux in CD section is 2 mWb. Given μr = 1000.
[Figure: a rectangular magnetic core with corners E (top left), C (top right), F (bottom left), D (bottom right) and a central limb B (top) to A (bottom). Coil B with N1 = 300 turns and current I1 and coil A with N2 = 400 turns and current I2 are wound on the central limb. The central limb has cross-section A1 = 6 cm² and the two outer limbs (E-F side and C-D side) have cross-section A2 = 3 cm² each.]
Given: I2 = 3 A, A1 = 6 cm², A2 = 3 cm², AB = CD = EF = 20 cm, BC = AD = BE = AF = 20 cm.

## Short answer

Central flux = 4 mWb. Reluctances: S_c = 2.653×10⁵ AT/Wb, S_o = 1.592×10⁶ AT/Wb. Total mmf = 4×10⁻³×2.653×10⁵ + 2×10⁻³×1.592×10⁶ = 4244 AT. Coil A mmf = 1200 AT. Coil B mmf = 3044 AT, so I = 3044/300 = 10.15 A.

## Answer

The central limb is common to both outer paths, which are in parallel. We find the mmf needed to drive the flux through the centre and one outer path and subtract the mmf of coil A.

**Reading the figure and assumptions**
- Central limb AB: 20 cm, area A1 = 6 cm² = 6×10⁻⁴ m², carries both coils (coil B: N1 = 300; coil A: N2 = 400, I2 = 3 A).
- Each outer path (B–C–D–A, and B–E–F–A) has length 20 + 20 + 20 = 60 cm and area A2 = 3×10⁻⁴ m².
- The two outer paths are identical, so each carries half of the centre flux. Φ in CD = 2 mWb, so **centre flux = 4 mWb**.
- μr = 1000. Both coils are assumed to magnetise in the same direction (aiding).

**Equivalent magnetic circuit**
```text
 F (coils) ---[S_centre]---+---[S_outer]---+
                           |               |
                           +---[S_outer]---+
```

**Step 1: reluctances**
```text
S_c = l/(μ0 μr A1) = 0.2/(4π×10⁻⁷×1000×6×10⁻⁴)
    = 2.653×10⁵ AT/Wb
S_o = 0.6/(μ0 μr A2) = 0.6/(4π×10⁻⁷×1000×3×10⁻⁴)
    = 1.592×10⁶ AT/Wb
```

**Step 2: total mmf required**
```text
F = Φ_c S_c + Φ_o S_o
  = 4×10⁻³ × 2.653×10⁵ + 2×10⁻³ × 1.592×10⁶
  = 1061 + 3183 = 4244 AT
```

**Step 3: coil B**
```text
N2 I2 = 400 × 3 = 1200 AT
N1 I1 = 4244 − 1200 = 3044 AT
I1    = 3044/300 = 10.15 A
```
(If coil A were opposing, I1 would be (4244 + 1200)/300 = 18.1 A.)

**Answer: I1 ≈ 10.15 A**

## Check yourself

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

1. If CD carries 2 mWb, what flux is in the central limb?
   A. 2 mWb
   B. 4 mWb
   C. 6 mWb
   D. 1 mWb
   Answer: B. The two identical outer paths are in parallel and share the centre flux equally, so centre = 2 x 2 = 4 mWb. (Key point: Flux in centre limb)

2. How is the total mmf found?
   A. Average of the drops
   B. Centre-limb drop plus one outer-path drop (parallel outer paths)
   C. Only the outer path drop
   D. Sum of both outer path drops and the centre
   Answer: B. Parallel branches have the same mmf drop, so F = phi_c S_c + phi_o S_o. (Key point: Circuit structure)

3. What are the drops in the centre limb and an outer path?
   A. 1061 AT and 3183 AT
   B. 3183 AT and 1061 AT
   C. 1061 AT and 1592 AT
   D. 4244 AT and 0
   Answer: A. Centre: 4 mWb x 2.653x10^5 = 1061 AT; outer: 2 mWb x 1.592x10^6 = 3183 AT; total 4244 AT. (Key point: Mmf value)

4. Coil A gives 400 x 3 = 1200 AT aiding. What is I1 in coil B (300 turns)?
   A. 14.1 A
   B. 18.1 A
   C. 10.15 A
   D. 4.0 A
   Answer: C. N1 I1 = 4244 - 1200 = 3044 AT, so I1 = 3044/300 = 10.15 A. (Key point: Coil B current)

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