Reluctance, MMF and Flux Density of a Wound Toroid: 16-Mechatronics-B10, December 2019, Q1

Question 1: Reluctance, MMF and Flux Density of a Wound Toroid (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Mean radius, r_av	25 cm = 0.25 m
Cross-sectional area, A	3 cm² = 3×10⁻⁴ m²
Turns, N	600
Coil current, I	1.5 A (dc)
Relative permeability, μ_r	1500

Find. (a) reluctance ℜ; (b) mmf ℱ and field intensity H; (c) flux Φ and flux density B.

[Figure not reproduced: Partially wound toroid, Figure 1. See the official exam paper or the cited reference text.]

Figure 1 — partially wound toroid (source figure): mean radius r_av, cross-sectional area a, current I in the N-turn coil.

Approach. Model the toroid as a single uniform magnetic circuit of mean length l = 2πr_av and apply Hopkinson's law (ℱ = NI = Φℜ) together with Ampère's circuital law (H = ℱ/l).

Mean magnetic path length. $$l=2\pi r_{av}=2\pi(0.25\ \text{m})=1.571\ \text{m}$$
(a) Reluctance. With μ = μ₀μ_r: $$\Re=\frac{l}{\mu_0\mu_r A}=\frac{1.571}{(4\pi\times10^{-7})(1500)(3\times10^{-4})} =\boxed{2.778\times10^{6}\ \text{A-t/Wb}}$$
(b) Magnetomotive force and field intensity. $$\mathcal{F}=NI=600(1.5)=\boxed{900\ \text{A-t}}$$ $$H=\frac{\mathcal{F}}{l}=\frac{900}{1.571}=\boxed{572.96\ \text{A/m}}$$
(c) Flux and flux density. By Hopkinson's law: $$\Phi=\frac{\mathcal{F}}{\Re}=\frac{900}{2.778\times10^{6}}=\boxed{3.24\times10^{-4}\ \text{Wb}}$$ $$B=\frac{\Phi}{A}=\frac{3.24\times10^{-4}}{3\times10^{-4}}=\boxed{1.08\ \text{T}}$$ (check: B = μ₀μ_rH = 1.08 T, consistent).

Quantity	Value
(a) Reluctance, ℜ	2.778×10⁶ A-t/Wb
(b) MMF, ℱ	900 A-t
(b) Field intensity, H	572.96 A/m
(c) Flux, Φ	3.24×10⁻⁴ Wb
(c) Flux density, B	1.08 T

Concept & Theory

Topic: Magnetic circuits — reluctance and Hopkinson's law.

Key relations: ℱ = NI = Φℜ (Hopkinson's law, the magnetic-circuit analogue of Ohm's law); ℜ = l/(μ₀μ_rA); H = ℱ/l (Ampère's law on the mean path); B = μ₀μ_rH.

Why this works: A toroid confines essentially all of its flux inside the core (negligible fringing/leakage), so a single mean flux path of length 2πr_av and uniform cross-section A gives an exact lumped-reluctance model, exactly analogous to a resistive circuit with MMF playing the role of EMF, flux the role of current, and reluctance the role of resistance.

Common pitfall: forgetting to multiply μ₀ by μ_r (using μ₀ alone overstates the reluctance by the permeability factor, here 1500×), and mixing cm²/m² when substituting the cross-sectional area.

Source: Chapman, Electric Machinery Fundamentals, Ch. 1 §1.2–1.3 (magnetic circuits, reluctance).

Question 1 of 5: Reluctance, MMF and Flux Density of a Wound Toroid

Question 1: Reluctance, MMF and Flux Density of a Wound Toroid (20 marks)