MMF: (\mathcalF = NI = 200 \times 2 = 400 \ \textA-turns) [ \Phi = \frac\mathcalF\mathcalR_c = \frac400398 \times 10^3 \approx 1.005 \ \textmWb ]
So: [ \mathcalR_eq, branches = \frac(\mathcalR_o + 2\mathcalR_y)2 = \frac530.5 + 132.62 = 331.55 \ \textkA-t/Wb ] Wait – (2\mathcalR_y = 132.6), so (\mathcalR_o + 2\mathcalR_y = 530.5+132.6 = 663.1). Half of that is kA-t/Wb.
Given: After fault, (\Phi_actual = 0.8\ \textmWb) at (NI=250). So total reluctance = (250 / 0.8\times10^-3 = 312.5 \ \textkA-t/Wb). Core reluctance alone = (497.4 \ \textkA-t/Wb). If total reluctance is lower than iron alone, that’s impossible. Therefore: The original core for design purposes. The fault increased the gap.
The center limb carries (\Phi_c). That flux splits into two paths, each with total reluctance (\mathcalR_branch = \mathcalR_o + 2\mathcalR_y). The center limb reluctance is in series with the parallel combination of the two branch reluctances. magnetic circuits problems and solutions pdf
Author: Electromagnetics Education Lab Date: April 2026 Abstract Magnetic circuits are the hidden backbone of motors, transformers, and relays. Yet, students often struggle because magnetic quantities (MMF, flux, reluctance) lack the intuitive feel of voltage and current. This paper bridges that gap using a three-pronged approach: (1) the Ohm’s law analogy for magnetic circuits, (2) real-world fault problems (air gaps, fringing, saturation), and (3) a mini design challenge . Each problem includes a full solution with commentary on common mistakes. By the end, you will be able to analyze complex series-parallel magnetic circuits with confidence. 1. The Great Analogy: Why Magnetic Circuits Feel Strange | Electrical Circuit | Magnetic Circuit | Symbol | |---|---|---| | Electromotive force (EMF), ( \mathcalE ) (V) | Magnetomotive force (MMF), ( \mathcalF = NI ) (A-turns) | ( \mathcalF ) | | Current, ( I ) (A) | Magnetic flux, ( \Phi ) (Wb) | ( \Phi ) | | Resistance, ( R = \fracl\sigma A ) ((\Omega)) | Reluctance, ( \mathcalR = \fracl\mu A ) (A-turns/Wb) | ( \mathcalR ) | | Ohm’s law: ( \mathcalE = I R ) | Hopkinson’s law: ( \mathcalF = \Phi \mathcalR ) | — |
Flux: [ \Phi = \frac4001.99\times 10^6 \approx 0.201 \ \textmWb ]
Let (\Phi_c) = flux in center limb, (\Phi_o) = flux in each outer limb. By KFL (Kirchhoff’s flux law): (\Phi_c = 2\Phi_o) MMF equation around center-outer loop: [ NI = \Phi_o (\mathcalR_c + 2\mathcalR_y + \mathcalR_o) \quad \text(wait – this is wrong because center flux splits) ] Better: MMF = (\Phi_c \mathcalR_c + \Phi_o (\mathcalR_o + 2\mathcalR_y)) – no, that’s inconsistent. MMF: (\mathcalF = NI = 200 \times 2
Total reluctance seen by MMF: [ \mathcalR_total = \mathcalR c + \mathcalR eq,branches = 132.6 + 331.55 = 464.15 \ \textkA-t/Wb ] MMF = (300 \times 1.5 = 450 \ \textA-turns) [ \Phi_c = \frac450464.15 \times 10^3 \approx 0.969 \ \textmWb ] Then (\Phi_o = \Phi_c / 2 = 0.4845 \ \textmWb)
Let’s find gap length that gives (\mathcalR total = 312.5\ \textkA-t/Wb): [ \mathcalR g = \mathcalR total - \mathcalR iron = 312.5 - 497.4 = -184.9 \ \text(negative → impossible) ] Conclusion: The core is saturating or the permeability has dropped. A better problem would give (\Phi_healthy) first.
Let’s correct the fault diagnosis realistically: So total reluctance = (250 / 0
Flux density in yokes = same as center limb area? Yokes have (A=6\ \textcm^2), but they carry (\Phi_c)? No – yokes carry the outer branch flux? Actually each yoke segment carries (\Phi_o) if symmetric. Check: At top yoke, flux from center splits: half to left outer, half to right outer. So yoke carries (\Phi_o). [ B_yoke = \frac0.4845\times 10^-36\times 10^-4 = 0.8075 \ \textT ] Desired flux (\Phi_des = 1.2 \ \textmWb) with (NI = 250 \ \textA-turns) (since (0.5 \times 500)).
Flux density: [ B = \frac\PhiA = \frac1.005\times 10^-35\times 10^-4 = 2.01 \ \textT ] Good – below saturation for typical iron. Solution 2 – With Air Gap (a) Core reluctance same as above: (\mathcalR_c \approx 398 \ \textkA-turns/Wb) Gap reluctance: [ \mathcalR g = \fracl_g\mu_0 A = \frac0.001(4\pi\times 10^-7)(5\times 10^-4) \approx 1.592 \times 10^6 \ \textA-turns/Wb ] Total reluctance: [ \mathcalR total = 3.98\times 10^5 + 1.592\times 10^6 = 1.99 \times 10^6 \ \textA-turns/Wb ]
Percent change from Problem 2: [ \frac0.232 - 0.2010.201 \times 100 \approx +15.4% ] Fringing reduces reluctance → increases flux. Ignoring fringing underestimates performance. Solution 4 – Series-Parallel Circuit Step 1 – Reluctances (all (\mu = 1000 \mu_0))