Question 5 of 7: Magnetic Circuit — Reluctance, MMF, Flux Density and Armature Force
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam — 04-BS-4 Electric Circuits and Power — undated sitting (internal
evidence places it at May 2019). 3 hours, closed book, one
double-sided aid sheet, approved Casio/Sharp calculator only. The exam instructs
"any five questions constitute a complete paper" — all seven are solved below as a
complete study resource.
Reference texts: Sadiku, Fundamentals of Electric Circuits
(7th ed.) — Ch. 2–4 (resistive circuits, node/mesh analysis, Thévenin/Norton,
maximum power transfer), Ch. 7 (first-order RC/RL transients), Ch. 9–10 (sinusoidal
steady state, phasors, AC power); Chapman, Electric Machinery Fundamentals
(5th ed.) — Ch. 1 (magnetic circuits, reluctance, fringing, force on an armature);
Sadiku, Elements of Electromagnetics — Ch. 5, 8 (magnetic energy and force);
Mano & Ciletti, Digital Design — Ch. 2–3 (Boolean algebra, combinational
logic design).
Question 5: Magnetic Circuit — Reluctance, MMF, Flux Density and Armature Force (20 marks)
Given. Horseshoe core mean path $\approx60\text{ cm}$, leg cross-section $5\text{ cm}\times5\text{ cm}$, armature mean length $30\text{ cm}$, armature thickness $6\text{ cm}$, two air gaps of $1\text{ mm}$ each (one under each leg), $\mu_r=2000$, $N=1000$ turns.
Check: the source figure gives the leg width ($5\text{ cm}$) and a $5\text{ cm}$ vertical dimension on the same drawing; this solution reads the leg (and gap) cross-section as the square $5\text{ cm}\times5\text{ cm}=25\text{ cm}^2$, and takes the armature depth as the same $5\text{ cm}$ (so $A_{arm}=6\times5=30\text{ cm}^2$). This assumption is corroborated below: it makes the air-gap flux density come out to the clean value $B=0.8\text{ T}$, which would be a coincidence under any other reading of the figure. It is also assumed both legs carry an identical $1\text{ mm}$ air gap to the armature (the drawing marks the gap once, but a flat armature bridging a two-legged core is only mechanically sensible if both gaps are equal).
Find. $\mathcal{R}$ of the core, the armature, and each air gap; the coil current for $\phi=2\text{ mWb}$; $B$ and $H$ in the gap; and the total force on the armature.
[Figure not reproduced: Figure 5 — horseshoe core + moveable relay armature (schematic, redrawn from the source figure). See the official exam paper.]
Approach. Model the loop as three reluctances in series (core, armature, and the two air gaps) sharing one flux $\phi$, apply $\mathcal{F}=NI=\phi\mathcal{R}_{total}$ for the MMF, then get $B$, $H$ directly from $\phi$ and the gap area, and the force from the magnetic energy stored in the (dominant) air gaps.
Part (a) — reluctance of each part. Using $\mathcal{R}=\ell/(\mu_0\mu_r A)$ for the ferromagnetic legs and $\mathcal{R}_g=g/(\mu_0 A)$ for each (non-magnetic) air gap: $$\mathcal{R}_{core}=\dfrac{0.60}{(4\pi\times10^{-7})(2000)(2.5\times10^{-3})}=\boxed{9.549\times10^{4}\ \text{A}\cdot\text{turn/Wb}}$$$$\mathcal{R}_{arm}=\dfrac{0.30}{(4\pi\times10^{-7})(2000)(3.0\times10^{-3})}=\boxed{3.979\times10^{4}\ \text{A}\cdot\text{turn/Wb}}$$$$\mathcal{R}_{gap,\text{each}}=\dfrac{0.001}{(4\pi\times10^{-7})(2.5\times10^{-3})}=\boxed{3.183\times10^{5}\ \text{A}\cdot\text{turn/Wb}}\quad(\times2\text{ gaps})$$
Part (b) — coil current for $\phi=2\text{ mWb}$. The loop is one series path, so the reluctances add: $$\mathcal{R}_{total}=\mathcal{R}_{core}+\mathcal{R}_{arm}+2\mathcal{R}_{gap}=(0.9549+0.3979+2\times3.183)\times10^{5}=7.719\times10^{5}\ \text{A}\cdot\text{turn/Wb}$$ and Ampère’s law for the magnetic circuit gives $$NI=\phi\,\mathcal{R}_{total}\ \Rightarrow\ I=\dfrac{\phi\,\mathcal{R}_{total}}{N}=\dfrac{(2\times10^{-3})(7.719\times10^{5})}{1000}=\boxed{1.544\text{ A}}$$ Note the two air gaps alone account for over 80% of the total reluctance — typical of relay/electromagnet designs, since $\mu_r$ makes the iron path almost free by comparison.
Part (c) — flux density and field intensity in the gap. The same flux crosses the gap’s cross-sectional area (no fringing assumed): $$B_{gap}=\dfrac{\phi}{A_{gap}}=\dfrac{2\times10^{-3}}{2.5\times10^{-3}}=\boxed{0.8\text{ T}}$$ and, since the gap is non-magnetic ($\mu_r=1$), $$H_{gap}=\dfrac{B_{gap}}{\mu_0}=\dfrac{0.8}{4\pi\times10^{-7}}=\boxed{6.366\times10^{5}\text{ A/m}}$$
Part (d) — total electromagnetic force on the armature. The magnetic-pressure (Maxwell stress) formula gives the pull at one gap as $F_{\text{gap}}=B_{gap}^2A_{gap}/(2\mu_0)$; the armature is pulled by both gaps at once, so the total force is twice that: $$F_{total}=2\times\dfrac{B_{gap}^2A_{gap}}{2\mu_0}=\dfrac{B_{gap}^2A_{gap}}{\mu_0}=\dfrac{(0.8)^2(2.5\times10^{-3})}{4\pi\times10^{-7}}=\boxed{1273\text{ N}}$$
