Question 5 of 7: Magnetic Circuit — Horseshoe Core with Air Gap and Armature
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams, 04-BS-4 Electric Circuits and Power — May 2016. Closed book;
one double-sided aid sheet permitted; any five of the seven questions constitute a
complete paper (all seven are answered here as a complete study resource).
Reference texts: Sadiku & Alexander, Fundamentals of Electric
Circuits, 6th ed. (circuit analysis, transients, AC steady state, magnetic circuits);
Boylestad & Nashelsky, Electronic Devices and Circuit Theory (rectifiers);
Mano, Digital Design (combinational logic).
Question 5: Magnetic Circuit — Horseshoe Core with Air Gap and Armature (20 marks)
Given. The fixed horseshoe core carries N=1000 turns split between its two legs, current i=1 A, relative permeability μr=2000. Figure 5 dimensions give a 60 cm mean top-arc length, 30 cm per leg, a uniform 5 cm×5 cm cross-section, and a 1 mm air gap under EACH leg (two gaps in series) to the moveable armature below.
Given data
Quantity
Value
μr
2000
N
1000 turns
i
1 A
Top arc (mean length)
60 cm
Each leg (mean length)
30 cm (×2)
Cross-section
5 cm × 5 cm
Air gap
1 mm (×2, one per leg)
Find. Total MMF; the reluctance of the iron path and of the air gaps; the air-gap flux, flux density and field intensity; and the total electromagnetic force on the armature.
Figure 5: Horseshoe core with air gap and relay armature
Check: the armature's own iron reluctance (a short, thick return
path directly beneath both gaps) is neglected relative to the iron legs/arc and the air gaps —
standard practice for this style of relay problem, and immaterial here since the sub-mm air gaps alone
already account for roughly 80% of the total reluctance.
Approach. Treat the loop as a single series magnetic circuit (magnetic Ohm's law): top arc + two legs + two air gaps, all carrying the same flux φ; solve for φ, then B and H in the gap; finally apply the co-energy expression for the force pulling the armature toward the core at each of the two gaps.
Total MMF (part a).
$$ \mathcal{F} = Ni = (1000)(1) = \boxed{1000\text{ A}\cdot\text{turns}}. $$
Reluctance of the iron path. Mean iron length and cross-section:
$$ \ell_{iron} = \ell_{arc}+2\ell_{leg} = 0.60+2(0.30) = 1.2\text{ m}, \qquad
A = (0.05)(0.05) = 2.5\times10^{-3}\text{ m}^2. $$
$$ R_{iron} = \frac{\ell_{iron}}{\mu_0\mu_rA} =
\frac{1.2}{(4\pi\times10^{-7})(2000)(2.5\times10^{-3})} \approx 1.910\times10^5\ \text{A}\cdot
\text{turns/Wb}. $$
Reluctance of the air gaps (part b). Each gap has the same
cross-section as the leg above it; the two gaps (one per leg) are in series with each other and with
the iron:
$$ R_{gap,each} = \frac{g}{\mu_0A} = \frac{10^{-3}}{(4\pi\times10^{-7})(2.5\times10^{-3})}
\approx 3.183\times10^5, \qquad R_{gap,total}=2R_{gap,each}\approx \boxed{6.366\times10^5\
\text{A}\cdot\text{turns/Wb}}. $$
$$ R_{total} = R_{iron}+R_{gap,total} \approx \boxed{8.276\times10^5\ \text{A}\cdot\text{turns/Wb}}
\quad(\text{air gaps} \approx 77\%\text{ of the total}). $$
Flux, flux density, field intensity in the gap (part c).
$$ \phi = \frac{\mathcal F}{R_{total}} = \frac{1000}{8.276\times10^5} \approx \boxed{1.208\text{ mWb}}. $$
$$ B_{gap} = \frac{\phi}{A} = \frac{1.208\times10^{-3}}{2.5\times10^{-3}} \approx \boxed{0.483\text{ T}},
\qquad H_{gap} = \frac{B_{gap}}{\mu_0} \approx \boxed{3.85\times10^5\text{ A/m}}. $$
Electromagnetic force on the armature (part d). Using the
Maxwell-stress (co-energy) result for the pull across one air gap, $F_{gap}=B_{gap}^2A/(2\mu_0)$, and
summing the two gaps (both pull the single rigid armature toward the core in the same sense):
$$ F_{total} = 2\cdot\frac{B_{gap}^2A}{2\mu_0} = \frac{B_{gap}^2A}{\mu_0} =
\frac{(0.483)^2(2.5\times10^{-3})}{4\pi\times10^{-7}} \approx \boxed{464.7\text{ N}}. $$
