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 2017. 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. Figure 5: a horseshoe (U-shaped) iron core, mean path 60 cm,
uniform 5×5 cm2 cross-section, carries the N-turn winding; a movable iron
armature bar (mean path between the two gap centres 30 cm, cross-section
6×5 cm2) closes the loop through TWO 1 mm air gaps (one under each leg,
same 5×5 cm2 area as the legs). The flux path is a single series loop: leg
→ gap → armature → gap → leg → top yoke → back to the first leg.
Given data
Quantity
Value
μr
2000
N (total)
1000 turns
i
1 A
Core mean length
60 cm
Leg / gap area
5×5 cm2
Armature mean length
30 cm
Armature area
6×5 cm2
Air gap (×2)
1 mm each
Find. Total MMF; the reluctance of the core, the armature, and each air gap;
the air-gap flux, flux density, and field intensity; and the total electromagnetic force on the
armature.
Figure 5: horseshoe core + armature, single series
magnetic loop with two 1 mm air gaps
Approach. Sum the series reluctances of the iron core, the iron armature, and
the two air gaps; MMF divided by total reluctance gives the flux (same flux everywhere in a series
loop); B, H follow directly in the gap, and the magnetic force on each gap face sums to the total
pull on the armature.
Total MMF (part a).
$$ \mathcal{F} = Ni = (1000)(1) = \boxed{1000\text{ A-t}}. $$
Reluctance of each part (part b). With
$\mu=\mu_r\mu_0=2000(4\pi\times10^{-7})=2.513\times10^{-3}$ H/m and
$A_{leg}=A_{gap}=(0.05)(0.05)=25\times10^{-4}\text{ m}^2$,
$A_{arm}=(0.06)(0.05)=30\times10^{-4}\text{ m}^2$:
$$ \mathcal{R}_{core} = \frac{L_{core}}{\mu A_{leg}} = \frac{0.60}{(2.513\times10^{-3})(25\times10^{-4})}
= \boxed{9.549\times10^{4}\text{ A-t/Wb}}, $$
$$ \mathcal{R}_{arm} = \frac{L_{arm}}{\mu A_{arm}} = \frac{0.30}{(2.513\times10^{-3})(30\times10^{-4})}
= \boxed{3.979\times10^{4}\text{ A-t/Wb}}, $$
$$ \mathcal{R}_{gap} = \frac{l_g}{\mu_0 A_{gap}} = \frac{0.001}{(4\pi\times10^{-7})(25\times10^{-4})}
= \boxed{3.183\times10^{5}\text{ A-t/Wb (each of the two gaps)}}. $$
Total reluctance and flux. All parts carry the same flux (one series loop),
so the reluctances add, including BOTH gaps:
$$ \mathcal{R}_{tot} = \mathcal{R}_{core}+\mathcal{R}_{arm}+2\mathcal{R}_{gap} =
9.549\times10^4+3.979\times10^4+2(3.183\times10^5) = 7.719\times10^5\text{ A-t/Wb}. $$
$$ \Phi = \frac{\mathcal{F}}{\mathcal{R}_{tot}} = \frac{1000}{7.719\times10^5} =
\boxed{1.296\text{ mWb}}. $$
Flux density and field intensity in the gap (part c).
$$ B_{gap} = \frac{\Phi}{A_{gap}} = \frac{1.296\times10^{-3}}{25\times10^{-4}} = \boxed{0.518\text{ T}},
\qquad H_{gap} = \frac{B_{gap}}{\mu_0} = \frac{0.518}{4\pi\times10^{-7}} =
\boxed{4.124\times10^{5}\text{ A/m}}. $$
Force on the armature (part d). Each gap pulls the armature toward its leg
with $F=B^2A/(2\mu_0)$; both gaps pull in the same closing direction, so the totals add:
$$ F_{each} = \frac{B_{gap}^2 A_{gap}}{2\mu_0} = \frac{(0.518)^2(25\times10^{-4})}{2(4\pi\times10^{-7})}
\approx 267.1\text{ N}, \qquad F_{tot} = 2F_{each} = \boxed{534.2\text{ N}}. $$
