Question 5 of 7: Magnetic Circuit — Relay Armature
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-4 — Electric Circuits and Power — National Exam, May 2014.
Closed book; one aid sheet permitted; Casio/Sharp approved calculator only.
Any five of the seven questions constitute a complete paper — every question is answered.
Reference texts: Sadiku & Alexander, Fundamentals of
Electric Circuits (DC/AC network analysis, Thevenin, resonance);
Chapman, Electric Machinery Fundamentals (magnetic circuits);
Boylestad, Electronic Devices and Circuit Theory (diode rectifiers);
Mano, Digital Design (combinational logic).
Question 5: Magnetic Circuit — Relay Armature (20 marks)
Given. A single series magnetic loop: flux travels up one leg,
across the horseshoe top, down the other leg, across two 1 mm air gaps in series,
and along the armature back to the start.
Given data
Quantity
Value
$\mu_r$ (core & armature)
2000
$N$ (total, both legs)
1000 turns
$i$
1 A
Leg/arc cross-section
5 cm × 6 cm = 30 cm²
Horseshoe mean path (legs+arc)
60 cm
Armature mean path
30 cm
Air gap (each of 2)
1 mm
Find. $\mathcal{F}=Ni$; the reluctance of the core, armature and
air gaps; $\phi$, $B_{gap}$, $H_{gap}$; and the total electromagnetic pull on the
armature.
[Figure not reproduced: Figure 5 — redrawn from the printed figure. The horizontal-arrow "5 cm" dimension is the leg thickness , not a leg length (used with the 6 cm depth for the cross-sectional area). With no separate leg-length dimension given anywhere in the figure, the printed "60 cm" is read as the total mean p. See the official exam paper.]
Check: the leg/arc cross-section (5 cm × 6 cm)
and the "60 cm total horseshoe path" reading are read from the printed figure, which gives no separate leg-length dimension. This reading is the only one that is
dimensionally complete; treated as an engineering assumption per the exam's own
"state assumptions" instruction.
Approach. Model the magnetic circuit as a single series loop of
reluctances (core + armature + two air gaps in series), compute the total mmf and
divide by total reluctance to get the flux, then convert to $B$ and $H$ in the gap and
to force via the Maxwell stress relation.
(a) Total mmf.
$$\boxed{\mathcal{F}=Ni=(1000)(1)=1000\text{ A}\cdot\text{t}}$$
(b) Reluctances. With $A=(0.05)(0.06)=3.0\times10^{-3}\text{ m}^2$
and $\mu_0=4\pi\times10^{-7}$ H/m:
$$\mathcal{R}_{core}=\dfrac{l_{core}}{\mu_0\mu_rA}
=\dfrac{0.60}{(4\pi{\times}10^{-7})(2000)(3.0{\times}10^{-3})}=7.958\times10^4
\text{ A}\cdot\text{t/Wb}$$
$$\mathcal{R}_{arm}=\dfrac{l_{arm}}{\mu_0\mu_rA}
=\dfrac{0.30}{(4\pi{\times}10^{-7})(2000)(3.0{\times}10^{-3})}=3.979\times10^4
\text{ A}\cdot\text{t/Wb}$$
$$\mathcal{R}_{gap,\,total}=\dfrac{2\,l_{gap}}{\mu_0A}
=\dfrac{2(0.001)}{(4\pi{\times}10^{-7})(3.0{\times}10^{-3})}=5.305\times10^5
\text{ A}\cdot\text{t/Wb}$$
$$\boxed{\mathcal{R}_{total}=\mathcal{R}_{core}+\mathcal{R}_{arm}+\mathcal{R}_{gap}
=6.499\times10^5\text{ A}\cdot\text{t/Wb}}$$
(the two 1 mm gaps together account for ≈82% of the total reluctance,
despite being under 0.3% of the flux path length — typical of magnetic circuits
with a working air gap.)
(c) Flux, flux density, field intensity in the gap.
$$\phi=\dfrac{\mathcal{F}}{\mathcal{R}_{total}}=\dfrac{1000}{6.499\times10^5}
\;\Rightarrow\;\boxed{\phi=1.539\times10^{-3}\text{ Wb}}$$
$$B_{gap}=\dfrac{\phi}{A}=\dfrac{1.539\times10^{-3}}{3.0\times10^{-3}}
\;\Rightarrow\;\boxed{B_{gap}=0.5129\text{ T}}$$
$$H_{gap}=\dfrac{B_{gap}}{\mu_0}
\;\Rightarrow\;\boxed{H_{gap}=4.082\times10^5\text{ A/m}}$$
(d) Electromagnetic force on the armature. Each gap pulls with
$F_{gap}=B_{gap}^2A/(2\mu_0)$; the two gaps act together on the same armature:
$$F_{total}=2\cdot\dfrac{B_{gap}^2A}{2\mu_0}=\dfrac{B_{gap}^2A}{\mu_0}
=\dfrac{(0.5129)^2(3.0{\times}10^{-3})}{4\pi{\times}10^{-7}}$$
$$\boxed{F_{total}=628.1\text{ N}}$$
