Magnetic Circuit — Reluctance and Air-Gap Flux: 04-BS-4, December 2015, Q5

Question 5: Magnetic Circuit — Reluctance and Air-Gap Flux (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A 3-limb, 2-window E-core (double-window laminated core) (left limb carries the N-turn coil; a 2 mm air gap is cut into the right limb). The right limb's dimensioning (49 mm + 2 mm gap + 49 mm = 100 mm) confirms the overall core height is 10 cm; the width dimensions (5 cm, 10 cm) locate the three limbs symmetrically. Cross-section A=100 mm² is stated as uniform and overrides any dimension read off the drawing.

Given data
A	μ_r (iron)	N	I	Core height	Air gap
100 mm²	2000	100 turns	1 A	10 cm	2 mm

Check: the figure gives only vertical dimensions numerically (49 mm×2 and 2 mm on the right limb, 10 cm overall); no top/bottom yoke thickness is separately dimensioned. Following the standard simplification for this class of problem, the mean path length of EACH vertical limb is taken as the full 10 cm core height (the right limb splitting into 98 mm iron + 2 mm gap), with yoke reluctance neglected. Because the sub-millimetre air gap dominates the right branch's reluctance by roughly 3 orders of magnitude, the boxed flux/field results are insensitive to reasonable alternate readings of the yoke length.

Find. MMF; the reluctance of each limb; the flux, flux density, and field intensity in the air gap.

(c) The analog (electrical-circuit) representation of the magnetic circuit is shown below: an MMF source in series with R_left, driving R_mid in parallel with (R_right + R_gap).

Analog (electrical-circuit) representation of the magnetic circuit: MMF source in series with R_left, driving R_mid in parallel with (R_right + R_gap).

Approach. Treat the magnetic circuit exactly like a DC resistive network (MMF↔voltage source, reluctance↔resistance, flux↔current): the left limb (with the coil) is in series with the parallel combination of the middle limb and the right limb+gap branch, then apply a flux divider to isolate the gap flux.

(a) Magnetomotive force.
$$\boxed{\mathcal{F} = NI = (100)(1) = 100\text{ A}\cdot\text{turns}}$$
(b) Reluctance of each part. $R=\ell/(\mu A)$, with $\mu=\mu_0\mu_r$ for iron and $\mu=\mu_0$ for the gap:
$$R_{left}=R_{mid}=\frac{0.100}{(4\pi\times10^{-7})(2000)(100\times10^{-6})} = \boxed{3.979\times10^{5}\text{ A}\cdot\text{t/Wb}}$$$$R_{right,iron}=\frac{0.098}{(4\pi\times10^{-7})(2000)(100\times10^{-6})} = 3.899\times10^{5}\text{ A}\cdot\text{t/Wb}$$$$R_{gap}=\frac{0.002}{(4\pi\times10^{-7})(1)(100\times10^{-6})} = \boxed{1.591\times10^{7}\text{ A}\cdot\text{t/Wb}}$$The gap alone accounts for about 97.6% of the right branch's total reluctance $R_{right}=R_{right,iron}+R_{gap}=1.630\times10^{7}$ A·t/Wb.
Total reluctance and total flux. R_mid is in parallel with R_right, in series with R_left:
$$R_{par} = \left(\frac{1}{R_{mid}}+\frac{1}{R_{right}}\right)^{-1} = 3.884\times10^{5}\text{ A}\cdot\text{t/Wb}$$$$R_{tot} = R_{left}+R_{par} = 7.863\times10^{5}\text{ A}\cdot\text{t/Wb}$$$$\phi_{total} = \frac{\mathcal{F}}{R_{tot}} = \frac{100}{7.863\times10^{5}} = 1.272\times10^{-4}\text{ Wb}$$
Flux through the gap (flux divider). Flux splits between the two parallel branches in inverse proportion to their own reluctance:
$$\phi_{gap} = \phi_{total}\cdot\frac{R_{mid}}{R_{mid}+R_{right}} = 1.272\times10^{-4}\cdot\frac{3.979\times10^{5}}{3.979\times10^{5}+1.630\times10^{7}}$$$$\boxed{\phi_{gap} = 3.030\times10^{-6}\text{ Wb}}$$
(d) Flux density and field intensity in the gap.
$$B_{gap} = \frac{\phi_{gap}}{A} = \frac{3.030\times10^{-6}}{100\times10^{-6}} = \boxed{0.0303\text{ T}}$$$$H_{gap} = \frac{B_{gap}}{\mu_0} = \frac{0.0303}{4\pi\times10^{-7}} = \boxed{24{,}108\text{ A}\cdot\text{t/m} \approx 24.1\text{ kA}\cdot\text{t/m}}$$

Final results — Question 5
Quantity	Value
MMF	100 A·t
R_left = R_mid	3.979×10⁵ A·t/Wb
R_right (iron+gap)	1.630×10⁷ A·t/Wb
φ_gap	3.030×10⁻⁶ Wb
B_gap	0.0303 T
H_gap	24,108 A·t/m

Topic: Magnetic circuit analysis by the reluctance/MMF analog.

Key relations: $\mathcal{F}=NI$, $R=\ell/(\mu A)$, $\phi=\mathcal{F}/R_{tot}$, flux divider $\phi_i=\phi_{total}\cdot R_{other}/(R_i+R_{other})$, $B=\phi/A$, $H=B/\mu$.

Why this works: Ampère's law and the solenoidal nature of B (no magnetic "charge") make the magnetic circuit exactly analogous to a resistive electrical network under the substitutions MMF↔EMF, flux↔current, reluctance↔resistance — so ordinary series/parallel reduction and current(flux)-divider reasoning apply unchanged.

Common pitfall: forgetting that an air gap, despite being geometrically tiny, dominates the reluctance of any branch it sits in (because μ₀ ≪ μ₀μ_r) — a 2 mm gap here outweighs 98 mm of iron by roughly 40×, so nearly all of the branch's MMF drop, and nearly all of the design sensitivity, is at the gap.

Source: Sadiku, Fundamentals of Electric Circuits, Ch. 13 (Magnetically Coupled Circuits, magnetic-circuit analog).

Question 5 of 7: Magnetic Circuit — Reluctance and Air-Gap Flux

Question 5: Magnetic Circuit — Reluctance and Air-Gap Flux (20 marks)