Magnetomotive Force, Flux and Inductances of a Gapped Two-Winding Magnetic Core: 04-BS-9, December 2019, Q7

Question 7: Magnetomotive Force, Flux and Inductances of a Gapped Two-Winding Magnetic Core (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.

Quantity	Value
Core relative permeability $\mu_r$	$2000$
Turns $N_L$, $N_R$; current $i$ (in $N_L$ only)	$1000$, $500$; $i=2$ A
Core external width $\times$ height	$70\times30$ cm
Arm (limb) thickness $\times$ depth	$5\times6$ cm $\Rightarrow A=0.003$ m$^2$
Air-gap length $\ell_g$ (left limb)	$1$ mm

Find. (a) the mmf. (b) flux $\phi$ and $B$ in the gap. (c) self-inductance of the left coil. (d) mutual inductance between the two coils.

Rectangular core (schematic, not to scale): single closed flux path, left limb carries N_L and the current i, right limb carries N_R; a 1mm air gap sits in the left limb's flux path.

Approach. Model the closed core as a single series magnetic circuit (core reluctance in series with the air-gap reluctance); mmf drives the same flux through both, so self- and mutual inductance follow directly from $N^2/R_{\text{tot}}$ and $N_LN_R/R_{\text{tot}}$.

Mean path and reluctances (needed for all four parts). Mean centreline dimensions are the external dimensions minus one arm thickness: $65\times25$ cm, giving mean path length $l_{\text{mean}}=2(0.65+0.25)=1.800$ m; the $1$mm gap is cut out of this path, leaving $l_{\text{core}}=1.799$ m of actual core material. Cross-section $A=(0.05)(0.06)=0.003$ m$^2$: $$R_{\text{core}}=\frac{l_{\text{core}}}{\mu_0\mu_rA}=\frac{1.799}{(4\pi\times10^{-7})(2000)(0.003)}=2.386\times10^5\ \text{A}\cdot\text{t/Wb}$$ $$R_{\text{gap}}=\frac{\ell_g}{\mu_0A}=\frac{0.001}{(4\pi\times10^{-7})(0.003)}=2.653\times10^5\ \text{A}\cdot\text{t/Wb}$$ $$R_{\text{tot}}=R_{\text{core}}+R_{\text{gap}}=5.039\times10^5\ \text{A}\cdot\text{t/Wb}$$ Note the $1$mm gap contributes MORE reluctance than the entire $1.8$m of high-permeability core — the classic "small gap, big effect" result.
Part (a) — magnetomotive force. $$\boxed{\mathcal F=N_Li=(1000)(2)=2000\ \text{A}\cdot\text{turns}}$$
Part (b) — flux and gap flux density. $$\phi=\frac{\mathcal F}{R_{\text{tot}}}=\frac{2000}{5.039\times10^5}$$ $$\boxed{\phi=3.969\times10^{-3}\ \text{Wb}}$$ $$B_{\text{gap}}=\frac{\phi}{A}=\frac{3.969\times10^{-3}}{0.003}$$ $$\boxed{B_{\text{gap}}=1.323\ \text{T}}$$
Part (c) — self-inductance of the left coil. All $N_L$ turns link the single flux loop, so $L=N\Phi/I=N^2/R_{\text{tot}}$: $$L_L=\frac{N_L^2}{R_{\text{tot}}}=\frac{(1000)^2}{5.039\times10^5}$$ $$\boxed{L_L=1.985\ \text{H}}$$
Part (d) — mutual inductance. The same flux $\phi$ (driven by the left coil) links all $N_R$ turns of the right coil, so $M=N_R\Phi/I=N_LN_R/R_{\text{tot}}$: $$M=\frac{N_LN_R}{R_{\text{tot}}}=\frac{(1000)(500)}{5.039\times10^5}$$ $$\boxed{M=0.9923\ \text{H}}$$

Quantity	Result
$R_{\text{core}}$, $R_{\text{gap}}$, $R_{\text{tot}}$	$2.386\times10^5$, $2.653\times10^5$, $5.039\times10^5$ A·t/Wb
Magnetomotive force	$2000$ A·turns
Flux $\phi$	$3.969\times10^{-3}$ Wb
Gap flux density $B_{\text{gap}}$	$1.323$ T
Self-inductance $L_L$	$1.985$ H
Mutual inductance $M$	$0.9923$ H

Concept & Theory

Topic. Magnetic-circuit (reluctance) analysis of a gapped, two-winding core.

Key relations. Reluctance $R=l/(\mu A)$; series reluctances add; mmf $\mathcal F=Ni$; $\phi=\mathcal F/R_{\text{tot}}$; self-inductance $L=N^2/R_{\text{tot}}$; mutual inductance $M=N_1N_2/R_{\text{tot}}$ when both windings share one flux path with no leakage.

Why this works. A single closed core with one flux loop is exactly the magnetic analogue of a series electrical circuit — reluctances (like resistances) add along the path, and every winding on that same loop links the identical flux, so both self- and mutual inductance reduce to simple ratios of $N^2$ or $N_1N_2$ over the same total reluctance.

Common pitfall. Neglecting the air gap's reluctance because it is "only 1mm" — here it is actually LARGER than the reluctance of the entire $1.8$m high-permeability core path, because $\mu_r=2000$ makes the core's reluctance-per-metre roughly $2000\times$ smaller than air's.

Source. Sadiku, Elements of Electromagnetics, Ch. 8 (magnetic circuits, inductance); Hayt & Buck, Ch. 9.

Question 7 of 7: Magnetomotive Force, Flux and Inductances of a Gapped Two-Winding Magnetic Core

Question 7: Magnetomotive Force, Flux and Inductances of a Gapped Two-Winding Magnetic Core (20 marks)