Magnetic Circuit with an Air Gap: 04-BS-4, December 2013, Q5

Question 5: Magnetic Circuit with an Air Gap (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. The symmetric double-window core of Figure 5 with uniform cross-section $A=100$ mm$^2$, relative permeability $\mu_r=2000$, $N=100$ turns carrying $I=1$ A ($\mu_0=4\pi\times10^{-7}$).

Given data / geometry
$A$	$\mu_r$	$N$	$I$	gap $\ell_g$	central iron	outer leg (each)
100 mm$^2$	2000	100	1 A	0.2 mm	49.8 mm	90 mm

Find. (a) the magnetomotive force; (b) the reluctance of each part of the magnetic circuit; (c) the analogous electric circuit; (d) the flux, flux density and field intensity in the air gap.

Figure 5 — Symmetric double-window core: central limb (coil N, air gap) with two outer return legs.

Flux driven up the central limb crosses the air gap and then splits equally between the two outer return legs. The magnetic circuit is therefore an MMF source in series with the central-iron and air-gap reluctances, feeding two equal outer-leg reluctances in parallel — exactly the resistive analogue drawn below.

(a) Magnetomotive force. $$\mathcal{F}=NI=100\times1=100\text{ A‑t}.$$
(b) Reluctances $\big(\mathcal{R}=\ell/(\mu_0\mu_rA)$ for iron, $\ell/(\mu_0A)$ for the gap$\big)$: $$\mathcal{R}_{core}=1.981\times10^{5},\ \mathcal{R}_{gap}=1.592\times10^{6},\ \mathcal{R}_{outer}=3.581\times10^{5}\ \text{(each)}.$$The two outer legs in parallel give $\mathcal{R}_{outer}/2=1.790\times10^{5}$, so $$\mathcal{R}_{tot}=\mathcal{R}_{core}+\mathcal{R}_{gap}+\tfrac12\mathcal{R}_{outer}=1.969\times10^{6}.$$ (Units: A‑t/Wb.)
(c) Analogous circuit. The reluctance network is shown below: it is Ohm's law for magnetics, $\mathcal{F}=\phi\,\mathcal{R}$.
Reluctance (magnetic-circuit) analogue: MMF F = NI drives φ through Rcore and Rgap, then splits between the two equal outer-leg reluctances.
(d) Air-gap quantities. The central flux is $$\boxed{\phi=\frac{\mathcal{F}}{\mathcal{R}_{tot}}=5.079\times10^{-5}\text{ Wb}=50.794\ \mu\text{Wb}.}$$ Since the gap shares the central cross-section, $$B_{gap}=\frac{\phi}{A}=0.508\text{ T},\qquad H_{gap}=\frac{B_{gap}}{\mu_0}=4.042\times10^{5}\text{ A/m}.$$

Final results
Quantity	Value
MMF $\mathcal{F}$	100 A‑t
$\mathcal{R}_{core},\mathcal{R}_{gap},\mathcal{R}_{outer}$	$1.981\times10^{5}$, $1.592\times10^{6}$, $3.581\times10^{5}$ A‑t/Wb
Air-gap flux $\phi$	50.794 µWb
$B_{gap}$	0.508 T
$H_{gap}$	$4.042\times10^{5}$ A/m

Check: mean magnetic path lengths are read from the labelled outer dimensions (central iron $=49.8$ mm from the two 24.9 mm halves; each outer leg $\approx90$ mm from the labelled width and height), with fringing neglected and the stated uniform $A=100$ mm$^2$ used everywhere. The air gap dominates the total reluctance (80.841 %), so the result is robust to reasonable path-length choices.

Topic: Magnetic circuits: reluctance, the electric analogue, and air-gap fields.

Key relations: MMF $\mathcal{F}=NI$; reluctance $\mathcal{R}=\ell/(\mu A)$; Hopkinson's law $\mathcal{F}=\phi\mathcal{R}$; series/parallel reluctances; $B=\phi/A$; $H=B/\mu_0$ in air.

Why this works: The magnetic circuit maps one-to-one onto a resistive circuit ($\mathcal{F}\!\leftrightarrow\!V$, $\phi\!\leftrightarrow\!I$, $\mathcal{R}\!\leftrightarrow\!R$), so series-parallel reduction gives the flux directly. Because air has no $\mu_r$, a sub-millimetre gap dominates the reluctance of an otherwise high-permeability core.

Common pitfall: Using $\mu_r$ for the air gap (it is $1$), or forgetting that the flux halves in each outer leg while the full flux crosses the gap.

Source: Sadiku & Alexander, Fundamentals of Electric Circuits, 6th ed., Ch. on magnetically coupled circuits; Chapman, Electric Machinery Fundamentals, Ch. 1 (magnetic circuits).

Question 5 of 7: Magnetic Circuit with an Air Gap

Question 5: Magnetic Circuit with an Air Gap (20 marks)