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

Question 5: Magnetic Circuit — MMF, 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. Double-window (two-window, E-I style) core: the coil ($N$ turns) is wound on the centre leg; a 0.2 mm air gap sits in the right outer leg only (the left outer leg is solid iron). All legs share the same uniform cross-section.

Find. MMF; $\mathcal{R}_{center}$, $\mathcal{R}_{left}$, $\mathcal{R}_{right,iron}$, $\mathcal{R}_{gap}$; the analog circuit; $\Phi_{gap}$, $B_{gap}$, $H_{gap}$.

Approach. Treat the core exactly like the DC bridge of Question 1: the coil-leg reluctance is a series element carrying the total flux, which then splits between the two outer legs (in parallel) back to the yoke. The left leg is all iron; the right leg is iron in series with the air gap.

1. a) Magnetomotive force. $$\mathcal{F} = NI = (100)(1\text{ A}) = \boxed{100\text{ A-turns}}$$
2. b) Reluctances. $\mathcal{R}=\dfrac{l}{\mu A}$, with $A=100\text{ mm}^2=1\times10^{-4}\text{ m}^2$. Centre and left legs share the same iron length (50 mm, no gap); the right leg splits into 49.8 mm of iron plus the 0.2 mm gap ($\mu_r=1$ in air): $$\mathcal{R}_{center}=\mathcal{R}_{left}=\frac{0.050}{(4\pi\times10^{-7})(2000)(10^{-4})}=1.989\times10^{5}\text{ A-t/Wb}$$ $$\mathcal{R}_{right,iron}=\frac{0.0498}{(4\pi\times10^{-7})(2000)(10^{-4})}=1.981\times10^{5}\text{ A-t/Wb}$$ $$\mathcal{R}_{gap}=\frac{0.0002}{(4\pi\times10^{-7})(1)(10^{-4})}=\boxed{1.592\times10^{6}\text{ A-t/Wb}}$$ The 0.2 mm gap alone contributes 89% of the right leg's total reluctance ($\mathcal{R}_{right}=\mathcal{R}_{right,iron}+\mathcal{R}_{gap}=1.790\times10^6$), which is the usual signature of an air-gapped magnetic circuit.
3. c) Analog circuit. The MMF source drives $\mathcal{R}_{center}$ in series with the parallel combination of $\mathcal{R}_{left}$ and $(\mathcal{R}_{right,iron}+\mathcal{R}_{gap})$ — identical topology to the electrical bridge of Question 1, with MMF↔voltage, flux↔current, reluctance↔resistance.
   Figure 5b: Reluctance-network analog of the magnetic circuit
4. d) Flux, flux density and field intensity in the gap. $$\mathcal{R}_{par}=\mathcal{R}_{left}\parallel\mathcal{R}_{right}=1.790\times10^{5}\text{ A-t/Wb}\ ,\quad \mathcal{R}_{tot}=\mathcal{R}_{center}+\mathcal{R}_{par}=3.780\times10^{5}\text{ A-t/Wb}$$ $$\Phi_{total}=\frac{\mathcal{F}}{\mathcal{R}_{tot}}=\frac{100}{3.780\times10^{5}}=2.646\times10^{-4}\text{ Wb}$$ Current-divider analog splits $\Phi_{total}$ between the two outer legs; the gap carries the entire right-branch flux: $$\Phi_{gap}=\Phi_{total}\cdot\frac{\mathcal{R}_{left}}{\mathcal{R}_{left}+\mathcal{R}_{right}}=\boxed{2.647\times10^{-5}\text{ Wb}}$$ $$B_{gap}=\frac{\Phi_{gap}}{A}=\frac{2.647\times10^{-5}}{1\times10^{-4}}=\boxed{0.2647\text{ T}}\ ,\qquad H_{gap}=\frac{B_{gap}}{\mu_0}=\boxed{2.106\times10^{5}\text{ A/m}}$$