Question 7 of 7: Magnetic Circuit — Gapped Core with Two Windings

Question 7: Magnetic Circuit — Gapped Core with Two Windings (15 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.

Find. (a) MMF. (b) flux $\phi$ and gap flux density $B$. (c) self inductance $L_L$. (d) mutual inductance $M$.

Approach. Model the whole loop as one series magnetic circuit (reluctances in series, since it is a single closed path, not parallel branches). Because the combined gap length ($3$ mm) is negligible next to the core's own path length, take the core's own material length as the full mean path minus the (tiny) gap length, then add each gap's own reluctance (its permeability is $\mu_0$, not $\mu_0\mu_r$, so it is not negligible there). Solve $\phi=\text{MMF}/R_{\text{tot}}$, then $L=N^2/R_{\text{tot}}$ and $M=N_LN_R/R_{\text{tot}}$.

1. Part (a) — magnetomotive force. Only the left winding carries current: $$\boxed{\mathcal F=N_Li=(100)(2)=200\ \text{A}\cdot\text{t}}$$
2. Part (b) — flux and gap flux density. Mean path length (core material, gaps' own extent dropped as instructed) $\ell_{\text{core}}=1.200-0.003=1.197$ m; cross-section $A=(0.05)(0.06)=3.0\times10^{-3}\ \text{m}^2$: $$R_{\text{core}}=\frac{\ell_{\text{core}}}{\mu_0\mu_rA}=\frac{1.197}{(4\pi\times10^{-7})(2000)(3.0\times10^{-3})}=1.588\times10^5\ \text{A}\cdot\text{t/Wb}$$ $$R_{\text{gap}}=\frac{0.003}{\mu_0A}=\frac{0.003}{(4\pi\times10^{-7})(3.0\times10^{-3})}=7.958\times10^5\ \text{A}\cdot\text{t/Wb}$$ $$R_{\text{tot}}=R_{\text{core}}+R_{\text{gap}}=9.545\times10^5\ \text{A}\cdot\text{t/Wb}$$ (the two thin air gaps dominate the total reluctance, even though they are a tiny fraction of the path length — $\mu_r=2000$ makes the core $2000\times$ "easier" per metre than air). Then: $$\phi=\frac{\mathcal F}{R_{\text{tot}}}=\frac{200}{9.545\times10^5}$$ $$\boxed{\phi=2.095\times10^{-4}\ \text{Wb}}$$ Since the cross-section is uniform all the way around the series loop, $B$ is the same in the gap as everywhere else: $$B=\frac{\phi}{A}=\frac{2.095\times10^{-4}}{3.0\times10^{-3}}$$ $$\boxed{B_{\text{gap}}=0.0698\ \text{T}=69.8\ \text{mT}}$$
3. Part (c) — self inductance of the left coil. $$L_L=\frac{N_L^2}{R_{\text{tot}}}=\frac{100^2}{9.545\times10^5}$$ $$\boxed{L_L=10.48\ \text{mH}}$$
4. Part (d) — mutual inductance. The same flux $\phi$ (driven entirely by the left winding) links every one of the right winding's turns, since both sit on the one shared loop: $$M=\frac{N_LN_R}{R_{\text{tot}}}=\frac{(100)(10)}{9.545\times10^5}$$ $$\boxed{M=1.048\ \text{mH}}$$ (consistency check: $M=L_L\cdot N_R/N_L=10.48\times(10/100)=1.048$ mH, matching.)