Question 5 of 7: Magnetic Circuit — Horseshoe Core with Armature and Air Gap

Question 5: Magnetic Circuit — Horseshoe Core with Armature and 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. Figure 5: a single horseshoe iron piece (both legs plus the top arc, one continuous member) with a $60$ cm mean path and a uniform $5\text{ cm}\times5\text{ cm}$ leg cross-section; a single $1$ mm air gap between the core's left leg and a separate armature bar; the armature has its own $30$ cm mean path (the horizontal span between the two leg contact points) and a $6\text{ cm}\times5\text{ cm}$ cross-section (its own height times the shared leg width); $\mu_r=2000$ for all iron, $N=1000$ turns total.

Find. The reluctance of the core, gap, and armature; the coil current for $\phi=2$ mWb; $B_{gap}$ and $H_{gap}$; and the total force on the armature.

Approach. This is a single series flux loop (core → gap → armature → back into the core), so the reluctances simply add; solve $\mathcal{F}=NI=\phi\mathcal{R}_{tot}$ for $I$, then $B=\phi/A$, $H=B/\mu_0$, and the Maxwell stress $F=B^2A/(2\mu_0)$ at the one air gap.

1. a) Reluctance of each part. $\mathcal{R}=\ell/(\mu A)$, with $A_{core}=A_{gap}=(0.05)(0.05)=0.0025\text{ m}^2$ and $A_{arm}=(0.06)(0.05)=0.0030\text{ m}^2$: $$\mathcal{R}_{core}=\frac{0.60}{\mu_0(2000)(0.0025)}=\boxed{9.549\times10^{4}\ \text{A}\cdot\text{t/Wb}}$$ $$\mathcal{R}_{gap}=\frac{0.001}{\mu_0(0.0025)}=\boxed{3.183\times10^{5}\ \text{A}\cdot\text{t/Wb}}$$ $$\mathcal{R}_{arm}=\frac{0.30}{\mu_0(2000)(0.0030)}=\boxed{3.979\times10^{4}\ \text{A}\cdot\text{t/Wb}}\ ,\qquad \mathcal{R}_{tot}=\mathcal{R}_{core}+\mathcal{R}_{gap}+\mathcal{R}_{arm}=\boxed{4.536\times10^{5}\ \text{A}\cdot\text{t/Wb}}$$ Even with a mean path over an order of magnitude shorter than the iron, the sub-mm air gap alone supplies about $70\%$ of the total reluctance — the usual signature of a gapped magnetic circuit.
2. b) Coil current. $$NI=\phi\,\mathcal{R}_{tot}\ \Rightarrow\ I=\frac{\phi\,\mathcal{R}_{tot}}{N}=\frac{(0.002)(4.536\times10^{5})}{1000}=\boxed{0.907\text{ A}}$$
3. c) Flux density and field intensity in the gap. $$B_{gap}=\frac{\phi}{A_{gap}}=\frac{0.002}{0.0025}=\boxed{0.800\text{ T}}\ ,\qquad H_{gap}=\frac{B_{gap}}{\mu_0}=\frac{0.800}{4\pi\times10^{-7}}=\boxed{6.366\times10^{5}\text{ A/m}}$$
4. d) Force on the armature. The magnetic pull across the single air gap (Maxwell stress tensor result for a gap of area $A_{gap}$): $$F=\frac{B_{gap}^2A_{gap}}{2\mu_0}=\frac{(0.800)^2(0.0025)}{2(4\pi\times10^{-7})}=\boxed{636.6\text{ N}}$$