Question 1 of 5: Toroidal Magnetic Circuit

Question 1: Toroidal Magnetic Circuit (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. A closed iron toroid carrying a magnetizing winding: mean radius \(r = 25\text{ cm}\), core cross-section \(A = 3\text{ cm}^2\), turns \(N = 600\), DC coil current \(I = 1.5\text{ A}\), relative permeability \(\mu_r = 1500\) (assumed constant — linear magnetics).

Find. (a) the reluctance \(\mathcal{R}\); (b) the mmf \(\mathcal{F}\) and field intensity \(H\); (c) the flux \(\Phi\) and flux density \(B\).

[Figure not reproduced: Figure 1 (redrawn): partially wound toroid — a uniform ferromagnetic ring of mean radius \(r\) and cross-section \(A\), excited by an \(N\)-turn coil carrying \(I\); the flux \(B\) circulates within the core. See the official exam paper.]

Approach. Treat the ring as a single uniform magnetic loop: mean path length \(\ell = 2\pi r\); apply the reluctance definition, then Hopkinson's law \(\Phi = \mathcal{F}/\mathcal{R}\) with \(\mathcal{F}=NI\).

1. Mean magnetic path length. The flux follows the mean circumference of the ring: $$\ell = 2\pi r = 2\pi(0.25) = 1.5708\ \text{m}.$$
2. (a) Reluctance of the core. With permeability \(\mu = \mu_0\mu_r\), $$\mathcal{R} = \frac{\ell}{\mu_0\mu_r A} = \frac{1.5708}{(4\pi\times10^{-7})(1500)(3\times10^{-4})}.$$ The denominator is \(\mu_0\mu_r A = 5.655\times10^{-7}\ \text{Wb/A}\cdot\text{t}\), so $$\boxed{\mathcal{R} = 2.78\times10^{6}\ \text{A}\cdot\text{t/Wb}.}$$
3. (b) Magnetomotive force and field intensity. The mmf is the ampere-turns of the coil: $$\mathcal{F} = NI = 600\times1.5 = 900\ \text{A}\cdot\text{t}.$$ The magnetic field intensity is the mmf per unit path length: $$H = \frac{\mathcal{F}}{\ell} = \frac{NI}{2\pi r} = \frac{900}{1.5708} = \boxed{573\ \text{A/m}.}$$
4. (c) Flux and flux density. Hopkinson's law gives the flux driven through the reluctance: $$\Phi = \frac{\mathcal{F}}{\mathcal{R}} = \frac{900}{2.78\times10^{6}} = 3.24\times10^{-4}\ \text{Wb} = 0.324\ \text{mWb}.$$ The flux density is the flux divided by the core area: $$B = \frac{\Phi}{A} = \frac{3.24\times10^{-4}}{3\times10^{-4}} = \boxed{1.08\ \text{T}.}$$ Cross-check: \(B = \mu_0\mu_r H = (4\pi\times10^{-7})(1500)(573) = 1.08\ \text{T}\) — consistent.