Magnetic Circuit of a Toroid: 22-Elec-A6 Power Systems and Machines, December 2018, Q1

Question 1: Magnetic Circuit of a Toroid (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 toroidal iron core wound with a magnetizing coil carrying DC.

Given data
Mean radius	$r_{av} = 25\ \text{cm} = 0.25\ \text{m}$
Cross-sectional area	$A = 3\ \text{cm}^2 = 3\times10^{-4}\ \text{m}^2$
Number of turns	$N = 600$
Coil current	$I = 1.5\ \text{A}$
Relative permeability	$\mu_r = 1500$

[Figure not reproduced: Figure 1 (redrawn): partially wound toroid; the dashed line is the mean magnetic path of length $\ell = 2\pi r_{av}$, and the flux density $B$ circulates around it. See the official exam paper.]

Find. The circuit reluctance $\mathcal{R}$, the magnetomotive force $\mathcal{F}$ and field intensity $H$, and the flux $\Phi$ and flux density $B$.

Approach. Treat the toroid as a single magnetic loop of uniform cross-section; the mean path length gives the reluctance, Ampere's law gives the mmf and $H$, and the magnetic Ohm's law $\Phi = \mathcal{F}/\mathcal{R}$ closes the problem.

Mean magnetic path length. The field follows the circular centre-line of the ring: $$\ell = 2\pi r_{av} = 2\pi(0.25) = 1.5708\ \text{m}.$$
Reluctance of the circuit (a). With permeability $\mu = \mu_0\mu_r$, $$\mathcal{R} = \frac{\ell}{\mu_0\mu_r A} = \frac{1.5708}{(4\pi\times10^{-7})(1500)(3\times10^{-4})}.$$ Evaluating the denominator, $\mu_0\mu_r A = 5.655\times10^{-7}\ \text{Wb/A}\cdot\text{t}$, so $$\boxed{\mathcal{R} = 2.78\times10^{6}\ \text{A}\cdot\text{t/Wb}.}$$
Magnetomotive force (b). The mmf is simply the ampere-turns of the coil: $$\mathcal{F} = N I = (600)(1.5) = 900\ \text{A}\cdot\text{t}.$$
Magnetic field intensity (b). By Ampere's law around the mean path, $$H = \frac{\mathcal{F}}{\ell} = \frac{NI}{\ell} = \frac{900}{1.5708} = 573\ \text{A/m}.$$
Flux (c). Using the magnetic Ohm's law, $$\Phi = \frac{\mathcal{F}}{\mathcal{R}} = \frac{900}{2.78\times10^{6}} = 3.24\times10^{-4}\ \text{Wb} = 0.324\ \text{mWb}.$$
Flux density (c). Dividing the flux by the core area, $$B = \frac{\Phi}{A} = \frac{3.24\times10^{-4}}{3\times10^{-4}} = 1.08\ \text{T}.$$ As a cross-check, $B = \mu_0\mu_r H = (4\pi\times10^{-7})(1500)(573) = 1.08\ \text{T}$, which agrees.

Final results — Question 1
Quantity	Symbol	Value
Reluctance	$\mathcal{R}$	$2.78\times10^{6}$ A·t/Wb
Magnetomotive force	$\mathcal{F}$	900 A·t
Field intensity	$H$	573 A/m
Flux	$\Phi$	0.324 mWb
Flux density	$B$	1.08 T

Topic: Magnetic circuits — reluctance, mmf and the magnetic Ohm's law.

Key relations: $\ell = 2\pi r_{av}$; $\mathcal{R} = \ell/(\mu_0\mu_r A)$; $\mathcal{F} = NI$; $H = \mathcal{F}/\ell$; $\Phi = \mathcal{F}/\mathcal{R}$; $B = \Phi/A = \mu H$.

Why this works: A toroid is the ideal magnetic circuit because the flux is confined to the ring and the cross-section is uniform, so a single lumped reluctance describes the whole loop. Reluctance plays the role of resistance, mmf the role of EMF, and flux the role of current, so $\Phi = \mathcal{F}/\mathcal{R}$ is exactly Ohm's law in magnetic form. Because the relative permeability is given as a constant (1500), the material is treated as linear and $B$ and $H$ are proportional — no B–H curve read is needed here.

Common pitfall: Using the toroid diameter or the outer radius instead of the mean radius for $\ell$, or forgetting to convert cm² to m² ($3\ \text{cm}^2 = 3\times10^{-4}\ \text{m}^2$, a factor of $10^4$). Either slip shifts every downstream answer.

Source: Chapman, Electric Machinery Fundamentals, §1.3–1.4 (magnetic circuits); Guru & Hiziroglu, ch. 2.

Question 1 of 5: Magnetic Circuit of a Toroid

Question 1: Magnetic Circuit of a Toroid (20 marks)