Irodov 3.337: Energy Read Off the Iron's Curve – Amit & Deepali Sinha Foundation

Problem Irodov 3.337

An iron core in the shape of a doughnut, with a round cross-section of radius $a = 3.0$ cm and a mean radius $b = 32$ cm, carries a winding of $N = 1000$ turns with a current $I = 1.0$ A. Using the basic magnetization curve of the iron shown in the figure, find the magnetic energy stored in the core. Take the field strength $H$ to be the same over the whole cross-section and equal to its value at the centre of the cross-section.

The iron doughnut seen from above, with its winding of N turns carrying I and its mean diameter 2b. Its cross-section, seen edge on as the band, is a circle of radius a. — The iron doughnut seen from above, with its winding of $N$ turns carrying $I$ and its mean diameter $2b$ . Its cross-section, seen edge on as the band, is a circle of radius $a$ .

Given

$a = 3.0$ cm
$b = 32$ cm
$N = 1000$
$I = 1.0$ A
the curve $B(H)$
$H$ uniform, its value at the centre of the section
$B$ read to about $0.02$ T (assumed)

Find

the magnetic energy

W

in the core

Solution

Step 1The field strength from the winding

By the symmetry of the doughnut, $\vect H$ runs along circles about its axis with the same magnitude all round each circle. The centre circle of the cross-section has radius $b$ and links the winding $N$ times, so the circulation theorem for $\vect H$ gives

\begin{aligned} H\cdot 2\pi b &= NI\\ H &= \frac{NI}{2\pi b} \end{aligned} \tag{1}

Numerically,

H = \frac{(1000)(1.0)}{2\pi\,(0.32)} = 497\text{ A/m} \approx 0.50\text{ kA/m}

Step 2The induction from the curve

The iron’s $B$ at this $H$ is read off the magnetization curve, at the point where the vertical through $H = 0.50$ kA/m meets it.

The given magnetization curve with the iron's working point (H, B): the dashed lines run from H = NI/2π b 0.50 kA/m up to the curve and across to the B axis. — The given magnetization curve with the iron’s working point $(H, B)$ : the dashed lines run from $H = NI/2\pi b \approx 0.50$ kA/m up to the curve and across to the $B$ axis.

Read to about $0.02$ T (assumed), the induction there is

B \approx 1.40\text{ T}

far up on the saturating part of the curve.

Step 3The work per unit volume along the curve

As the current rises from zero, the induction in every part of the core rises along the curve, and raising it by $dB$ costs the source $H\,dB$ per unit volume. Up to the working point this is the area between the curve and the $B$ axis, which is the rectangle $BH$ less the area under the curve:

w = \int_0^B H\,dB = BH - \int_0^H B\,dH \tag{2}

For the area under the curve we read it, to about $0.02$ T (assumed), at $H = 0.04$ , $0.05$ , $0.057$ , $0.10$ , $0.20$ , $0.30$ and $0.40$ kA/m, where $B = 0.10$ , $0.50$ , $0.80$ , $1.00$ , $1.19$ , $1.29$ and $1.36$ T, and sum the trapezoids up to the working point. Numerically,

\begin{aligned} BH &= (1.40)(497) = 696\text{ J/m}^3\\ \int_0^H B\,dH &\approx 549\text{ J/m}^3\\ w &\approx 696 - 549 \approx 148\text{ J/m}^3 \end{aligned}

Step 4The energy delivered to the core

With $H$ the same throughout the cross-section, so is $w$ , and the energy is $w$ times the volume of the doughnut: its cross-section $\pi a^2$ times the length $2\pi b$ of the centre circle (Pappus). With (2), numerically,

2\pi^2a^2b\int_0^B H\,dB = 2\pi^2\,(0.030)^2\,(0.32)\,(148)

W = 2\pi^2a^2b\int_0^B H\,dB \approx 0.8\text{ J}

(3)

This is the magnetic energy the source delivers to the core in bringing it up the curve; one figure is what the readings on the steep rise support. The density $BH/2$ , which would give $\pi^2a^2b\,BH = 2.0$ J here, holds only for a medium with $B$ proportional to $H$ .

Check

Units: $H\,dB$ is $(\text{A/m})\cdot\text{T} = (\text{A/m})(\text{V s/m}^2) = \text{J/m}^3$ , an energy density, and $2\pi^2a^2b = 5.7\times10^{-3}$ m $^3$ is the volume. Bounds: the curve lies above the straight chord from the origin to the working point almost all the way (at $0.10$ kA/m the chord is at $0.28$ T and the curve at $1.00$ T), so the area under it exceeds the triangle $BH/2$ , and by (2) $w$ must come out below $BH/2 = 348$ J/m $^3$ , and $148$ J/m $^3$ does. The working point gives $\mu = B/\mu_0H \approx 2.2\times10^3$ , on the falling side of the iron’s permeability, as it should be at $0.50$ kA/m.

Remark

The energy depends on the path through the curve, not only on the working point. Switching the current off, the iron returns along a lower, descending branch and keeps some remanent induction, so less than the $0.8$ J delivered comes back; the difference is the hysteresis loss. Of the $148$ J/m $^3$ , only about a third goes in on the steep rise to $1$ T; the rest is spent in the saturating tail, where each further tenth of a tesla costs a large $H$ .