Irodov 3.284: A Load Line Meets the Iron – Amit & Deepali Sinha Foundation

Problem Irodov 3.284

A thin iron ring of mean diameter $d = 50$ cm carries a winding of $N = 800$ turns with a current $I = 3.0$ A. The ring is cut across, leaving a gap of width $b = 2.0$ mm. Neglecting the spreading of the flux at the edges of the gap, and using the basic magnetization curve of the iron shown in the figure, find the permeability of the iron in these conditions.

The cut iron ring (permeability μ) seen from above, with its winding of N turns carrying I, the gap b (widened for clarity) and the mean diameter d. The closed lines are B. — The cut iron ring (permeability $\mu$ ) seen from above, with its winding of $N$ turns carrying $I$ , the gap $b$ (widened for clarity) and the mean diameter $d$ . The closed lines are $\vect B$ .

Given

$d = 50$ cm
$N = 800$
$I = 3.0$ A
$b = 2.0$ mm
the curve $B(H)$
no spreading at the gap edges
$H$ read to about $0.01$ kA/m, $B$ to about $0.02$ T (assumed)

Find

the permeability

\mu

of the iron

Solution

Step 1The circulation of H round the ring

With no spreading at the edges, the flux tube keeps the ring’s cross-section across the gap, and the normal component of $\vect B$ is continuous at each face, so the induction $B$ is the same in the iron and in the gap. In the gap the field strength is $B/\mu_0$ ; in the iron it is $H$ , unknown, tied to $B$ by the curve. The contour along the mean circle runs $\pi d - b$ in the iron and $b$ in the gap, and links the winding $N$ times:

H(\pi d - b) + \frac{B}{\mu_0}\,b = NI \tag{1}

Step 2The load line

Solving (1) for $B$ , and dropping $b$ beside $\pi d$ (it is $0.13$ percent of the path):

\begin{aligned} B &= \frac{\mu_0 NI}{b} - \frac{\mu_0(\pi d - b)}{b}\,H\\[2pt] &\approx \frac{\mu_0 NI}{b} - \frac{\mu_0\pi d}{b}\,H \end{aligned}

Numerically, the intercept and the slope (per kA/m, the unit of the curve’s axis) are

\begin{aligned} \frac{\mu_0 NI}{b} &= \frac{(4\pi\times10^{-7})(800)(3.0)}{2.0\times10^{-3}} = 1.508\text{ T}\\[2pt] \frac{\mu_0\pi d}{b} &= \frac{(4\pi\times10^{-7})\,\pi\,(0.50)}{2.0\times10^{-3}}\times10^{3} = 0.987\text{ T per kA/m} \end{aligned}

so the circulation demands, with $B$ in tesla and $H$ in kA/m,

B \approx 1.51 - 0.987\,H

(2)

Step 3Where the line meets the curve

The iron’s state must satisfy (2) and lie on the magnetization curve at the same time, so it is the intersection of the straight line with the curve.

The given magnetization curve with the line (gold), falling from 1.51 T on the B axis. They cross at the iron's working point, whose coordinates are marked on the axes. — The given magnetization curve with the line (2) (gold), falling from $1.51$ T on the $B$ axis. They cross at the iron’s working point, whose coordinates are marked on the axes.

Read off the figure, $H$ to about $0.01$ kA/m and $B$ to about $0.02$ T (assumed):

H \approx 0.26\text{ kA/m}, \qquad B \approx 1.25\text{ T}

As a test of the reading, the line (2) at that $H$ gives

B = 1.508 - 0.987 \times 0.26 = 1.251\text{ T}

which agrees with the $B$ read on the curve.

Step 4The permeability

With the working point known, the permeability is $\mu = B/(\mu_0 H)$ . Numerically,

\frac{B}{\mu_0 H} = \frac{1.25}{(4\pi\times10^{-7})(260)}

\mu = \frac{B}{\mu_0 H} \approx 3.8\times10^{3}

(3)

Reading $H$ to $\pm 0.01$ kA/m leaves $\mu$ uncertain by about $4$ percent, so the second figure of the result is only just significant.

Check

The working point lies on the curve independently of the line: the curve passes through $1.19$ T at $0.20$ kA/m and $1.29$ T at $0.30$ kA/m, and six tenths of the way between them, at $0.26$ kA/m, it is at $1.25$ T, the value read. And $\mu \approx 3.8\times10^3$ sits on the falling side of the iron’s permeability, past its peak of about $10^4$ at the knee, as it must at $H = 0.26$ kA/m.

Remark

The $2$ mm gap takes about five sixths of the ampere-turns although it is $0.13$ percent of the path. With ideal iron ( $\mu \to \infty$ ) the gap alone would set $B = \mu_0NI/b = 1.51$ T, and the real iron lowers that only to $1.25$ T: the gap, not the iron, mostly decides the field. The same construction, a straight line from the circuit against a curve from the material, is how a transistor’s operating point or a magnet’s working point is found.