Irodov 3.283: Iron Is Most Permeable at the Knee – Amit & Deepali Sinha Foundation

Problem Irodov 3.283

The figure shows the basic magnetization curve, induction $B$ against field strength $H$ , of commercially pure iron. Using this curve, plot the permeability $\mu$ of the iron as a function of $H$ . At what value of $H$ is the permeability greatest, and what is that greatest value $\mu_{\max}$ ?

Given

the curve $B(H)$ of the figure
$H$ read to about $0.01$ kA/m, $B$ to about $0.02$ T (assumed)

Find

\mu(H)

· the

H

of greatest

\mu

\mu_{\max}

Solution

Step 1Permeability as the slope of a chord

For iron on its basic magnetization curve, $B = \mu\mu_0 H$ defines the permeability at each point of the curve:

\mu = \frac{B}{\mu_0 H} \tag{1}

Geometrically, $B/H$ is the slope of the straight line from the origin to the point $(H, B)$ of the curve, so $\mu$ is that slope divided by $\mu_0$ .

Step 2The permeability point by point

Reading pairs $(H, B)$ off the curve and applying (1), with $\mu_0 = 4\pi\times10^{-7}$ H/m and $H$ converted to A/m:

\begin{aligned} H &= 0.04\text{ kA/m},\ B = 0.10\text{ T}: & \mu &= \frac{0.10}{(4\pi\times10^{-7})(40)} \approx 2.0\times10^{3}\\[2pt] H &= 0.05\text{ kA/m},\ B = 0.50\text{ T}: & \mu &\approx 8.0\times10^{3}\\[2pt] H &= 0.06\text{ kA/m},\ B = 0.83\text{ T}: & \mu &\approx 1.1\times10^{4}\\[2pt] H &= 0.10\text{ kA/m},\ B = 1.00\text{ T}: & \mu &\approx 8.0\times10^{3}\\[2pt] H &= 0.20\text{ kA/m},\ B = 1.19\text{ T}: & \mu &\approx 4.7\times10^{3}\\[2pt] H &= 0.40\text{ kA/m},\ B = 1.36\text{ T}: & \mu &\approx 2.7\times10^{3}\\[2pt] H &= 0.80\text{ kA/m},\ B = 1.47\text{ T}: & \mu &\approx 1.5\times10^{3} \end{aligned}

Plotting these against $H$ gives the curve $\mu(H)$ below the magnetization curve.

Top, the given B(H) with the tangent from the origin (dashed) touching it at the knee. Bottom, the permeability μ = B/(μ_0 H) computed from it, sharing the H axis; the dotted drop joins the tangent point to the peak of μ. — Top, the given $B(H)$ with the tangent from the origin (dashed) touching it at the knee. Bottom, the permeability $\mu = B/(\mu_0 H)$ computed from it, sharing the $H$ axis; the dotted drop joins the tangent point to the peak of $\mu$ .

\mu

rises steeply to a peak at the knee, then falls.

Step 3Where the permeability is greatest

By (1), $\mu$ is largest where the chord from the origin is steepest. Turning a line about the origin until it last touches the curve, it becomes the tangent from the origin, and it touches at the top of the steep rise, where the curve begins to bend over:

H \approx 0.06\text{ kA/m}

(2)

On the curve that point is read as $H \approx 0.057$ kA/m and $B \approx 0.80$ T. Numerically, by (1),

\frac{B}{\mu_0 H} = \frac{0.80}{(4\pi\times10^{-7})(57)} \approx 1.1\times10^{4}

On the steep rise $H$ is read only to about $0.01$ kA/m, about a sixth of $H$ itself, so this reading fixes $\mu_{\max}$ only to about twenty percent, one significant figure:

\mu_{\max} \approx 1\times10^{4}

(3)

Check

The chord picture agrees with the readings of Step 2. At the tangent point the line from the origin has slope $0.80/0.057 \approx 14$ T per kA/m, and every other reading has a smaller ratio $B/H$ : $10$ at $0.05$ and $0.10$ kA/m, $2.5$ at $0.04$ kA/m, $1.8$ at $0.80$ kA/m. At large $H$ the iron saturates, $B$ hardly grows, and $\mu \approx B_{\mathrm{sat}}/(\mu_0 H)$ falls roughly as $1/H$ : from $0.40$ to $0.80$ kA/m it drops from $2.7$ to $1.5$ thousand, close to half.

Remark

The permeability of iron is not a constant of the material but a function of the field: a few thousand at the foot of the curve ( $2\times10^{3}$ at $0.04$ kA/m), $10^{4}$ at the knee, and falling toward $1$ deep in saturation. A transformer or an electromagnet core is therefore run near the knee, where a given number of ampere-turns buys the most induction; pushing it further into saturation spends current for little gain.