Irodov 3.282: A Millimeter Gap Takes the Lead – Amit & Deepali Sinha Foundation

Problem Irodov 3.282

A ring-shaped (toroidal) iron core of mean radius $R = 250$ mm carries a winding of $N = 1000$ turns in total. The core is cut across, leaving a gap of width $b = 1.00$ mm. When a current $I = 0.85$ A flows in the winding, the magnetic induction in the gap is $B = 0.75$ T. Neglecting the spreading of the flux at the edges of the gap, 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 and the gap of width b. The closed lines are B, the same field in the iron and across the gap. — The cut iron ring (permeability $\mu$ ) seen from above, with its winding of $N$ turns carrying $I$ and the gap of width $b$ . The closed lines are $\vect B$ , the same field in the iron and across the gap.

Given

$R = 250$ mm
$N = 1000$
$b = 1.00$ mm
$I = 0.85$ A
$B = 0.75$ T in the gap

Find

the permeability

\mu

of the iron

Solution

Step 1The same B in the iron and in the gap

With no spreading at the edges, the flux tube keeps the core’s cross-section through the gap. The normal component of $\vect B$ is continuous across each face of the cut, and here it is the whole field, so the induction is $B$ everywhere along the ring. The field strength $\vect H$ is not the same: in the iron and in the gap it is

H_{\mathrm{Fe}} = \frac{B}{\mu\mu_0}, \qquad H_0 = \frac{B}{\mu_0} \tag{1}

The ring with its field strength: H in the iron is a short stub, H_0 spans the gap. The true ratio H_Fe/H_0 = 1/μ is far smaller than drawn. — The ring with its field strength: $\vect H$ in the iron is a short stub, $\vect H_0$ spans the gap. The true ratio $H_{\mathrm{Fe}}/H_0 = 1/\mu$ is far smaller than drawn.

Step 2Circulation of H round the ring

Take the contour along the mean circle of the ring, a path of length $2\pi R - b$ in the iron and $b$ across the gap. It links the winding $N$ times, so the circulation of $\vect H$ along it equals $NI$ :

H_{\mathrm{Fe}}\,(2\pi R - b) + H_0\,b = NI

Substituting (1) and multiplying through by $\mu_0$ :

\frac{B\,(2\pi R - b)}{\mu} + Bb = \mu_0 NI \tag{2}

Step 3Solving for the permeability

Moving $Bb$ across in (2) isolates the iron’s term:

\frac{B\,(2\pi R - b)}{\mu} = \mu_0 NI - bB

so $\mu$ is the ratio of $B\,(2\pi R - b)$ to $\mu_0NI - bB$ . To evaluate it, note that the gap is $1$ mm against a mean circumference $2\pi R = 1.57$ m, so $b$ is negligible beside $2\pi R$ in the numerator (it changes the result by less than $0.1$ percent), which leaves $\mu \approx 2\pi RB/(\mu_0NI - bB)$ . The two terms of the denominator are both near $10^{-3}$ T m, so neither may be dropped there:

\begin{aligned} \mu_0 NI &= (4\pi\times10^{-7})(1000)(0.85) = 1.068\times10^{-3}\text{ T m}\\ bB &= (1.00\times10^{-3})(0.75) = 0.750\times10^{-3}\text{ T m}\\ 2\pi RB &= 2\pi(0.250)(0.75) = 1.178\text{ T m} \end{aligned}

Numerically, the ratio is

\mu = \frac{1.178}{(1.068 - 0.750)\times10^{-3}} = \frac{1.178}{0.318\times10^{-3}} \approx 3.7\times10^{3}

\mu = \frac{B\,(2\pi R - b)}{\mu_0 NI - bB} \approx 3.7\times10^{3}

(3)

Check

The ampere-turns add back to $NI$ . With $\mu = 3.7\times10^3$ in (1), the iron’s share and the gap’s share are

\begin{aligned} H_{\mathrm{Fe}}\,(2\pi R - b) &= \frac{0.75}{(3.7\times10^3)(1.257\times10^{-6})}\,(1.570) \approx 253\text{ A}\\ H_0\,b &= \frac{0.75}{1.257\times10^{-6}}\,(1.00\times10^{-3}) \approx 597\text{ A} \end{aligned}

and their sum, $850$ A, is $NI = 1000 \times 0.85$ A. As a limit, $\mu \to \infty$ in (3) needs $\mu_0 NI = bB$ , so even ideal iron could give at most $B = \mu_0NI/b = 1.07$ T here; the measured $0.75$ T sits below that, as it must.

Remark

The gap, one part in about $1600$ of the path, takes $597$ of the $850$ ampere-turns. That is why a gapped core is used where a stable inductance is wanted: the field is set mostly by the gap and hardly by the iron’s $\mu$ , which varies with the field and the temperature. The other side of the same coin is that this measurement of $\mu$ is delicate, because it rests on the difference $\mu_0NI - bB$ of two nearly equal numbers: a 1 percent error in $B$ moves $\mu$ by about $3.4$ percent.