Irodov 3.330: A Wire Threading Every Turn – Amit & Deepali Sinha Foundation

Problem Irodov 3.330

A toroidal (doughnut) coil of $N$ turns has a rectangular cross-section: its inner radius is $a$ , its outer radius is $b$ , and the side of the cross-section parallel to the axis has length $h$ . An infinitely long straight wire runs along the axis of the coil. The whole system is immersed in a uniform magnetic medium of permeability $\mu$ . Find the mutual inductance of the coil and the wire.

The toroidal coil of N turns with rectangular section of height h (a sector cut away to show it), and the straight wire along its axis. — The toroidal coil of $N$ turns with rectangular section of height $h$ (a sector cut away to show it), and the straight wire along its axis.

The section through the axis: the wire and the core's two cross-sections, inner radius a, outer radius b, height h. — The section through the axis: the wire and the core’s two cross-sections, inner radius $a$ , outer radius $b$ , height $h$ .

Given

$N$ turns
inner radius $a$ , outer radius $b$
section side $h$ along the axis
wire along the axis
uniform medium of permeability $\mu$

Find

the mutual inductance

L_{12}

Solution

Step 1The wire’s field inside the core

Let a current $I$ flow along the wire. By symmetry its field is the same all round a circle of radius $r$ about the axis and runs along that circle, and in the medium the circulation of $\vect H$ equals the current enclosed:

H\cdot 2\pi r = I

With $B = \mu\mu_0 H$ in the uniform medium:

B(r) = \frac{\mu\mu_0 I}{2\pi r}

(1)

These circles run round the ring, so inside the core the field crosses every turn’s rectangular section square on.

Step 2The flux through one turn

The same section with a current I in the wire (sense chosen); the strip of height h and width dr at the radius r. — The same section with a current $I$ in the wire (sense chosen); the strip of height $h$ and width $dr$ at the radius $r$ .

On a strip of the section of height $h$ and width $dr$ at the radius $r$ , the field (1) is constant, so

d\Phi_1 = B(r)\,h\,dr = \frac{\mu\mu_0 I h}{2\pi}\,\frac{dr}{r}

Summing the strips across the section, from $r = a$ to $r = b$ :

\Phi_1 = \frac{\mu\mu_0 I h}{2\pi}\int_a^b\frac{dr}{r} = \frac{\mu\mu_0 I h}{2\pi}\,\ln\frac{b}{a}

Step 3The mutual inductance

Every one of the $N$ turns has the same section and catches the same flux $\Phi_1$ , so the flux linked by the whole coil is $N\Phi_1$ . Per unit current in the wire:

L_{12} = \frac{N\Phi_1}{I} = \frac{\mu\mu_0 N h}{2\pi}\,\ln\frac{b}{a}

(2)

Check

Reciprocity, $L_{21} = L_{12}$ , the other way round. A current $I'$ in the coil gives inside the core, by the circulation theorem round a circle of radius $r$ that links all $N$ turns, $B' = \mu\mu_0 N I'/(2\pi r)$ , and nothing outside. The wire’s circuit closes far away, so any surface it bounds is crossed once by the whole core, and the flux through it is the core’s section flux:

\Phi_{21} = \int_a^b\frac{\mu\mu_0 N I'}{2\pi r}\,h\,dr = \frac{\mu\mu_0 N h}{2\pi}\,\ln\frac{b}{a}\;I',

which is $L_{12}I'$ with $L_{12}$ from (2). Units: $\mu_0 h$ is in henries.

Remark

The wire passes through the hole of the doughnut and so links every turn once: the coil and the wire together are a transformer whose secondary has $N$ turns and whose primary has one. That is the current transformer used to measure a large current without breaking the circuit, a toroid clamped round the conductor, and since only the flux matters the reading does not depend on where the wire passes through the hole. The permeability $\mu$ multiplies $L_{12}$ directly, which is why such cores are made of iron.