Irodov 3.320: A Square Section, a Logarithm – Amit & Deepali Sinha Foundation

Problem Irodov 3.320

A toroidal coil of $N$ turns is wound on a ring whose cross-section is a square of side $a$ ; the inner radius of the ring is $b$ . The inside of the coil is filled with a uniform paramagnetic material of permeability $\mu$ . Find the inductance of the coil.

The toroidal coil, a sector cut away to show its square section (hatched) of side a; b is the inner radius, from the axis to the inner wall. The strokes are the turns; the core inside them has permeability μ. — The toroidal coil, a sector cut away to show its square section (hatched) of side $a$ ; $b$ is the inner radius, from the axis to the inner wall. The strokes are the turns; the core inside them has permeability $\mu$ .

Given

$b$ (inner radius)
square section, side $a$
$N$ turns
core of permeability $\mu$ , uniform

Find

L

Solution

Step 1The field inside the core

The coil from above: B runs round the ring, and a circle of radius r inside the core links all N turns. The current's sense is not given; it is drawn one way (sense chosen), and reversing it reverses B only. — The coil from above: $\vect B$ runs round the ring, and a circle of radius $r$ inside the core links all $N$ turns. The current’s sense is not given; it is drawn one way (sense chosen), and reversing it reverses $\vect B$ only.

By symmetry the field runs round the ring and its magnitude depends only on the distance $r$ from the axis. The circulation theorem for $\vect H$ counts the free current only, and a circle of radius $r$ inside the core threads all $N$ turns, each carrying $I$ :

H\cdot 2\pi r = N I

The core is uniform with permeability $\mu$ , so $B = \mu\mu_0 H$ :

B = \frac{\mu\mu_0 N I}{2\pi r},\qquad b < r < b + a

(1)

A circle inside the hole encloses no current, and one beyond the outer wall encloses each turn’s current twice, up the inner wall and down the outer one, for a total of zero: the field outside the core is zero.

Step 2The flux through one turn

The element of the section: a strip of height a and width dr on the hatched square. — The element of the section: a strip of height $a$ and width $dr$ on the hatched square.

Each turn wraps the square section. The field (1) crosses the section square to it and is constant on a strip of height $a$ and width $dr$ at the radius $r$ , so integrating across the section from $b$ to $b + a$ :

\Phi = \int_b^{b+a}\frac{\mu\mu_0 N I}{2\pi r}\,a\,dr = \frac{\mu\mu_0 N I a}{2\pi}\,\ln\frac{b+a}{b}

Step 3The inductance

All $N$ turns link this flux, so the flux linkage is $N\Phi$ and the inductance is $N\Phi/I$ :

L = \frac{\mu\mu_0 N^2 a}{2\pi}\,\ln\!\left(1 + \frac{a}{b}\right)

(2)

Check

A thin ring, $a \ll b$ : then $\ln(1 + a/b) \approx a/b$ and (2) becomes

L \approx \frac{\mu\mu_0 N^2 a^2}{2\pi b}

which is the long-solenoid result $\mu\mu_0 N^2 S/\ell$ with the section $S = a^2$ and the length $\ell = 2\pi b$ of the ring bent round. Units: $\mu_0 N^2 a$ is $(\text{H/m})\cdot\text{m} = \text{H}$ .

Remark

A paramagnetic has $\mu$ above 1 by only a few parts in $10^4$ or $10^5$ (aluminium about $1.00002$ ), so filling the coil with one raises its inductance by that tiny fraction: the factor $\mu$ in (2) matters in practice only for an iron core, where it reaches thousands. Because the core fills the whole space the field occupies, $\vect H$ is set by the winding alone and the core simply multiplies $\vect B$ everywhere by $\mu$ .