Irodov 3.242: A Logarithm from a Rectangle – Amit & Deepali Sinha Foundation

Problem Irodov 3.242

A toroidal solenoid is wound on a ring whose cross-section is a rectangle: the ring is a flat annulus of height $h = 5.0$ cm, and the ratio of its outer diameter to its inner diameter is $\eta = 1.6$ . The winding has $N = 1000$ turns in total and carries a current $I = 1.7$ A. Find the magnetic flux through the rectangular cross-section of the ring.

The toroidal solenoid of rectangular section, a sector cut away to show the section (hatched) of height h; the strokes are the turns of the winding. — The toroidal solenoid of rectangular section, a sector cut away to show the section (hatched) of height $h$ ; the strokes are the turns of the winding.

Given

$I = 1.7$ A
$N = 1000$
$\eta = 1.6$ (outer to inner diameter)
$h = 5.0$ cm
non-magnetic core, $\mu = 1$ (assumed)

Find

\Phi

through the cross-section

Solution

Step 1The field inside the core

The ring from above: B runs round the ring, and a circle of radius r inside the core links all N turns; B and the current are drawn for one sense (sense chosen). — The ring from above: $\vect B$ runs round the ring, and a circle of radius $r$ inside the core links all $N$ turns; $\vect B$ and the current are drawn for one sense (sense chosen).

By symmetry the field inside runs round the ring, with a magnitude that depends only on the distance $r$ from the axis. A circle of radius $r$ inside the core links all $N$ turns, so Ampere’s law gives

B\cdot 2\pi r = \mu_0 N I

B = \frac{\mu_0 N I}{2\pi r}

(1)

Outside the core the enclosed current is zero and so is the field. Inside, $B$ is not uniform: it is stronger near the inner wall.

Step 2The flux through the section

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

Let the inner and outer radii be $a$ and $b$ , so $b/a = \eta$ (the ratio of the diameters). The field (1) crosses the section square to it and is constant on a strip of height $h$ and width $dr$ , so

d\Phi = B\,h\,dr = \frac{\mu_0 N I h}{2\pi}\,\frac{dr}{r}

Integrating across the section:

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

Numerically,

\Phi = \frac{(4\pi\times 10^{-7}\text{ H/m})(1000)(1.7\text{ A})(0.050\text{ m})}{2\pi}\,\ln 1.6

\Phi = \frac{\mu_0 N I h}{2\pi}\,\ln\eta = 8.0\ \mu\text{Wb}

(2)

Check

For a thin ring, $\eta = 1 + \delta$ with $\delta \ll 1$ , $\ln\eta \approx \delta = (b - a)/a$ and (2) becomes $\mu_0 N I\,h(b - a)/(2\pi a)$ : the uniform field $\mu_0 n I$ , with $n = N/(2\pi a)$ turns per unit length, times the section’s area $h(b - a)$ , as for a straight solenoid. Units: $\mu_0 N I$ is in Wb/m, times a length gives Wb.

Remark

Using the field at the mean radius would give $\Phi \approx \mu_0 N I h (b - a)/\bigl(\pi(a + b)\bigr)$ , which for $\eta = 1.6$ is $2(\eta - 1)/(\eta + 1) = 0.462$ against $\ln\eta = 0.470$ , under $2\%$ low. The mean-radius estimate is low because $1/r$ is convex, and its relative error is only of second order in $(b - a)/(a + b)$ , about a third of that ratio squared.