Irodov 3.225: Half a Tube, Field Along Its Edges – Amit & Deepali Sinha Foundation

Problem Irodov 3.225

A long straight conductor has the cross-section of a thin half-ring of radius $R$ : it is half of a thin-walled tube cut lengthwise along a diameter. A current $I$ flows along the conductor. Find the magnetic induction at a point $O$ on the axis of the half-ring, the center of curvature of its cross-section.

Cross-section of the conductor: a thin half-ring of radius R about the point O, which lies on the line joining its two edges. The current runs along the conductor, normal to the page. — Cross-section of the conductor: a thin half-ring of radius $R$ about the point $O$ , which lies on the line joining its two edges. The current runs along the conductor, normal to the page.

Given

$I$
$R$
current spread evenly over the half-ring (assumed)

Find

\vect B

O

Solution

Step 1Slice the half-ring into thin wires

Take the current out of the page. The current is spread evenly over the half-ring, whose cross-section spans the angle $\pi$ , so a strip of it that subtends $d\psi$ at $O$ is a long thin wire carrying

dI = \frac{I}{\pi}\,d\psi

Two strips symmetric about the axis, at ψ either side of it, each a wire carrying current out of the page. Their fields at O are perpendicular to their radii: the vertical parts cancel, the horizontal parts add. — Two strips symmetric about the axis, at $\psi$ either side of it, each a wire carrying current out of the page. Their fields at $O$ are perpendicular to their radii: the vertical parts cancel, the horizontal parts add.

Every strip is a distance $R$ from $O$ . The field of a long straight wire follows from Ampere’s law on a circle of radius $R$ about it, on which, by symmetry, $dB$ is tangential and the same all round:

dB\cdot 2\pi R = \mu_0\,dI

so, with the strip’s current,

dB = \frac{\mu_0\,dI}{2\pi R} = \frac{\mu_0 I}{2\pi^2 R}\,d\psi \tag{1}

directed perpendicular to the radius from the strip to $O$ .

Step 2Only the part along the edges survives

Measure $\psi$ from the half-ring’s axis of symmetry, the radius to its lowest point, so that $\psi$ runs from $-\pi/2$ to $\pi/2$ . The field (1) of the strip at $\psi$ makes the angle $\psi$ with the line through the two edges, since it is perpendicular to a radius that makes $\psi$ with the axis. The strip at $-\psi$ gives the mirror image: the components along the axis cancel in pairs, and the components along the edge line add. The total is the integral of those:

B = \int_{-\pi/2}^{\pi/2}\frac{\mu_0 I}{2\pi^2 R}\cos\psi\,d\psi = \frac{\mu_0 I}{2\pi^2 R}\Big[\sin\psi\Big]_{-\pi/2}^{\pi/2} = \frac{\mu_0 I}{2\pi^2 R}\cdot 2

Step 3The induction at

O

The whole half-ring with the current out of the page (sense chosen): B at O lies along the line through the edges, here to the left. — The whole half-ring with the current out of the page (sense chosen): $\vect B$ at $O$ lies along the line through the edges, here to the left.

B = \frac{\mu_0 I}{\pi^2 R}

(2)

$\vect B$ lies along the line joining the two edges, in the sense the right-hand rule gives: with the current out of the page and the half-ring below $O$ , $\vect B$ points to the left; reversing the current reverses it.

Check

A complete thin tube carrying $2I$ would be two such halves, an upper and a lower; the upper half gives the mirror image of this field, pointing the other way along the same line, so the total at the axis is zero, as Ampere’s law demands for the inside of a tube. The units: $\mu_0 I/R$ is tesla, and $\pi^2$ is a pure number.

Remark

Compare a single thin wire carrying all of $I$ at the half-ring’s lowest point, a distance $R$ from $O$ : it would give $\mu_0 I/(2\pi R)$ . Spreading the same current over the half-ring keeps every element at distance $R$ but tilts their fields away from the edge line, and the result is smaller by the factor $2/\pi \approx 0.64$ , the mean of $\cos\psi$ over the half-ring.