Irodov 3.244: Turn Moments Add Along the Chord – Amit & Deepali Sinha Foundation

Problem Irodov 3.244

A thin wire carrying a current $I = 0.8$ A is wound tightly, in $N = 500$ turns, over half of a torus: a tube of circular cross-section bent into a half ring, as in the figure. The diameter of the tube’s cross-section is $d = 5.0$ cm. Find the magnetic moment of the winding.

The half torus wound tightly along its whole length, with the sense of the current I marked on both sides of each end turn and d the diameter of the tube's cross-section. — The half torus wound tightly along its whole length, with the sense of the current $I$ marked on both sides of each end turn and $d$ the diameter of the tube’s cross-section.

Given

$I = 0.8$ A
$d = 5.0$ cm
$N = 500$ turns over half a torus

Find

the magnetic moment

p_m

of the winding

Solution

Step 1The moment of one turn

Each turn is a circle of diameter $d$ lying in a plane through the torus axis, so its normal points along the tube’s center line at that place. Its moment is the current times the area it encloses.

p_1 = I\,\frac{\pi d^2}{4} \tag{1}

By the right-hand rule the moment of every turn points along the center line, all in the same sense round the ring; for the current drawn that sense runs from the left end, over the crown, to the right end.

Step 2The moment of the turns in an angle

d\varphi

The half torus seen along its axis: the dashed center line of radius R, the moments dp of the turns at three angles, each along the local tangent, and their resultant p_m parallel to the chord. — The half torus seen along its axis: the dashed center line of radius $R$ , the moments $d\vect{p}$ of the turns at three angles, each along the local tangent, and their resultant $\vect{p}_m$ parallel to the chord.

Let $R$ be the radius of the center line, and measure the angle $\varphi$ at the center $O$ from the chord, with $x$ along the chord toward the right end and $y$ toward the crown. The $N$ turns are spread evenly over the angle $\pi$ , so an angle $d\varphi$ holds

dN = \frac{N}{\pi}\,d\varphi

turns. In the drawn sense the tangent at $\varphi$ is $\sin\varphi\,\vect{i} - \cos\varphi\,\vect{j}$ , so with (1) these turns carry the moment

d\vect{p} = p_1\,dN\,(\sin\varphi\,\vect{i} - \cos\varphi\,\vect{j}) = \frac{N I d^2}{4}\,(\sin\varphi\,\vect{i} - \cos\varphi\,\vect{j})\,d\varphi \tag{2}

Step 3Adding the moments as vectors

Integrating (2) over the half ring:

\vect{p}_m = \frac{N I d^2}{4}\left(\vect{i}\int_0^{\pi}\sin\varphi\,d\varphi - \vect{j}\int_0^{\pi}\cos\varphi\,d\varphi\right) = \frac{N I d^2}{2}\,\vect{i}

The resultant lies along the chord joining the centers of the two end faces, pointing from the left end toward the right one for the current drawn. Numerically,

p_m = \frac{(500)(0.8\text{ A})(0.050\text{ m})^2}{2}

p_m = \frac{N I d^2}{2} = 0.50 \text{ A}\cdot\text{m}^2

(3)

Check

Had the moments all been parallel, as in a straight coil, they would add to $N p_1 = N I\pi d^2/4 = 0.79$ A $\cdot$ m $^2$ . The ratio of (3) to that is $2/\pi$ , exactly the ratio of the chord $2R$ to the length $\pi R$ of the center line, as it must be when each element of length points along the curve and the elements add up to the chord. The units are those of a current times an area, A $\cdot$ m $^2$ .

Remark

Because the integral of the unit tangent along any curve is the vector from its start to its end, an evenly wound coil bent into any shape has the moment of its turns per unit length times $p_1$ times the vector joining its two ends. Bent into a full torus the ends meet and the moment vanishes, which is why a toroidal coil has no external dipole field.