Irodov 3.245: A Spiral Is a Stack of Rings – Amit & Deepali Sinha Foundation

Problem Irodov 3.245

A thin insulated wire is coiled into a flat spiral of $N = 100$ closely packed turns, the innermost of radius $a = 50$ mm and the outermost of radius $b = 100$ mm; the wire carries a current $I = 8$ mA, fed in at the outer end and taken out from the inner end by a straight lead across the turns. Find:

the magnetic induction at the centre of the spiral;
the magnetic moment of the spiral carrying this current.

The flat spiral (a few of its turns drawn): the current I enters at the outermost turn and leaves from the inner end by a straight lead running across the turns. — The flat spiral (a few of its turns drawn): the current $I$ enters at the outermost turn and leaves from the inner end by a straight lead running across the turns.

Given

$N = 100$
$I = 8$ mA
$a = 50$ mm
$b = 100$ mm
turns tight, so evenly spread from $a$ to $b$

Find

(a)

B

at the centre; (b)

p_m

Solution

Step 1The turns as rings

The turns between r and r + dr, inside the annulus a < r < b, act as one ring of radius r carrying I dN; its field d B at the centre is out of the page for the current drawn. — The turns between $r$ and $r + dr$ , inside the annulus $a < r < b$ , act as one ring of radius $r$ carrying $I\,dN$ ; its field $d\vect B$ at the centre is out of the page for the current drawn.

The turns are tight, so they fill the annulus $a < r < b$ evenly, $N/(b - a)$ of them per unit radius. Each turn is nearly a circle, its radius growing by only $(b - a)/N$ per turn, so the turns between $r$ and $r + dr$ act as one ring of radius $r$ carrying $I\,dN$ , with

dN = \frac{N}{b - a}\,dr \tag{1}

All these rings carry the current in the same sense round the centre. The return lead runs straight across the turns, close to a radius; an exactly radial lead gives no field at the centre (each element is parallel to its radius), so the lead’s field there is negligible beside that of the turns, and the two leads outside the spiral, close together, enclose no area and add no moment.

Step 2The field at the centre

A ring of radius $r$ carrying $I\,dN$ gives $\mu_0 I\,dN/(2r)$ at its centre, along the axis (every element of the ring is square to its radius). With (1):

dB = \frac{\mu_0 I}{2r}\,\frac{N\,dr}{b - a}

Integrating over the annulus:

B = \frac{\mu_0 N I}{2(b - a)}\int_a^b \frac{dr}{r} = \frac{\mu_0 N I}{2(b - a)}\,\ln\frac{b}{a}

Numerically,

B = \frac{(4\pi\times 10^{-7}\text{ H/m})(100)(8\times 10^{-3}\text{ A})}{2\,(0.050\text{ m})}\,\ln 2

B = \frac{\mu_0 N I\ln(b/a)}{2(b - a)} = 7.0\ \mu\text{T}

(2)

$\vect B$ points along the axis, out of the page for the current drawn.

Step 3The magnetic moment

A ring of radius $r$ carrying $I\,dN$ has the moment $I\,\pi r^2\,dN$ , along the axis, and all the rings’ moments point the same way. With (1):

p_m = \frac{\pi N I}{b - a}\int_a^b r^2\,dr = \frac{\pi N I}{b - a}\cdot\frac{b^3 - a^3}{3}

Since $b^3 - a^3 = (b - a)(a^2 + ab + b^2)$ , the factor $b - a$ cancels. Numerically,

p_m = \frac{\pi\,(8\times 10^{-3}\text{ A})(100)}{3}\left[(0.050)^2 + (0.050)(0.100) + (0.100)^2\right]\text{ m}^2

p_m = \frac{\pi N I}{3}\left(a^2 + ab + b^2\right) = 15\ \text{mA}\cdot\text{m}^2

(3)

$\vect p_m$ points the same way as $\vect B$ at the centre.

Check

For a narrow spiral, $b = a + \delta$ with $\delta \ll a$ , $\ln(b/a) \approx \delta/a$ and (2) becomes $\mu_0 N I/(2a)$ , the field of $N$ turns bunched on one circle; likewise (3) becomes $\pi N I\cdot 3a^2/3 = N I\pi a^2$ , the moment of $N$ turns of area $\pi a^2$ . Units: $\mu_0 I$ over a length is a field, and a current times an area is a moment.

Remark

The two answers weight the turns oppositely: the field at the centre counts each ring as $1/r$ , the moment as $r^2$ . The moment is therefore set by the outer turns, $\pi N I\,(a^2 + ab + b^2)/3$ lying between $N I\pi a^2$ and $N I\pi b^2$ (here $0.58$ of the second), while the field at the centre is set by the inner ones; a spiral filled in to $a \to 0$ keeps a finite moment but its central field grows without bound as $\ln(b/a)$ .