Irodov 3.306: Inner Turns Count for Less – Amit & Deepali Sinha Foundation

Problem Irodov 3.306

A flat spiral of wire has a large number $N$ of turns lying tightly side by side; the radius of its outermost turn is $a$ . It lies in a uniform magnetic field perpendicular to its plane, and the induction varies as $B = B_0\sin\omega t$ , with $B_0$ and $\omega$ constant. Find the amplitude of the emf induced in the spiral.

Given

$N \gg 1$
$a$
$B = B_0\sin\omega t$ normal to the plane
turns spread evenly from the centre to $a$ (assumed)

Find

\mathcal{E}_m

Solution

Step 1How many turns lie at each radius

The turns are wound tightly side by side, so they are spread evenly across the radius; they are taken to fill it from the centre out to $a$ (assumed: the answer needs the inner radius, and a tight winding leaves no hole). With $N$ large, each turn is very nearly a circle, and the number of turns with radii between $r$ and $r + dr$ is

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

The ring between r and r + dr holds N dr/a turns, each a circle of radius r enclosing the area π r^2. — The ring between $r$ and $r + dr$ holds $N\,dr/a$ turns, each a circle of radius $r$ enclosing the area $\pi r^2$ .

Step 2The flux linked by the whole spiral

The emf of the spiral is the sum of the emfs of its turns, all in series, so it is set by the total flux linkage: each turn of radius $r$ links $B\pi r^2$ . Summing over the rings with (1):

\begin{aligned} \Psi &= \int_0^a B\,\pi r^2\,\frac{N}{a}\,dr = \frac{\pi B N}{a}\cdot\frac{a^3}{3} \end{aligned}

\Psi = \frac{\pi a^2 N}{3}\,B

(2)

Step 3The emf and its amplitude

Faraday’s law with $B = B_0\sin\omega t$ in (2):

\mathcal{E}_i = -\frac{d\Psi}{dt} = -\frac{\pi a^2 N}{3}\,\omega B_0\cos\omega t

The amplitude is the coefficient of $\cos\omega t$ :

\mathcal{E}_m = \frac{1}{3}\pi a^2 N\omega B_0

(3)

Check

Compared with $N$ turns all of radius $a$ , which would give $\pi a^2 N\omega B_0$ , the spiral gives a third, as the average of $r^2$ requires; with one turn at each of $N$ radii the sum $\sum k^2 \approx N^3/3$ gives the same factor. Units: $\text{m}^2\cdot\text{s}^{-1}\cdot\text{T} = \text{Wb/s} = \text{V}$ .

Remark

A flat pancake coil wound from the centre therefore acts like $N$ turns of the effective area $\pi a^2/3$ , not $\pi a^2$ . If the turns left a hole of radius $a_0$ , the factor would become $(a^3 - a_0^3)/\bigl(3a^2(a - a_0)\bigr)$ times $\pi a^2$ per turn, rising toward $\pi a^2$ as the winding narrows to a single band at the rim.