Irodov 3.236: A Short Coil Falls Short – Amit & Deepali Sinha Foundation

Problem Irodov 3.236

A solenoid wound in a single layer has length $l$ , cross-sectional radius $R$ and $n$ turns per unit length, and carries a current $I$ . What is the magnetic induction at the centre of the coil?

The coil cut along its axis (dots and crosses: the current I in the turns, its sense chosen), its radius R, and the centre O on the axis, a distance l/2 from either end. — The coil cut along its axis (dots and crosses: the current $I$ in the turns, its sense chosen), its radius $R$ , and the centre $O$ on the axis, a distance $l/2$ from either end.

Given

$l$
$R$
$n$ turns per unit length
$I$

Find

B

at the centre

O

Solution

Step 1The field of one turn on its axis

A turn is a circular loop of radius $R$ . By symmetry the field on its axis is axial: the transverse parts of $d\vect B$ from opposite elements cancel, and Biot and Savart give, at axial distance $z$ from the loop,

B_{\mathrm{loop}} = \frac{\mu_0 I R^2}{2\left(R^2 + z^2\right)^{3/2}} \tag{1}

Every turn’s field at $O$ points the same way along the axis, so the contributions add as numbers.

Step 2The turns in a slice

dz

The turns in a slice of width dz at axial distance z from the centre O: together they are one ring carrying the current nI dz (sense chosen). — The turns in a slice of width $dz$ at axial distance $z$ from the centre $O$ : together they are one ring carrying the current $nI\,dz$ (sense chosen).

Measure $z$ along the axis from $O$ . The slice between $z$ and $z + dz$ holds $n\,dz$ turns, so it acts as one loop carrying $nI\,dz$ ; by (1)

dB = \frac{\mu_0 n I R^2\,dz}{2\left(R^2 + z^2\right)^{3/2}} \tag{2}

Step 3Adding the slices over the length

The winding runs from $z = -l/2$ to $z = l/2$ . Integrating (2):

\begin{aligned} B &= \frac{\mu_0 n I R^2}{2}\int_{-l/2}^{l/2} \frac{dz}{\left(R^2 + z^2\right)^{3/2}} = \frac{\mu_0 n I R^2}{2}\left[\frac{z}{R^2\sqrt{R^2 + z^2}}\right]_{-l/2}^{l/2}\\ &= \frac{\mu_0 n I}{2}\cdot\frac{2\,(l/2)}{\sqrt{R^2 + l^2/4}} = \frac{\mu_0 n I\,l}{\sqrt{4R^2 + l^2}} \end{aligned}

Dividing numerator and denominator by $l$ :

B = \frac{\mu_0 n I}{\sqrt{1 + (2R/l)^2}}

(3)

The field at $O$ lies along the axis, in the sense the right-hand rule gives from the current in the turns (left to right for the current drawn).

Check

For a long coil, $l \gg R$ , the root tends to $1$ and (3) becomes $\mu_0 n I$ , the field inside an infinite solenoid. For a short one, $l \ll R$ , the root is about $2R/l$ and $B \approx \mu_0 (nl) I/(2R)$ : the field at the centre of $nl$ turns bunched into one loop, $\mu_0 I/(2R)$ per turn. Dimensionally $\mu_0 n I$ is a field and the root is a pure number.

Remark

The factor dividing $\mu_0 nI$ in (3) is $1/\cos\theta$ , where $\theta$ is the angle between the axis and the line from $O$ to the rim of an end turn ( $\tan\theta = 2R/l$ ), so $B = \mu_0 nI\cos\theta$ . Even a coil three radii long, as drawn, reaches only $1/\sqrt{1 + 4/9} = 0.83$ of $\mu_0 nI$ , and for the centre to come within $1\%$ of the infinite value the coil must be about $14$ radii long.