Irodov 3.237: Half the Field at the Mouth – Amit & Deepali Sinha Foundation

Problem Irodov 3.237

A straight solenoid, long enough to be treated as infinite in one direction, has cross-sectional radius $R$ , $n$ turns per unit length, and carries a steady current $I$ . Let $x$ be the distance of a point on the axis from the solenoid’s end, measured along the axis.

Find the magnetic induction $B$ on the axis as a function of $x$ , and sketch $B$ against $x/R$ .
Find the distance $x_0$ to the point on the axis where $B$ differs by $\eta = 1\%$ from its value deep inside the solenoid.

The end of the very long solenoid cut along its axis (dots and crosses: the current I in the turns, its sense chosen; the winding runs on to the left), its radius R, and an axis point P at the distance x from the end. — The end of the very long solenoid cut along its axis (dots and crosses: the current $I$ in the turns, its sense chosen; the winding runs on to the left), its radius $R$ , and an axis point $P$ at the distance $x$ from the end.

Given

$R$
$n$ turns per unit length
$I$
$\eta = 1\%$

Find

(a)

B(x)

on the axis and its plot against

x/R

; (b)

x_0

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 cancelling, and Biot and Savart give, at axial distance $u$ from the loop,

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

All the turns send their fields at $P$ the same way along the axis.

Step 2The ring of turns in a slice

ds

The turns in a slice of width ds at depth s inside the end: one ring carrying nI ds, at the distance x + s from P (sense chosen). — The turns in a slice of width $ds$ at depth $s$ inside the end: one ring carrying $nI\,ds$ , at the distance $x + s$ from $P$ (sense chosen).

Take $x > 0$ outside the solenoid and $x < 0$ inside it, and let $s \ge 0$ be the depth of a slice behind the end. The slice holds $n\,ds$ turns, a loop carrying $nI\,ds$ at the distance $u = x + s$ from $P$ , so by (1)

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

Step 3Adding all the slices behind the end

The winding fills $0 \le s < \infty$ . Integrating (2) with $u = x + s$ running from $x$ to infinity:

\begin{aligned} B &= \frac{\mu_0 n I R^2}{2}\int_{x}^{\infty} \frac{du}{\left(R^2 + u^2\right)^{3/2}} = \frac{\mu_0 n I R^2}{2}\left[\frac{u}{R^2\sqrt{R^2 + u^2}}\right]_{x}^{\infty}\\ &= \frac{\mu_0 n I R^2}{2}\left(\frac{1}{R^2} - \frac{x}{R^2\sqrt{x^2 + R^2}}\right) \end{aligned}

B = \frac{\mu_0 n I}{2}\left(1 - \frac{x}{\sqrt{x^2 + R^2}}\right)

(3)

Here $x > 0$ outside the solenoid and $x < 0$ inside it; $\vect B$ lies along the axis in the sense the right-hand rule gives from the current in the turns.

Step 4The plot of

B

against

x/R

Deep inside, $x \to -\infty$ , the bracket in (3) tends to $2$ , so the field there is

B_0 = \mu_0 n I \tag{4}

In units of $B_0$ and of $R$ , (3) reads

\frac{B}{B_0} = \frac{1}{2}\left(1 - \frac{x/R}{\sqrt{(x/R)^2 + 1}}\right)

Above, the end of the solenoid with its computed B lines, for the current drawn (sense chosen); below, on the same abscissa, B/B_0 against x/R: it falls from 1 deep inside to 1/2 at the end and toward 0 outside. — Above, the end of the solenoid with its computed $\vect B$ lines, for the current drawn (sense chosen); below, on the same abscissa, $B/B_0$ against $x/R$ : it falls from $1$ deep inside to $1/2$ at the end and toward $0$ outside.

The curve passes through $1/2$ at the end, $x = 0$ , and is antisymmetric about that point: $B(x)/B_0 + B(-x)/B_0 = 1$ . One radius inside it is already $0.85$ ; one radius outside, $0.15$ .

Step 5Where the field is within

\eta

B_0

The point lies inside, at $x = -x_0$ . With (3) and (4), the condition $B = (1 - \eta)B_0$ is

\frac{1}{2}\left(1 + \frac{x_0}{\sqrt{x_0^2 + R^2}}\right) = 1 - \eta

Isolating the fraction and squaring:

\begin{aligned} \frac{x_0}{\sqrt{x_0^2 + R^2}} &= 1 - 2\eta\\ x_0^2 &= (1 - 2\eta)^2\left(x_0^2 + R^2\right)\\ x_0^2\left[1 - (1 - 2\eta)^2\right] &= (1 - 2\eta)^2 R^2\\ 4\eta(1 - \eta)\,x_0^2 &= (1 - 2\eta)^2 R^2 \end{aligned}

Numerically, for $\eta = 0.01$ ,

x_0 = \frac{(0.98)\,R}{2\sqrt{(0.01)(0.99)}}

x_0 = \frac{(1 - 2\eta)\,R}{2\sqrt{\eta(1 - \eta)}} \approx 4.9\,R

(5)

For small $\eta$ this is close to $R/(2\sqrt\eta) = 5R$ .

Check

At the end, $x = 0$ , (3) gives $\mu_0 nI/2$ , half the field deep inside, as the two-half-solenoids argument requires (two equal semi-infinite pieces placed end to end make the infinite one). Far outside, $x \gg R$ , the bracket is $1 - (1 - R^2/2x^2) = R^2/2x^2$ , so $B \approx \mu_0 n I R^2/(4x^2)$ : an inverse-square fall, as from a magnetic pole sitting at the end of a long solenoid. In (5), $\eta \to 0$ sends $x_0$ to infinity, and $\eta = 1/2$ gives $x_0 = 0$ , the end.

Remark

The approach to $B_0$ inside is slow: the relative deficit $(B_0 - B)/B_0$ falls only as $R^2/(4x^2)$ , so the field is within $1\%$ only five radii in and within $0.1\%$ about sixteen radii in. A solenoid that must give a uniform field over its middle therefore has to be many radii longer than the region used, which is the result 3.236 expresses for a coil of finite length.