Irodov 3.238: A Strip Is Also a Wire – Amit & Deepali Sinha Foundation

Problem Irodov 3.238

A thin conducting strip of width $h = 2.0$ cm is wound edge to edge, in a single layer, round a cylinder of radius $R = 2.5$ cm, forming a very long straight solenoid. A steady current $I = 5.0$ A flows along the strip. Find the magnetic induction inside and outside the solenoid as a function of the distance $r$ from its axis.

The coil from the side: the strip of width h wound edge to edge on a cylinder of radius R (the seams between turns drawn on the near side), the current I running along the strip (its sense and the winding's hand chosen). — The coil from the side: the strip of width $h$ wound edge to edge on a cylinder of radius $R$ (the seams between turns drawn on the near side), the current $I$ running along the strip (its sense and the winding’s hand chosen).

Given

$h = 2.0$ cm
$R = 2.5$ cm
$I = 5.0$ A
turns edge to edge
very long coil

Find

B(r)

for

r < R

and for

r > R

Solution

Step 1The pitch of a winding made of a strip

The winding cut along a generator and laid flat (axis horizontal, circumference 2π R vertical): the seams make the angle α with the circumference, one turn of the strip is shaded, p is the pitch and h the strip's width. Angle and pitch exaggerated. — The winding cut along a generator and laid flat (axis horizontal, circumference $2\pi R$ vertical): the seams make the angle $\alpha$ with the circumference, one turn of the strip is shaded, $p$ is the pitch and $h$ the strip’s width. Angle and pitch exaggerated.

Cut the cylinder along a generator and lay it flat. One turn of the strip then runs once across the height $2\pi R$ while advancing the pitch $p$ along the axis, so its edges make an angle $\alpha$ with the circumference, where

\tan\alpha = \frac{p}{2\pi R} \tag{1}

Neighbouring seams are a horizontal distance $p$ apart, and the strip’s width $h$ is their distance measured square to them:

h = p\cos\alpha \tag{2}

Eliminating $p$ between (1) and (2):

2\pi R\tan\alpha\,\cos\alpha = h

\sin\alpha = \frac{h}{2\pi R}

(3)

Step 2The current splits into a ring current and an axial current

A line drawn on the cylinder parallel to the axis (a generator), of length $L$ , crosses $L/p$ turns of the strip, each carrying $I$ ; so the current circulating round the axis, per unit length of the solenoid, is the current per pitch, $I/p$ . Taking $p$ from (2) and $\sin\alpha$ from (3):

i_\varphi = \frac{I}{p} = \frac{I}{h}\cos\alpha = \frac{I}{h}\sqrt{1 - \left(\frac{h}{2\pi R}\right)^2} \tag{4}

A circle round the axis on the cylinder, on the other hand, crosses the strip exactly once, so the whole current $I$ also flows along the axis, spread over the surface.

Step 3The field inside

The two parts act independently. The circulating part (4) is an ideal long solenoid: by symmetry its field is axial, uniform inside and zero outside, so Ampere’s law round a rectangle of length $\ell$ straddling the winding gives

B\,\ell = \mu_0 i_\varphi\,\ell

The axial part gives nothing inside: by symmetry its field circles the axis, and an Amperian circle of radius $r < R$ encloses none of it,

B_\varphi\cdot 2\pi r = 0

So for $r < R$ the field is the axial one, $B = \mu_0 i_\varphi$ with (4); numerically,

B = \frac{(4\pi\times 10^{-7}\text{ H/m})(5.0\text{ A})}{0.020\text{ m}}\sqrt{1 - \left(\frac{0.020\text{ m}}{2\pi\,(0.025\text{ m})}\right)^2}

B = \frac{\mu_0 I}{h}\sqrt{1 - \left(\frac{h}{2\pi R}\right)^2} = 0.31\text{ mT}

(5)

for every $r < R$ , directed along the axis.

Step 4The field outside

Outside, the circulating part gives no field, and the axial current $I$ acts as a straight wire on the axis: by symmetry its field circles the axis with the same magnitude all round, and an Amperian circle of radius $r > R$ encloses all of it:

B\cdot 2\pi r = \mu_0 I

Numerically,

\frac{\mu_0 I}{2\pi} = (2\times 10^{-7}\text{ H/m})(5.0\text{ A})

B = \frac{\mu_0 I}{2\pi r} = \frac{1.0\ \mu\text{T}\cdot\text{m}}{r}

(6)

for $r > R$ , its lines being circles round the axis.

Check

For a winding of thin wire, $h \to 0$ with $1/h$ turns per unit length, the root in (5) tends to $1$ and $B \to \mu_0 I/h$ , the familiar $\mu_0 n I$ . Here $h/2\pi R = 0.127$ , so the root is $0.992$ : the tilt lowers the inside field by less than $1\%$ . Outside, (6) at $r = R$ gives $40\ \mu$ T, about an eighth of the field inside, and it falls as $1/r$ like a wire’s.

Remark

Just inside the sheet $\vect B$ is axial with magnitude $\mu_0 I\cos\alpha/h$ ; just outside it is circular with magnitude $\mu_0 I/(2\pi R) = \mu_0 I\sin\alpha/h$ by (3). Their ratio is $\tan\alpha$ , so the jump in $\vect B$ across the sheet is a vector square to the strip’s direction, as a surface current along the strip demands. Any helical winding, wire or strip, carries this axial current, and its weak outside field is that of one straight wire.