Irodov 3.224: The Field of a Missing Strip – Amit & Deepali Sinha Foundation

Problem Irodov 3.224

A long thin-walled tube of radius $R$ has a narrow slit of width $h$ cut along its whole length, and a current $I$ flows along the tube. Assuming $h \ll R$ , find the magnetic induction inside the tube.

Cross-section of the tube of radius R with its slit of width h (drawn much wider than to scale); the current I runs along the tube, drawn out of the page (sense chosen). — Cross-section of the tube of radius $R$ with its slit of width $h$ (drawn much wider than to scale); the current $I$ runs along the tube, drawn out of the page (sense chosen).

Given

$I$
$R$
$h \ll R$
field point far from the slit compared with $h$ (assumed)

Find

B

inside the tube

Solution

Step 1Fill the slit, then take the strip away

The current spreads evenly round the wall, a current per unit width of the wall

i = \frac{I}{2\pi R - h} \approx \frac{I}{2\pi R}

the approximation using $h \ll R$ . The slit tube carries exactly what a complete tube with this $i$ all round would carry, plus a strip of width $h$ in the slit carrying the opposite current

I' = ih = \frac{Ih}{2\pi R} \tag{1}

By superposition the field inside is the field of the complete tube plus the field of that strip.

Step 2The complete tube gives nothing inside

For a complete tube $\vect B$ is, by symmetry, along circles about the axis and the same all round each circle; Ampere’s law on a circle of radius $\rho < R$ , which encloses no current, gives

B\cdot 2\pi\rho = \mu_0 I_{\text{enclosed}} = 0

so the complete tube contributes $B = 0$ everywhere inside.

Step 3The strip acts as a thin wire

The strip in the slit as a wire carrying I' into the page (opposite to the wall's current), and B at an inside point P a distance r from the slit: perpendicular to r, in the sense the right-hand rule gives for I'. — The strip in the slit as a wire carrying $I'$ into the page (opposite to the wall’s current), and $\vect B$ at an inside point $P$ a distance $r$ from the slit: perpendicular to $r$ , in the sense the right-hand rule gives for $I'$ .

Seen from a point $P$ inside whose distance $r$ from the slit is large compared with $h$ , the strip is a long straight wire carrying (1), whose field is

B = \frac{\mu_0 I'}{2\pi r} = \frac{\mu_0}{2\pi r}\cdot\frac{Ih}{2\pi R}

B = \frac{\mu_0 h I}{4\pi^2 R r}

(2)

Here $r$ is the distance from the slit. $\vect B$ is perpendicular to the plane through the slit and $P$ , circling the slit in the sense the right-hand rule gives for a current opposite to $I$ .

Check

As $h \to 0$ the tube closes and (2) goes to zero, the field inside a complete tube. The units are those of $\mu_0 I/r$ times the pure number $h/R$ , tesla. At the far side of the tube, $r = 2R$ , the field is $\mu_0 I h/(8\pi^2 R^2)$ , smaller than the field $\mu_0 I/(2\pi R)$ just outside the wall by the factor $h/(4\pi R)$ .

Remark

This is the method of the missing piece, the magnetic twin of the uniformly charged sphere with a small hole: the complete body gives a simple field (here zero inside), and the defect acts as a small source of the opposite sign. Inside the slit tube the whole field is that of the absent strip.