Irodov 3.240: A Flux That Ignores the Radius – Amit & Deepali Sinha Foundation

Problem Irodov 3.240

A long straight wire of circular cross-section carries a steady current $I = 10$ A. Consider the flat strip that runs along the wire from its axis to its surface (half of an axial section). Find the magnetic flux through this strip per metre of the wire’s length.

A stretch of the long round conductor from the side, carrying the current I (sense chosen): the shaded strip S, between the axis and the surface over a length of 1 m, is the half section the flux is asked through. — A stretch of the long round conductor from the side, carrying the current $I$ (sense chosen): the shaded strip $S$ , between the axis and the surface over a length of $1$ m, is the half section the flux is asked through.

Given

$I = 10$ A
current uniform over the section (assumed)
$\mu = 1$ (assumed)

Find

\Phi

through

S

, per metre of length

Solution

Step 1The field inside the wire

By symmetry the field lines inside are circles round the axis with $B$ the same all round each. A circle of radius $r < R$ , with $R$ the wire’s radius, encloses the fraction $r^2/R^2$ of the current, assumed uniform over the section, so Ampere’s law gives

B\cdot 2\pi r = \mu_0 I\,\frac{r^2}{R^2}

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

(1)

The field lines cross the strip $S$ square to it, since $S$ lies along a radius.

Step 2The flux through the strip

The section end-on, the current out of the page (sense chosen): the strip S is the heavy radius, the element dr on it at the distance r from the axis, and the B line through the element, a circle. — The section end-on, the current out of the page (sense chosen): the strip $S$ is the heavy radius, the element $dr$ on it at the distance $r$ from the axis, and the $\vect B$ line through the element, a circle.

Cut $S$ into strips of width $dr$ and length $l$ parallel to the axis. The field (1) crosses each one normally and is uniform over it, so

d\Phi = B\,l\,dr = \frac{\mu_0 I\,l}{2\pi R^2}\,r\,dr

Integrating from the axis to the surface:

\Phi = \frac{\mu_0 I\,l}{2\pi R^2}\int_0^R r\,dr = \frac{\mu_0 I\,l}{2\pi R^2}\cdot\frac{R^2}{2}

Per unit length, numerically,

\frac{\Phi}{l} = \frac{(4\pi\times 10^{-7}\text{ H/m})(10\text{ A})}{4\pi}

\frac{\Phi}{l} = \frac{\mu_0 I}{4\pi} = 1.0\ \mu\text{Wb/m}

(2)

Check

Units: $\mu_0 I$ is in $\text{H}\cdot\text{A/m} = \text{Wb/m}$ , a flux per length, as (2) must be. The average of the field over the strip is half its surface value $\mu_0 I/(2\pi R)$ , and the strip’s width is $R$ : $(\mu_0 I/4\pi R)\cdot R = \mu_0 I/4\pi$ , the same.

Remark

The radius cancels: a thinner wire has a stronger field, its surface value $\mu_0 I/(2\pi R)$ growing as $1/R$ , over a strip narrower in proportion. The same internal field gives every straight round wire the same internal self-inductance per unit length, $\mu_0/8\pi$ , whatever its thickness: it is this flux with each line weighted by the fraction $r^2/R^2$ of the current it encircles.