Irodov 3.258: A Strip Pulled as Many Filaments – Amit & Deepali Sinha Foundation

Problem Irodov 3.258

A long thin straight wire carries a direct current $I_1$ . Beside it, in the same plane and parallel to it, lies a long thin flat strip of width $b$ carrying a direct current $I_2$ in the same direction. The gap between the wire and the near edge of the strip is $a$ . Find the magnetic force between the two conductors per unit length.

The thin wire with current I_1 and, in the same plane, the long flat strip of width b with current I_2 in the same direction; a is the gap from the wire to the strip's near edge. — The thin wire with current $I_1$ and, in the same plane, the long flat strip of width $b$ with current $I_2$ in the same direction; $a$ is the gap from the wire to the strip’s near edge.

Given

$I_1$ , $I_2$ , parallel
gap $a$ to the near edge
strip width $b$
$I_2$ spread evenly across the strip (assumed)

Find

the force between the conductors per unit length

Solution

Step 1The strip as a sheet of filaments

The strip is thin and uniform, so its current is spread evenly across its width (assumed). Cut it into filaments parallel to the wire; the one of width $dx$ at the distance $x$ from the wire carries

dI_2 = \frac{I_2}{b}\,dx \tag{1}

A filament of the strip, of width dx at the distance x from the wire, carrying the current dI_2 parallel to I_1. — A filament of the strip, of width $dx$ at the distance $x$ from the wire, carrying the current $dI_2$ parallel to $I_1$ .

Step 2The force on one filament

At the distance $x$ the wire’s field follows from Ampere’s law on a circle round the wire, on which $B$ is tangential and the same all round:

B\cdot 2\pi x = \mu_0 I_1

In the plane of the strip this field is normal to the strip, so the filament carrying $dI_2$ feels per unit length the force

dF = dI_2\,B = \frac{\mu_0 I_1}{2\pi x}\cdot\frac{I_2}{b}\,dx \tag{2}

The currents are parallel, so every filament is attracted toward the wire: all the elementary forces point the same way, along the strip’s plane, and add as numbers.

Step 3Summing across the strip

Integrating (2) from the near edge, $x = a$ , to the far edge, $x = a + b$ :

F_1 = \frac{\mu_0 I_1 I_2}{2\pi b}\int_a^{a+b}\frac{dx}{x} = \frac{\mu_0 I_1 I_2}{2\pi b}\Big[\ln x\Big]_a^{a+b}

F_1 = \frac{\mu_0 I_1 I_2}{2\pi b}\ln\left(1 + \frac{b}{a}\right)

(3)

The force is an attraction, in the plane of the conductors and across them: the wire and the strip are pulled toward each other.

Check

For a narrow strip, $b \ll a$ , the logarithm is $b/a$ to first order and (3) becomes $\mu_0 I_1 I_2/(2\pi a)$ , the force between two thin parallel wires a distance $a$ apart, as it should. For a wide strip, at a fixed total current $I_2$ , the force falls as $(\ln b)/b$ : the current per unit width falls as $1/b$ while the integral of the $1/x$ field grows only as $\ln b$ , so the far filaments add little.

Remark

As the gap $a$ closes, the force grows without bound, but only logarithmically: the field of the thin wire diverges as $1/x$ at the strip’s edge and its integral as $\ln a$ . For a real wire of radius $r_0$ the gap cannot fall below $r_0$ , so the divergence is cut off. The force is weighted toward the near edge: half of the logarithm $\ln(1 + b/a)$ comes from the filaments with $x < [a(a+b)]^{1/2}$ , the geometric mean of the two edge distances, which always lies nearer the wire than the strip’s middle.