Irodov 3.321: Field Trapped Between Two Tapes – Amit & Deepali Sinha Foundation

Problem Irodov 3.321

A double tape line is made of two long, thin, flat metal tapes of equal width $b$ , lying one directly above the other at the separation $h$ ; a current $I$ flows along one tape and returns along the other. The separation is much smaller than the width, $b/h = 50$ . Find the inductance per unit length of the line.

The double tape line, cut off by break lines (the tapes run on): two tapes of width b at the separation h (exaggerated), the current I going along one and returning along the other. — The double tape line, cut off by break lines (the tapes run on): two tapes of width $b$ at the separation $h$ (exaggerated), the current $I$ going along one and returning along the other.

Given

$b/h = 50$
tapes of width $b$ , separation $h \ll b$
go and return currents $I$

Find

L_1

, the inductance per unit length

Solution

Step 1Each tape as a current sheet

Since $h \ll b$ , near the middle of the line each tape looks like an infinite plane carrying the surface current $i = I/b$ per unit width; the edges, where this fails, are a fraction of order $h/b$ of the field region and are neglected. By symmetry the field of one sheet is parallel to it, square to the current and reversed across it, so Ampere’s law on a rectangle of length $\ell$ straddling the sheet gives

\begin{align*} 2B'\ell &= \mu_0 i\,\ell \\ B' &= \frac{\mu_0 i}{2} = \frac{\mu_0 I}{2b} \tag{1}\end{align*}

Step 2The two sheets together

The tapes end on, with opposite currents: the two sheets' fields add in the gap and cancel outside it. — The tapes end on, with opposite currents: the two sheets’ fields add in the gap and cancel outside it.

The currents are opposite, so in the gap the two fields (1) point the same way and add, while outside the pair they are opposite and cancel:

B = 2B' = \frac{\mu_0 I}{b}\ \text{in the gap},\qquad B = 0\ \text{outside}

(2)

Step 3The flux per unit length

The circuit runs along one tape and back along the other, so the flux that links it crosses the gap: a rectangle of height $h$ and unit length, standing between the tapes. With (2) uniform over it:

\Phi_1 = B\,h\cdot 1 = \frac{\mu_0 I h}{b}

Step 4The inductance per unit length

Dividing the flux by the current, $L_1 = \Phi_1/I$ . Numerically, with $h/b = 1/50$ :

L_1 = \frac{4\pi\times 10^{-7}\text{ H/m}}{50}

L_1 = \frac{\mu_0 h}{b} \approx 25\text{ nH/m}

(3)

Check

The field energy per unit length gives the same. The energy density $B^2/2\mu_0$ with (2), times the gap’s volume $bh$ per unit length:

W_1 = \frac{1}{2\mu_0}\left(\frac{\mu_0 I}{b}\right)^2 bh = \frac{\mu_0 h}{2b}\,I^2

which is $L_1 I^2/2$ with (3). Units: $\mu_0$ is H/m and $h/b$ is a pure number.

Remark

The inductance is proportional to the area of the gap per unit length, $h$ , divided by the width that spreads the current, $b$ : bringing the go and return conductors close and making them wide squeezes the magnetic field into a thin layer that holds little flux. That is why printed circuit boards route fast signals as a trace over a ground plane, and why power is fed through laminated busbars: tens of nanohenries per metre instead of the microhenry or so per metre of two separated wires.