Irodov 3.329: A Logarithm Couples Wire and Frame – Amit & Deepali Sinha Foundation

Problem Irodov 3.329

A long straight wire and a rectangular wire frame with sides $a$ and $b$ lie in one plane. The side $b$ of the frame is parallel to the wire and is the side nearest to it, at a distance $l$ from it. Find the mutual inductance of the wire and the frame.

The long wire and the frame in one plane: the side b parallel to the wire at the distance l, the side a across. The arrow on the wire marks a chosen sense. — The long wire and the frame in one plane: the side $b$ parallel to the wire at the distance $l$ , the side $a$ across. The arrow on the wire marks a chosen sense.

Given

frame sides $a$ and $b$
side $b$ parallel to the wire, at distance $l$
wire and frame coplanar
vacuum around them, $\mu = 1$ (assumed)

Find

the mutual inductance

L_{12}

Solution

Step 1The field of the wire

Let a current $I$ flow in the wire in the sense drawn. By symmetry the field is the same all round a circle of radius $r$ centred on the wire and runs along it, so the circulation theorem gives

B\cdot 2\pi r = \mu_0 I

B(r) = \frac{\mu_0 I}{2\pi r}

(1)

In the plane of the frame the field is normal to the plane, into the page on the frame’s side for the sense drawn, and it weakens across the frame as $1/r$ .

Step 2The flux through the frame

The wire's field over the frame, its glyphs thinning out as 1/r; the strip of width dr at the distance r runs the full height b of the frame. — The wire’s field over the frame, its glyphs thinning out as $1/r$ ; the strip of width $dr$ at the distance $r$ runs the full height $b$ of the frame.

The field (1) is constant along any line parallel to the wire, so the frame is cut into strips of height $b$ and width $dr$ , each crossed square on by the field:

d\Phi = B(r)\,b\,dr = \frac{\mu_0 I b}{2\pi}\,\frac{dr}{r}

Summing the strips from the near side, $r = l$ , to the far side, $r = l + a$ :

\Phi = \frac{\mu_0 I b}{2\pi}\int_l^{l + a}\frac{dr}{r} = \frac{\mu_0 I b}{2\pi}\,\ln\frac{l + a}{l}

Step 3The mutual inductance

The mutual inductance is the flux through the frame per unit current in the wire, $L_{12} = \Phi/I$ :

L_{12} = \frac{\mu_0 b}{2\pi}\,\ln\left(1 + \frac{a}{l}\right)

(2)

Check

For a narrow frame, $a \ll l$ , $\ln(1 + a/l) \approx a/l$ and (2) becomes $\mu_0 ab/(2\pi l)$ : the field $\mu_0 I/(2\pi l)$ at the near side times the whole area $ab$ , per unit current, as it must be when the field hardly varies across the frame. Units: $\mu_0$ in H/m times the length $b$ gives henries.

Remark

The wire’s own circuit closes far away, so “the flux through the wire’s circuit” is never computed here; the coefficient is found the easy way round, from the flux of the wire through the frame, and reciprocity, $L_{12} = L_{21}$ , says it serves both ways. Slide the frame across so that the wire runs along its middle and $L_{12}$ falls to zero: the field points into the page on one side and out on the other, and the two halves cancel.