Irodov 3.344: Perfect Coupling Gives the Geometric Mean

Problem Irodov 3.344

Two solenoids have the same length and nearly the same cross-sectional area, and one is pushed fully inside the other. Their inductances are $L_1$ and $L_2$ . Find their mutual inductance.

The two solenoids of equal length in longitudinal section, one inside the other on a common axis: L_1 outside, L_2 inside. The radii are drawn apart for clarity. The winding senses are not given; both are drawn the same way. — The two solenoids of equal length in longitudinal section, one inside the other on a common axis: $L_1$ outside, $L_2$ inside. The radii are drawn apart for clarity. The winding senses are not given; both are drawn the same way.

Given

$L_1$
$L_2$
equal lengths
almost equal cross-sections
solenoids long beside their radius (assumed)

Find

the mutual inductance

L_{12}

Solution

Step 1The two self-inductances

Let the common length be $l$ , the common cross-section $S$ , and the numbers of turns $N_1$ and $N_2$ . For a solenoid long beside its radius (assumed), symmetry makes the field inside axial and uniform and the field outside negligible, so the circulation of $\vect B$ round a rectangle with one side of length $l$ inside, threaded by all $N$ turns carrying $I$ , gives

Bl = \mu_0 NI

The flux $BS$ threads all $N$ turns, so the linkage per unit current is

\begin{aligned} L_1 &= \frac{\mu_0 N_1^2 S}{l}\\ L_2 &= \frac{\mu_0 N_2^2 S}{l} \end{aligned} \tag{1}

Step 2The mutual inductance

A current $I_1$ in the first solenoid makes the uniform field $\mu_0N_1I_1/l$ over the common cross-section, so its flux $\mu_0N_1I_1S/l$ threads each of the $N_2$ turns of the second. The mutual inductance is that linkage per unit $I_1$ :

L_{12} = \frac{N_2}{I_1}\,\frac{\mu_0N_1I_1S}{l} = \frac{\mu_0N_1N_2S}{l} \tag{2}

Step 3In terms of the self-inductances

Multiplying the two inductances (1) gives the square of (2):

L_1L_2 = \frac{\mu_0^2N_1^2N_2^2S^2}{l^2} = L_{12}^2

Taking the root, with both windings in the same sense as drawn:

L_{12} = \sqrt{L_1L_2}

(3)

If one winding is reversed, each flux enters the other against its positive direction, and the mutual inductance changes sign to $-\sqrt{L_1L_2}$ with the same magnitude.

Check

Two identical solenoids, $L_1 = L_2 = L$ , give $L_{12} = L$ . Then, joined in series in the same sense, they act as one solenoid of $2N$ turns, whose inductance by (1) is $4L$ , and indeed $L_1 + L_2 + 2L_{12} = 4L$ . Units: $\sqrt{\text{H}\cdot\text{H}} = \text{H}$ .

Remark

In general $|L_{12}| \le \sqrt{L_1L_2}$ , with equality only when every line of each coil’s flux threads every turn of the other. If the inner solenoid’s cross-section $S_2$ is a little smaller than $S_1$ , the same steps give $L_{12} = \sqrt{L_1L_2}\,\sqrt{S_2/S_1}$ , slightly less. The ratio $L_{12}/\sqrt{L_1L_2}$ is the coupling coefficient, close to $1$ in a well-built transformer.