Irodov 3.315: The Diameter Drops Out – Amit & Deepali Sinha Foundation

Problem Irodov 3.315

A solenoid of length $l_0 = 100$ cm is to be wound from thin wire so that its inductance is $L = 1.0$ mH. Its cross-sectional diameter is much smaller than its length. What length of wire is needed?

The solenoid side on: length l_0, cross-sectional diameter D much smaller than l_0. — The solenoid side on: length $l_0$ , cross-sectional diameter $D$ much smaller than $l_0$ .

Given

$l_0 = 100$ cm
$L = 1.0$ mH
diameter $\ll l_0$
every turn of the same diameter (assumed, thin wire)

Find

the wire length

l

Solution

Step 1Inductance of a long solenoid

Call the cross-sectional diameter $D$ and the number of turns $N$ . Since $D \ll l_0$ , the field inside is that of a long solenoid, uniform and axial, $B = \mu_0 N I/l_0$ (Ampere’s law round a rectangle with one side inside along the axis, the outside field negligible). The flux linked with all $N$ turns, per unit current, is the inductance:

L = \frac{N B S}{I} = \mu_0\frac{N^2}{l_0}\cdot\frac{\pi D^2}{4} \tag{1}

Step 2Trading turns and diameter for wire length

Each turn uses one circumference of wire, so the wire length is

l = N\pi D \tag{2}

The product $ND$ in (1) is exactly what (2) fixes. Substituting $N^2D^2 = l^2/\pi^2$ :

L = \frac{\mu_0\pi}{4 l_0}\cdot\frac{l^2}{\pi^2} = \frac{\mu_0 l^2}{4\pi l_0}

Step 3The length of wire

Solving for $l$ and substituting $l_0 = 1.00$ m, $L = 1.0\times 10^{-3}$ H and $\mu_0 = 4\pi\times 10^{-7}$ H/m:

l = \left(\frac{4\pi\times 1.00\times 1.0\times 10^{-3}}{4\pi\times 10^{-7}}\right)^{1/2}\text{m} = \left(1.0\times 10^{4}\right)^{1/2}\text{m}

l = \sqrt{\frac{4\pi l_0 L}{\mu_0}} \approx 1.0\times 10^{2}\ \text{m}

(3)

Check

Units: $l_0L/\mu_0$ is m $\cdot$ H divided by H/m, m $^2$ , so $l$ is a length. A sample coil: with $D = 2.0$ cm the 100 m of wire makes $N = l/(\pi D) \approx 1.6\times 10^3$ turns, and (1) gives $L = 4\pi\times 10^{-7}\times(1.6\times 10^3)^2\times 3.1\times 10^{-4}/1.0 \approx 1.0$ mH, as required.

Remark

The inductance depends on the wire only through its length, $L = \mu_0 l^2/(4\pi l_0)$ : quadratic in the wire, because doubling the wire doubles both the flux each turn sees and the number of turns linking it. The result holds only while the coil stays long; a short, fat coil wound from the same wire has less inductance, since its field spreads out and weakens past the ends.