Irodov 3.316: Inductance Weighed and Measured – Amit & Deepali Sinha Foundation

Problem Irodov 3.316

A solenoid of length $l$ is wound with copper wire (resistivity $\rho$ , density $\rho_0$ ) of total mass $m$ , and the resistance of the winding is $R$ . The solenoid’s diameter is much smaller than its length. Find its inductance.

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

Given

length $l$
wire mass $m$
winding resistance $R$
copper: resistivity $\rho$ , density $\rho_0$ (table values)
diameter $\ll l$
every turn of the same diameter (assumed, thin wire)

Find

the inductance

L

Solution

Step 1Inductance in terms of the wire length

Call the solenoid’s diameter $D$ , its number of turns $N$ , and the length of the wire $l_{\mathrm{w}}$ . Since $D \ll l$ , the field inside is that of a long solenoid, $B = \mu_0 N I/l$ , uniform over the cross-section $\pi D^2/4$ , and the flux linked with all $N$ turns per unit current is

L = \frac{NBS}{I} = \mu_0\frac{N^2}{l}\cdot\frac{\pi D^2}{4} \tag{1}

Each turn takes one circumference of wire, so the wire length is $l_{\mathrm{w}} = N\pi D$ . Putting $N^2D^2 = l_{\mathrm{w}}^2/\pi^2$ into (1):

L = \frac{\mu_0\pi}{4l}\cdot\frac{l_{\mathrm{w}}^2}{\pi^2} = \frac{\mu_0 l_{\mathrm{w}}^2}{4\pi l} \tag{2}

Step 2The wire length from its mass and resistance

Let $A$ be the wire’s cross-sectional area. The resistance and the mass of the wire are, with the table values $\rho$ and $\rho_0$ for copper,

R = \frac{\rho\, l_{\mathrm{w}}}{A}, \qquad m = \rho_0\, l_{\mathrm{w}} A

Multiplying the two eliminates the unknown area:

mR = \rho\rho_0\, l_{\mathrm{w}}^2 \quad\Longrightarrow\quad l_{\mathrm{w}}^2 = \frac{mR}{\rho\rho_0} \tag{3}

Step 3The inductance

Substituting (3) into (2):

L = \frac{\mu_0}{4\pi}\,\frac{mR}{l\,\rho\,\rho_0}

(4)

Check

Units: $mR/(\rho\rho_0)$ is kg $\,\Omega$ over $\Omega\,$ m $\,\cdot\,$ kg/m $^3$ , which is m $^2$ , the square of the wire length as (3) says; $\mu_0/(4\pi)$ in H/m times m $^2$ over the length $l$ gives henries. A sample: $m = 10$ g of copper ( $\rho = 1.7\times 10^{-8}\ \Omega\,$ m, $\rho_0 = 8.9\times 10^3$ kg/m $^3$ ) with $R = 1.0\ \Omega$ is $l_{\mathrm{w}} \approx 8.1$ m of wire, and on $l = 0.10$ m it gives $L \approx 10^{-7}\times 66/0.10 \approx 66\ \mu$ H.

Remark

For a fixed mass of copper, more resistance means a thinner and therefore longer wire, and the inductance grows in proportion to $R$ . That is the trade every coil designer meets: a large inductance from a given amount of copper costs resistance, so the ratio $L/R = \mu_0 m/(4\pi l\rho\rho_0)$ , which sets the coil’s time constant, is fixed by the copper’s mass and the coil’s length alone.