Irodov 3.318: Only the Copper per Metre Counts – Amit & Deepali Sinha Foundation

Problem Irodov 3.318

A straight solenoid of length $l = 1.0$ m carries a single-layer winding of copper wire (resistivity $1.7\times 10^{-8}\ \Omega\cdot$ m, density $8.9\times 10^{3}$ kg/m $^3$ ) of total mass $m = 1.0$ kg. The diameter of the solenoid is much smaller than its length. Find its time constant $\tau$ , the ratio $L/R$ of its inductance $L$ to its resistance $R$ .

Given

$l = 1.0$ m
$m = 1.0$ kg
diameter $\ll l$
copper, $\rho = 1.7\times 10^{-8}\ \Omega\cdot$ m (table value)
copper, $\rho_0 = 8.9\times 10^{3}$ kg/m $^3$ (table value)
turns touching, insulation neglected (assumed)

Find

\tau = L/R

Solution

Step 1The inductance of the long solenoid

Let the solenoid have $N$ turns and diameter $D$ , so its cross-section is $S = \pi D^2/4$ . It is long, so its field is the uniform $B = \mu_0 N I/l$ and every turn links the flux $BS$ . The flux linkage $N B S$ divided by $I$ is the inductance:

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

Step 2The resistance of the winding

Let the wire have diameter $d$ . The wire is $N$ turns of length $\pi D$ each, with cross-section $\pi d^2/4$ , so with the table resistivity $\rho$ of copper:

R = \rho\,\frac{N\pi D}{\pi d^2/4} = \frac{4\rho N D}{d^2} \tag{2}

Step 3The ratio, for a close winding

Dividing (1) by (2):

\frac{L}{R} = \frac{\mu_0 N^2\pi D^2}{4l}\cdot\frac{d^2}{4\rho N D} = \frac{\mu_0\pi N D d^2}{16\,\rho\, l}

In a single layer the turns lie side by side (assumed touching, insulation neglected), so the wire’s diameter fills the length $N$ times, $N d = l$ . Putting $N = l/d$ :

\tau = \frac{\mu_0\pi D d}{16\,\rho} \tag{3}

Step 4The product

Dd

from the mass

The copper’s volume is the wire length $N\pi D$ times its cross-section, and its density is $\rho_0$ (table value). With $N = l/d$ again:

m = \rho_0\, N\pi D\,\frac{\pi d^2}{4} = \rho_0\,\frac{l}{d}\,\frac{\pi^2 D d^2}{4} = \frac{\pi^2\rho_0\, l\, D d}{4}

Solving for the product:

D d = \frac{4m}{\pi^2\rho_0\, l} \tag{4}

Step 5The time constant

Substituting (4) into (3); the factors of $\pi$ and 4 cancel in the box:

\tau = \frac{\mu_0\pi}{16\,\rho}\cdot\frac{4m}{\pi^2\rho_0\, l}

Numerically,

\tau = \bigl(10^{-7}\text{ H/m}\bigr)\,\frac{1.0\text{ kg}}{(1.7\times 10^{-8}\ \Omega\cdot\text{m})(8.9\times 10^{3}\text{ kg/m}^3)(1.0\text{ m})}

\tau = \frac{\mu_0}{4\pi}\,\frac{m}{\rho\,\rho_0\, l} \approx 0.66\text{ ms}

(5)

Check

Units in (3): $\mu_0$ is H/m and $Dd/\rho$ is $\text{m}^2/(\Omega\cdot\text{m}) = \text{m}/\Omega$ , so $\tau$ is in $\text{H}/\Omega = \text{s}$ . In the answer, $\rho\rho_0 l$ is $\Omega\cdot\text{kg}/\text{m}$ , and $\mu_0 m$ over it is again $\text{H}/\Omega$ . Doubling the mass on the same length doubles $\tau$ , as (4) doubles $Dd$ .

Remark

The diameter of the solenoid and the gauge of the wire have both dropped out: only the mass of copper per unit length, $m/l$ , sets the time constant. Even a full kilogram of copper on a metre gives under a millisecond, which is why an inductor meant to hold its current for long needs an iron core to multiply $L$ , or a superconducting winding to remove $R$ .