Irodov 3.309: A Copper Turn Round a Solenoid – Amit & Deepali Sinha Foundation

Problem Irodov 3.309

A long straight solenoid has cross-sectional diameter $d = 5$ cm and $n = 20$ turns per centimetre of its length. A single round turn of copper wire (resistivity $\rho = 1.7\times 10^{-8}\ \Omega\,$ m) with cross-sectional area $S = 1.0$ mm $^2$ is fitted tightly over its winding. The current in the solenoid’s winding grows at the constant rate $\dot{I} = 100$ A/s. Neglecting the inductance of the turn, find the current in it.

Cross-section of the solenoid, diameter d, with the copper turn of wire section S fitted on its winding (outer circle). The winding's current I grows; its sense is not given, drawn counter-clockwise. — Cross-section of the solenoid, diameter $d$ , with the copper turn of wire section $S$ fitted on its winding (outer circle). The winding’s current $I$ grows; its sense is not given, drawn counter-clockwise.

Given

$d = 5$ cm
$n = 20$ cm $^{-1} = 2.0\times 10^{3}$ m $^{-1}$
$S = 1.0$ mm $^2$
$\dot{I} = 100$ A/s
copper, $\rho = 1.7\times 10^{-8}\ \Omega\,$ m (table value)
inductance of the turn neglected

Find

the current

I_{\mathrm{t}}

in the turn

Solution

Step 1The emf round the turn

Inside a long solenoid the field is uniform and axial, $B = \mu_0 n I$ (Ampere’s law on a rectangle with one side inside along the axis), and outside it is negligible. The turn sits on the winding, so it encloses the whole cross-section $\pi d^2/4$ , and by Faraday’s law the emf round it is the rate of change of that flux:

|\mathcal{E}_i| = \frac{d\Phi}{dt} = \frac{\pi d^2}{4}\,\frac{d}{dt}\bigl(\mu_0 n I\bigr)

|\mathcal{E}_i| = \frac{\pi d^2}{4}\,\mu_0 n\dot{I}

(1)

Step 2The resistance of the turn

The turn is a copper wire of length $\pi d$ and cross-section $S$ , so with copper’s resistivity $\rho$ (table value)

R_{\mathrm{t}} = \rho\,\frac{\pi d}{S} \tag{2}

Step 3The current

With the turn’s inductance neglected, the current is the emf over the resistance. Dividing (1) by (2):

I_{\mathrm{t}} = \frac{|\mathcal{E}_i|}{R_{\mathrm{t}}} = \frac{\pi d^2\mu_0 n\dot{I}/4}{\rho\pi d/S}

Numerically, with $\mu_0 = 4\pi\times 10^{-7}$ H/m:

I_{\mathrm{t}} = \frac{4\pi\times 10^{-7}\times 2.0\times 10^{3}\times 1.0\times 10^{-6}\times 0.05\times 100}{4\times 1.7\times 10^{-8}}\ \text{A}

I_{\mathrm{t}} = \frac{\mu_0 n S d\,\dot{I}}{4\rho} \approx 0.18\ \text{A}

(3)

The current circulates opposite to the solenoid’s own current (Lenz): its field opposes the growth of $\vect B$ , so it runs clockwise as seen with the growing $\vect B$ pointing at the viewer.

The induced current in the turn, clockwise as seen with the growing B out of the page. — The induced current in the turn, clockwise as seen with the growing $\vect B$ out of the page.

Check

Units: $\mu_0 n\dot I$ is $\dot B$ in T/s; times $Sd$ in m $^3$ it is V m, and over $\rho$ in $\Omega\,$ m it is V/ $\Omega$ , amperes. Neglecting the inductance is sound: a turn of radius 2.5 cm in wire of radius $(S/\pi)^{1/2} \approx 0.56$ mm has a self-inductance of order $\mu_0 (d/2)\ln(4d/(S/\pi)^{1/2}) \approx 10^{-7}$ H, and its resistance (2) is $2.7\times 10^{-3}\ \Omega$ , so its time constant is below $10^{-4}$ s. Within a fraction of a millisecond of a steady ramp, the current is the steady value found.

Remark

The current grows with the diameter, not with its square: a larger turn encloses flux in proportion to $d^2$ but its wire, and so its resistance, grows as $d$ . A thick ring of copper slipped over an alternating-current solenoid gets warm for exactly this reason, and a copper sleeve round a transformer core acts as a shorted turn that wastes power.