Irodov 3.293: A Connector Running into Weaker Field – Amit & Deepali Sinha Foundation

Problem Irodov 3.293

A long straight wire carries a steady current $I$ . In the same plane lies a $\Pi$ -shaped conductor: two parallel rails perpendicular to the wire, joined at their ends nearer the wire by a crosspiece parallel to it. A straight connector of length $l$ and resistance $R$ , parallel to the wire, lies across the rails and slides away from the wire with a constant velocity $v$ , closing a rectangular loop. The resistance of the $\Pi$ -shaped conductor and the self-inductance of the loop are negligible. Find the current induced in the loop as a function of the distance $r$ from the wire to the connector.

The long wire with current I and the Π-shaped conductor beside it, closed end nearest the wire; the connector of resistance R and length l, a distance r from the wire, slides away at v. — The long wire with current $I$ and the $\Pi$ -shaped conductor beside it, closed end nearest the wire; the connector of resistance $R$ and length $l$ , a distance $r$ from the wire, slides away at $\vect v$ .

Given

$I$
$l$
$R$ (connector only)
$v$ constant
$r$ (wire to connector)

Find

I_{\mathrm{ind}}(r)

Solution

Step 1The field of the wire

Ampere’s circuital law is applied round a circle of radius $r$ centered on the wire, on which the field has constant magnitude and runs along the circle. In the plane of the loop it is normal to the plane, into the page on the loop’s side for the current drawn.

\begin{aligned} B\cdot 2\pi r &= \mu_0 I \\ B(r) &= \frac{\mu_0 I}{2\pi r} \end{aligned} \tag{1}

Step 2The emf of the moving connector

Only the connector moves. It is parallel to the wire, so all of it is at the same distance $r$ and in the same field (1); its velocity, its length and the field are mutually perpendicular. Its motional emf is therefore

\mathcal{E}_i = B(r)\,l\,v = \frac{\mu_0 I l v}{2\pi r} \tag{2}

The wire's field B, into the page on the loop's side and thinning out with distance, and the induced current I_ind circulating counter-clockwise, up the connector, so that its own field opposes the growth of the inward flux. — The wire’s field $\vect B$ , into the page on the loop’s side and thinning out with distance, and the induced current $I_{\mathrm{ind}}$ circulating counter-clockwise, up the connector, so that its own field opposes the growth of the inward flux.

Step 3The induced current

The only resistance in the loop is the connector’s, $R$ , and self-inductance is negligible, so Ohm’s law applied to (2) gives the current:

I_{\mathrm{ind}} = \frac{\mathcal{E}_i}{R} = \frac{\mu_0 I l v}{2\pi R\,r}

(3)

The current falls off as $1/r$ . As the connector moves away the loop’s area grows, so the inward flux grows even though the field at the connector weakens; by Lenz’s rule the induced current circulates counter-clockwise in the figure, flowing up the connector, along $\vect v\times\vect B$ .

Check

The same emf from the flux. Let the closed end be a distance $r_0$ from the wire. The flux through the loop, summed over strips of area $l\,dr'$ parallel to the wire, and its rate of change as $r$ grows at $dr/dt = v$ , are

\begin{aligned} \Phi &= \int_{r_0}^{r}\frac{\mu_0 I}{2\pi r'}\,l\,dr' = \frac{\mu_0 I l}{2\pi}\ln\frac{r}{r_0} \\ \frac{d\Phi}{dt} &= \frac{\mu_0 I l}{2\pi r}\,v, \end{aligned}

which is (2); $r_0$ drops out. Dimensions: $\mu_0 I/r$ is a field in tesla, and $\text{T}\cdot\text{m}\cdot(\text{m/s})/\Omega = \text{V}/\Omega = \text{A}$ .

Remark

To keep the connector at constant speed an external force must balance the Ampere force on it, $I_{\mathrm{ind}}\,l\,B(r)$ , which with (1) and (3) falls as $1/r^2$ . Its power $I_{\mathrm{ind}}lBv = \mathcal{E}_i I_{\mathrm{ind}}$ is exactly the Joule heat in $R$ : the work done pulling the connector against the field is what heats it.