Irodov 3.301: A Connector in a Graded Field – Amit & Deepali Sinha Foundation

Problem Irodov 3.301

A long straight wire carries a current $I_0$ . Two further wires run parallel to it, in one plane with it and on the same side, at distances $a$ and $b$ from it ( $b > a$ ); at one end they are joined through a resistor $R$ . A straight connector, perpendicular to the wires, slides without friction along them at a constant velocity $v$ directed away from the resistor. The current $I_0$ , seen with the long wire below the other two and the resistor on the left, flows to the left, and the connector moves to the right. The resistances of the wires, the connector and the sliding contacts are negligible, and so is the self-inductance of the circuit. Find:

the magnitude and the direction of the current induced in the connector;
the force needed to keep the connector’s velocity constant.

The long wire with current I_0 to the left; above it the two wires at distances a and b, joined at the left through R, and the connector sliding to the right at v. — The long wire with current $I_0$ to the left; above it the two wires at distances $a$ and $b$ , joined at the left through $R$ , and the connector sliding to the right at $\vect v$ .

Given

$I_0$
$a$
$b$
$R$
$v = \text{const}$
no friction

Find

(a)

I

and its direction; (b)

F

Solution

Step 1The field across the connector

Ampere’s circuital law round a circle of radius $r$ centred on the long wire, on which the field has constant magnitude and runs along the circle, gives the field at distance $r$ . For $I_0$ to the left, the right-hand rule makes it point into the page above the wire, where the circuit is.

B(r) = \frac{\mu_0 I_0}{2\pi r} \tag{1}

The field varies along the connector, from $r = a$ to $r = b$ , so the emf is summed element by element.

Step 2The motional emf

An element $dr$ of the connector moving at $v$ across the field $B(r)$ contributes $vB\,dr$ to the emf, by $(\vect v\times\vect B)\cdot d\vect r$ . Summing over the connector with (1):

\begin{aligned} \mathcal{E}_i &= \int_a^b v\,\frac{\mu_0 I_0}{2\pi r}\,dr \\ &= \frac{\mu_0 I_0 v}{2\pi}\ln\frac{b}{a} \end{aligned} \tag{2}

Step 3The current

Every resistance but $R$ is negligible, and so is the self-inductance, so Ohm’s law for the whole circuit gives the current directly from (2):

I = \frac{\mathcal{E}_i}{R} = \frac{\mu_0 I_0 v}{2\pi R}\ln\frac{b}{a}

(3)

With $\vect v$ to the right and $\vect B$ into the page, $\vect v\times\vect B$ points away from the long wire, so the current in the connector flows from the wire at distance $a$ to the wire at distance $b$ , away from the long wire. Lenz’s rule agrees: the circuit’s area, and with it the inward flux, grows, so the current circulates counter-clockwise, which is upward in the connector.

The wire's field B, into the page and thinning as 1/r; the induced current I, up the connector and counter-clockwise round the circuit; the Ampere force F_A on the connector, opposing v. — The wire’s field $\vect B$ , into the page and thinning as $1/r$ ; the induced current $I$ , up the connector and counter-clockwise round the circuit; the Ampere force $\vect F_{\mathrm{A}}$ on the connector, opposing $\vect v$ .

Step 4The Ampere force on the connector

The current in the connector sits in the field, so each element $dr$ feels the force $I\,d\vect r\times\vect B$ , of magnitude $IB\,dr$ . For a current away from the wire and $\vect B$ into the page this points to the left, against $\vect v$ , on every element. Summing with (1):

F_{\mathrm{A}} = \int_a^b I\,\frac{\mu_0 I_0}{2\pi r}\,dr = I\,\frac{\mu_0 I_0}{2\pi}\ln\frac{b}{a} \tag{4}

Step 5The force that keeps

v

constant

With no friction and constant velocity, the applied force balances $F_{\mathrm{A}}$ , so it points along $\vect v$ with the magnitude (4). Substituting (3):

F = \frac{v}{R}\left(\frac{\mu_0 I_0}{2\pi}\ln\frac{b}{a}\right)^2

(5)

The force is directed along $\vect v$ , away from the resistor.

Check

Energy: the applied force does work at the rate $Fv$ , and the resistor dissipates $I^2R$ . With (3) and (5) both equal $(v^2/R)\bigl(\mu_0 I_0\ln(b/a)/2\pi\bigr)^2$ , as they must, since nothing else stores or absorbs energy. When the two wires coincide, $b \to a$ , the logarithm vanishes and so do the current and the force: a connector of zero length cuts no field lines.

Remark

In a uniform field the emf would be $Blv$ with $l = b - a$ ; here $B$ falls as $1/r$ along the connector, and the length is replaced by the logarithm: the integral $\int dr/r$ weights each element by the field it moves through. The force goes as the square of $I_0$ , so reversing $I_0$ reverses the field, the emf and the current, but the drag on the connector still opposes the motion.