Irodov 3.304: Which Way Round Each Loop – Amit & Deepali Sinha Foundation

Problem Irodov 3.304

Four plane figures made of thin wire lie in the plane of the page, in a uniform magnetic field directed into the page, away from the reader:

a circle with a straight wire along one diameter, joined to the circle at both ends;
two concentric circles, each cut by a small gap on the same side, the ends at the gaps joined by two short parallel wires so that the whole forms a single closed loop;
two concentric closed circles joined by one short radial wire;
a figure eight: a large and a small circle made of one wire, which crosses itself at the waist without contact.

The field begins to decrease. Find the directions of the currents induced in these conductors.

The four conductors in the plane of the page: (a) a circle with a diameter, (b) two broken circles joined into one loop, (c) two closed circles with a radial link, (d) a figure eight crossing itself without contact.

Given

$\vect B$ into the page, uniform
$B$ decreasing

Find

the sense of the induced current in (a), (b), (c), (d)

Solution

Step 1The rule for a single loop

Take the normal $\vect n$ into the page, along $\vect B$ , so that the positive sense of circulation about $\vect n$ is clockwise as the reader sees the page. A loop enclosing a flat area $S$ then carries the flux $\Phi = BS$ , and Faraday’s law gives its emf in that sense:

\mathcal{E}_i = -\frac{d\Phi}{dt} = -S\,\frac{dB}{dt} > 0 \tag{1}

since $dB/dt < 0$ . So in every simple loop the induced current runs clockwise, making a field into the page inside it: Lenz’s rule, opposing the loss of inward flux.

Step 2(a) The circle with a diameter

The diameter cuts the disc into two half discs of area $S/2$ , each bounded by a half circle of resistance $r$ and the diameter, of resistance $r_d$ . Let $I_1$ and $I_2$ be the clockwise currents of the upper and lower half circles; the diameter, shared by both loops in opposite directions, carries $I_1 - I_2$ . Kirchhoff’s loop rule round each half, with the emf (1) of area $S/2$ :

\begin{aligned} I_1 r + (I_1 - I_2)\,r_d &= -\frac{S}{2}\frac{dB}{dt} \\ I_2 r + (I_2 - I_1)\,r_d &= -\frac{S}{2}\frac{dB}{dt} \end{aligned}

Subtracting the second line from the first:

(I_1 - I_2)(r + 2r_d) = 0

So $I_1 = I_2$ , whatever the diameter’s resistance.

(a) Clockwise round the circle; no current in the diameter.

Step 3(b) Two broken circles forming one loop

Follow the single loop: having gone clockwise round the outer circle, it passes along a joining wire to the inner circle and runs round that one in the opposite sense, counter-clockwise, before returning to the start. The flux it links is therefore the outer area minus the inner one, $S_1 - S_2$ (the thin strip between the two joining wires adds nothing worth counting), and by (1):

\mathcal{E}_i = -(S_1 - S_2)\,\frac{dB}{dt} > 0

The emf is positive for the clockwise traversal of the outer circle, because $S_1 > S_2$ .

(b) Clockwise in the outer circle, counter-clockwise in the inner.

Step 4(c) Two closed circles and a radial link

Each circle is a closed loop of its own, and by (1) each carries a clockwise current. The link touches each circle at a single point, and no closed path runs through it: a current entering one circle through the link could leave that circle only back through the link. So charge conservation leaves it with no current.

Step 5(d) The figure eight

The crossing at the waist makes the single wire run round the two lobes in opposite senses: clockwise round one means counter-clockwise round the other. The net flux linked is the large lobe’s area minus the small lobe’s, $S_{\mathrm{L}} - S_{\mathrm{s}}$ , and the emf (1) is positive for the clockwise traversal of the large lobe:

\mathcal{E}_i = -(S_{\mathrm{L}} - S_{\mathrm{s}})\,\frac{dB}{dt} > 0

(d) Clockwise in the large lobe, counter-clockwise in the small.

The induced currents while B (into the page) decreases: (a) clockwise round the circle, none in the diameter; (b) clockwise outside, counter-clockwise inside; (c) clockwise in both circles, none in the link; (d) clockwise in the large lobe, counter-clockwise in the small. — The induced currents while $\vect B$ (into the page) decreases: (a) clockwise round the circle, none in the diameter; (b) clockwise outside, counter-clockwise inside; (c) clockwise in both circles, none in the link; (d) clockwise in the large lobe, counter-clockwise in the small.

Check

In (b) and (d) the current is one and the same in both parts of the wire, and it does run clockwise in one part and counter-clockwise in the other; in each case the larger enclosed area sets the sign. Shrinking the inner circle of (b) to nothing leaves a single clockwise circle, and making the two lobes of (d) equal leaves zero net emf and no current, both as the rule requires.

Remark

The induced emf belongs to the whole circuit, not to a piece of wire: the diameter in (a) and the link in (c) carry no current although the field changes around them, while in (b) and (d) the inner part of the loop runs against its own local tendency because the outer part wins. A figure eight in a uniform changing field is the principle of a twisted pair: equal lobes cancel any uniform pickup.