Irodov 3.300: Inductance Drops Out of the Charge – Amit & Deepali Sinha Foundation

Problem Irodov 3.300

A long straight wire carries a constant current $I$ . A square wire frame of side $a$ lies in the same plane, two of its sides parallel to the wire; its far side lies along an axis $OO'$ , parallel to the wire at a distance $b$ from it, so the frame occupies the strip from $b - a$ to $b$ . The frame has inductance $L$ and resistance $R$ . The frame is turned through $180^\circ$ about $OO'$ , ending in the same plane on the other side of the axis. Find the electric charge that flows through the frame.

The wire with current I, the square frame of side a (solid) whose far side lies on the axis OO' at distance b from the wire, and the frame after the half turn about OO' (dashed). — The wire with current $I$ , the square frame of side $a$ (solid) whose far side lies on the axis $OO'$ at distance $b$ from the wire, and the frame after the half turn about $OO'$ (dashed).

Given

$a$
$I$
$L$
$R$
$b$ , wire to the axis
turn through $180^\circ$

Find

q

Solution

Step 1The field of the wire

Ampere’s circuital law round a circle of radius $r$ centred on the wire, on which the field has constant magnitude and runs along the circle, gives the field in the plane of the frame. It points into the page on the frame’s side, on both sides of the axis.

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

The frame in its first position, from r = b - a to r = b, cut into strips parallel to the wire: the strip of width dr at distance r has area a dr. The glyphs mark the wire's field, into the page. — The frame in its first position, from $r = b - a$ to $r = b$ , cut into strips parallel to the wire: the strip of width $dr$ at distance $r$ has area $a\,dr$ . The glyphs mark the wire’s field, into the page.

Step 2The flux before the turn

The field is constant along a strip parallel to the wire, so the strips of area $a\,dr$ are summed across the frame. With the frame’s normal $\vect n$ taken into the page at the start and (1):

\Phi_1 = \int_{b-a}^{b}\frac{\mu_0 I}{2\pi r}\,a\,dr

\Phi_1 = \frac{\mu_0 I a}{2\pi}\ln\frac{b}{b-a}

(2)

Step 3The flux after the turn

After the half turn about $OO'$ the frame occupies $b \le r \le b + a$ , and its normal $\vect n$ , carried round with it, now points out of the page, against the field. The same strip sum therefore enters with a minus sign:

\Phi_2 = -\int_{b}^{b+a}\frac{\mu_0 I}{2\pi r}\,a\,dr

\Phi_2 = -\frac{\mu_0 I a}{2\pi}\ln\frac{b+a}{b}

(3)

Step 4The circuit equation and the charge

The frame’s current $i$ obeys Ohm’s law with both emfs, the induced one and the self-induced one:

Ri = -\frac{d\Phi}{dt} - L\frac{di}{dt} \tag{4}

Integrating (4) over the whole turn, with $q = \int i\,dt$ , and noting that no current flows before the turn or after it, so the change $\Delta i$ is zero:

\begin{aligned} Rq &= -(\Phi_2 - \Phi_1) - L\,\Delta i \\ q &= \frac{\Phi_1 - \Phi_2}{R} \end{aligned} \tag{5}

Step 5The charge

Substituting (2) and (3) into (5), the two logarithms add:

q = \frac{\mu_0 I a}{2\pi R}\left(\ln\frac{b}{b-a} + \ln\frac{b+a}{b}\right)

q = \frac{\mu_0 I a}{2\pi R}\ln\frac{b+a}{b-a}

(6)

The charge does not depend on the inductance $L$ , nor on how fast or how unevenly the frame is turned. Being positive, it flows in the positive sense about $\vect n$ , clockwise in the frame as first drawn: that sense makes a field into the page, opposing the loss of flux, as Lenz’s rule requires.

Check

For a frame far from the wire, $b \gg a$ , the logarithm is about $2a/b$ and $q \approx 2\cdot\mu_0 I a^2/(2\pi b R)$ : twice the flux $B(b)\,a^2$ of a nearly uniform field, divided by $R$ , the familiar result for flipping a loop in a uniform field. As $b \to a$ the near side reaches the wire, the flux diverges logarithmically and so does $q$ . Units: $\mu_0 I$ is $\text{T}\cdot\text{m}$ , times $a$ gives Wb, and $\text{Wb}/\Omega = \text{C}$ .

Remark

The turn moves the frame into a weaker field and reverses it there, and both effects add to the change of flux, which is why the two logarithms add rather than cancel. A straight translation of the frame by $a$ away from the wire, with no turn, would give only $\Phi_1$ minus the positive flux at the new place: the logarithm becomes $\ln\bigl(b^2/((b-a)(b+a))\bigr)$ , a much smaller charge.