Irodov 3.294: Two Sides in Unequal Fields – Amit & Deepali Sinha Foundation

Problem Irodov 3.294

A long straight wire carries a steady current $I$ . A square wire frame of side $a$ lies in the same plane, with two of its sides parallel to the wire, and the nearer of those sides a distance $x$ from it. The frame moves directly away from the wire, without turning, at a constant speed $v$ . Find the emf induced in the frame as a function of $x$ .

The long wire with current I and the square frame of side a beside it, its near side at distance x, translating away at v. The wire's field B points into the page on this side, thinning out with distance. — The long wire with current $I$ and the square frame of side $a$ beside it, its near side at distance $x$ , translating away at $\vect v$ . The wire’s field $\vect B$ points into the page on this side, thinning out with distance.

Given

$a$
$I$
$v$
$x$ , the distance from the wire to the near side

Find

\mathcal{E}_i(x)

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 frame it points into the page on the frame’s side, for the current direction drawn.

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

The frame cut into strips parallel to the wire: the strip of width dr at distance r has area a dr and field B(r). The induced current I_ind circulates clockwise, its own field into the page. — The frame cut into strips parallel to the wire: the strip of width $dr$ at distance $r$ has area $a\,dr$ and field $B(r)$ . The induced current $I_{\mathrm{ind}}$ circulates clockwise, its own field into the page.

Step 2The flux through the frame

The field is constant along a strip parallel to the wire, so the frame is cut into such strips, each of area $a\,dr$ , and their fluxes summed from the near side at $r = x$ to the far side at $r = x + a$ , using (1).

\begin{aligned} \Phi &= \int_{x}^{x+a} B(r)\,a\,dr = \frac{\mu_0 I a}{2\pi}\int_{x}^{x+a}\frac{dr}{r} \end{aligned}

\Phi(x) = \frac{\mu_0 I a}{2\pi}\ln\frac{x+a}{x}

(2)

Step 3The rate of change of the flux

As the frame moves, only $x$ changes, at the rate $dx/dt = v$ . Differentiating (2) by the chain rule:

\begin{aligned} \frac{d\Phi}{dt} &= \frac{d\Phi}{dx}\,\frac{dx}{dt} = \frac{\mu_0 I a}{2\pi}\left(\frac{1}{x+a} - \frac{1}{x}\right)v \\ &= -\frac{\mu_0 I a}{2\pi}\,\frac{a}{x(x+a)}\,v \end{aligned}

The flux falls as the frame recedes, as it must, since the frame moves into a weaker field.

Step 4The induced emf

Faraday’s law gives the emf as minus the rate of change of the flux:

\mathcal{E}_i = -\frac{d\Phi}{dt} = \frac{\mu_0 I a^2 v}{2\pi\,x(x+a)}

(3)

By Lenz’s rule the induced current opposes the loss of inward flux, so it circulates clockwise in the figure, adding field into the page inside the frame.

Check

The same emf from the motional picture, with no flux at all. The two sides parallel to the wire move at $v$ across the field, each carrying an emf $B a v$ ; the two sides perpendicular to the wire move along their own length and carry none. Round the loop the near and far emfs oppose each other, so with (1) the net is

\begin{aligned} a v\left[B(x) - B(x+a)\right] &= \frac{\mu_0 I a v}{2\pi}\left(\frac{1}{x} - \frac{1}{x+a}\right) \\ &= \frac{\mu_0 I a^2 v}{2\pi\,x(x+a)}, \end{aligned}

which is (3). Dimensions: $\mu_0 I/r$ is a field in tesla, and $B\,a\,v$ is $\text{T}\cdot\text{m}\cdot\text{m/s} = \text{V}$ .

Remark

Far from the wire, $x \gg a$ , the emf falls as $1/x^2$ , not as $1/x$ like the field itself: what drives the current is the difference between the fields at the two sides, $a\,dB/dr$ , and for a field going as $1/r$ that gradient goes as $1/r^2$ . A frame dragged through a uniform field gets no emf at all, however fast it moves.