Irodov 3.305: Two Squares Pulling Against Each Other

Problem Irodov 3.305

A single plane wire loop is bent into two squares of sides $a = 20$ cm and $b = 10$ cm that meet at one vertex, where the wire crosses itself without contact, so that the loop has the shape of a figure eight. The loop lies in a uniform magnetic field perpendicular to its plane, whose induction varies as $B = B_0\sin\omega t$ , with $B_0 = 10$ mT and $\omega = 100$ s $^{-1}$ . The wire has resistance $\rho = 50$ m $\Omega$ /m per unit length; the inductance of the loop is negligible. Find the amplitude of the induced current.

The loop: a square of side a and a square of side b, meeting vertex to vertex, the wire crossing itself without contact at the meeting point. — The loop: a square of side $a$ and a square of side $b$ , meeting vertex to vertex, the wire crossing itself without contact at the meeting point.

Given

$a = 20$ cm
$b = 10$ cm
$B = B_0\sin\omega t$ normal to the loop
$B_0 = 10$ mT
$\omega = 100$ s $^{-1}$
$\rho = 50$ m $\Omega$ /m
inductance negligible

Find

I_m

Solution

Step 1The net flux through the crossed loop

Follow the wire once round. Because it crosses itself at the meeting point, a traversal that goes clockwise round the large square goes counter-clockwise round the small one. With the normal fixed by the traversal of the large square, the small square’s area therefore counts negatively:

\Phi = B\,(a^2 - b^2) = B_0(a^2 - b^2)\sin\omega t \tag{1}

One traversal of the wire: clockwise round the square of area a^2, counter-clockwise round the square of area b^2, so their fluxes enter with opposite signs. — One traversal of the wire: clockwise round the square of area $a^2$ , counter-clockwise round the square of area $b^2$ , so their fluxes enter with opposite signs.

Step 2The emf and its amplitude

Faraday’s law, applied to (1):

\mathcal{E}_i = -\frac{d\Phi}{dt} = -\omega B_0(a^2 - b^2)\cos\omega t

Its amplitude is

\mathcal{E}_m = \omega B_0(a^2 - b^2) \tag{2}

Step 3The resistance and the current

The wire runs round both squares once, so its length is $4a + 4b$ and its resistance follows from $\rho$ . With the inductance negligible, the current is in phase with the emf, and its amplitude is (2) over that resistance; $a^2 - b^2$ factors as $(a - b)(a + b)$ :

\begin{aligned} R &= 4\rho\,(a + b) \\ I_m &= \frac{\omega B_0(a - b)(a + b)}{4\rho\,(a + b)} \end{aligned}

Numerically,

I_m = \frac{(100\text{ s}^{-1})(10\times 10^{-3}\text{ T})(0.20 - 0.10)\text{ m}}{4\,(50\times 10^{-3}\ \Omega/\text{m})}

I_m = \frac{\omega B_0(a - b)}{4\rho} = 0.50\text{ A}

(3)

Check

Separately: $\mathcal{E}_m = 100\cdot 0.010\cdot(0.040 - 0.010) = 3.0\times 10^{-2}$ V and $R = 0.050\cdot 1.2 = 0.060\ \Omega$ , whose ratio is $0.50$ A. Equal squares, $a = b$ , give zero, as they must: the two halves of the figure eight then link equal and opposite fluxes. Units: $\text{s}^{-1}\cdot\text{T}\cdot\text{m}/(\Omega/\text{m}) = \text{T}\,\text{m}^2\,\text{s}^{-1}/\Omega = \text{V}/\Omega = \text{A}$ .

Remark

Had the wire touched at the meeting point instead of crossing, the two squares would be separate loops sharing one junction, each driven by its own full area, and a single loop going the same way round both squares would link $a^2 + b^2$ ; the crossing is what makes the linked area $a^2 - b^2$ . The sides enter only through their difference: lengthening both by the same amount $\Delta$ raises the net flux and the wire’s resistance in the same proportion, $a + b + 2\Delta$ over $a + b$ , and leaves the current unchanged.