Irodov 3.255: Only the Two Parallel Sides Count – Amit & Deepali Sinha Foundation

Problem Irodov 3.255

A long straight wire carries a current $I_0 = 5.0$ A. A square frame of side $a = 8.0$ cm, carrying a current $I = 0.90$ A, lies in the same plane as the wire. The frame’s axis, the line through the midpoints of two opposite sides, is parallel to the wire at a distance $\eta a$ from it, with $\eta = 1.5$ . Find:

the Ampere force on the frame;
the mechanical work needed to turn the frame through $180^\circ$ about this axis while both currents are kept constant.

The long wire with current I_0 and the square frame of side a, carrying I, in the same plane; the frame's axis (dashed) runs parallel to the wire at the distance η a. — The long wire with current $I_0$ and the square frame of side $a$ , carrying $I$ , in the same plane; the frame’s axis (dashed) runs parallel to the wire at the distance $\eta a$ .

Given

$I = 0.90$ A
$I_0 = 5.0$ A
$a = 8.0$ cm
axis at $\eta a$ from the wire, $\eta = 1.5$
for (b), the frame starts with its near side’s current parallel to $I_0$ (assumed)

Find

(a) the force

F

on the frame; (b) the work

A

to turn it through

180^\circ

Solution

Step 1The force on each side

The wire’s field at a distance $r$ follows from Ampere’s law on a circle round the wire, on which by symmetry $B$ is tangential and the same all round:

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

In the plane of the frame this field is normal to the plane. The two sides parallel to the wire lie at the distances

r_1 = \eta a - \frac{a}{2}, \qquad r_2 = \eta a + \frac{a}{2} \tag{2}

The wire's field B on the frame (glyph density falling as 1/r), drawn for the frame current running parallel to I_0 along the near side: the force F_1 on the near side pulls toward the wire, the force F_2 on the far side, half as large, pushes away. — The wire’s field $\vect B$ on the frame (glyph density falling as $1/r$ ), drawn for the frame current running parallel to $I_0$ along the near side: the force $\vect F_1$ on the near side pulls toward the wire, the force $\vect F_2$ on the far side, half as large, pushes away.

Each of those two sides has length $a$ and sits in a uniform field, so by $\vect F = I\,\vect l\times\vect B$ with (1) the magnitudes are

F_1 = \frac{\mu_0 I_0 I a}{2\pi r_1}, \qquad F_2 = \frac{\mu_0 I_0 I a}{2\pi r_2} \tag{3}

The currents in these sides run opposite ways, so the forces are opposite: the side whose current is parallel to $I_0$ is attracted, the other repelled. The two sides perpendicular to the wire carry opposite currents through the same field profile, so their forces, along the wire, cancel exactly.

Step 2The net force

The near side is in the stronger field, so the net force is the difference of (3), directed along the line from the frame to the wire or away from it:

F = F_1 - F_2 = \frac{\mu_0 I_0 I a}{2\pi}\left(\frac{1}{r_1} - \frac{1}{r_2}\right) = \frac{\mu_0 I_0 I a}{2\pi}\cdot\frac{r_2 - r_1}{r_1 r_2}

With (2), $r_2 - r_1 = a$ and $r_1 r_2 = (\eta^2 - 1/4)a^2$ :

F = \frac{\mu_0 I_0 I a^2}{2\pi(\eta^2 - 1/4)a^2}

in which $a^2$ cancels. Numerically,

F = \frac{2(4\pi\times10^{-7}\text{ H/m})(5.0\text{ A})(0.90\text{ A})}{\pi\,(4(1.5)^2 - 1)}

F = \frac{2\mu_0 I I_0}{\pi(4\eta^2 - 1)} = 0.45\ \mu\text{N}

(4)

The frame is attracted to the wire when the current in its near side is parallel to $I_0$ , as drawn, and repelled when it is antiparallel.

Step 3The flux of the wire’s field through the frame

At constant currents the work of the Ampere forces on a circuit equals its current times the change of the flux through it, so the flux is what we need. Take a strip of width $dr$ and length $a$ at the distance $r$ from the wire.

The strip dr of the frame at the distance r from the wire, where the wire's field B is uniform across the strip; the near side is at η a - a/2. — The strip $dr$ of the frame at the distance $r$ from the wire, where the wire’s field $\vect B$ is uniform across the strip; the near side is at $\eta a - a/2$ .

Integrating (1) over the frame between the limits (2):

\Phi = \int_{r_1}^{r_2}\frac{\mu_0 I_0}{2\pi r}\,a\,dr = \frac{\mu_0 I_0 a}{2\pi}\ln\frac{r_2}{r_1} = \frac{\mu_0 I_0 a}{2\pi}\ln\frac{2\eta + 1}{2\eta - 1} \tag{5}

Its sign is taken relative to the frame’s own normal, the one given by its current through the right-hand rule. In the starting position (assumed) the near side’s current is parallel to $I_0$ : the frame’s own field inside it then points the same way as the wire’s, and the flux through it is $+\Phi$ .

Step 4The work to turn the frame over

Turning the frame through $180^\circ$ about its axis swaps its two parallel sides: it ends in the same place with its normal reversed, so the flux through it goes from $+\Phi$ to $-\Phi$ . At constant current the Ampere forces do the work $I\,\Delta\Phi$ , and the mechanical work is its negative:

A = -I\,(\Phi_{\text{final}} - \Phi_{\text{initial}}) = -I\,(-\Phi - \Phi) = 2I\Phi

With (5), and numerically,

A = \frac{(4\pi\times10^{-7}\text{ H/m})(0.080\text{ m})(0.90\text{ A})(5.0\text{ A})}{\pi}\,\ln\frac{4}{2}

A = \frac{\mu_0 a I I_0}{\pi}\ln\frac{2\eta + 1}{2\eta - 1} = 0.10\ \mu\text{J}

(6)

This is the work done against the Ampere forces in turning the frame out of its stable position; started from the other position, the forces would do this work instead and $A$ would be $-0.10\ \mu$ J.

Check

Far from the wire, $\eta \gg 1$ , the frame is a small dipole of moment $p_m = Ia^2$ in the field (1), and the force on a dipole along the gradient is $p_m\,dB/dr = Ia^2\mu_0 I_0/(2\pi r^2)$ at $r = \eta a$ , that is $\mu_0 I I_0/(2\pi\eta^2)$ . The answer (4) gives $2\mu_0 I I_0/(4\pi\eta^2)$ , the same. Likewise $\ln[(2\eta+1)/(2\eta-1)] \to 1/\eta$ , so $A \to 2 p_m B$ at the frame, the work to reverse a dipole in a field.

Remark

The force on the frame depends on $\eta$ alone, not on the size of the frame, because the wire’s field has no length scale of its own: shrink the whole arrangement and every force on a side stays the same. The flux, and so the work, does carry a length, $a$ , and grows with the frame.