Irodov 3.251: The Rest of the Wire Pushes – Amit & Deepali Sinha Foundation

Problem Irodov 3.251

A thin wire carries a current $I = 8.0$ A. Find the magnitude and direction of the force per unit length acting on the wire at its point $O$ when the wire is bent into each of the following shapes.

A closed loop made of a semicircle of radius $R = 10$ cm and the diameter joining its ends; the current runs clockwise, and $O$ is the midpoint of the diameter.
A hairpin: two long parallel straight wires a distance $l = 20$ cm apart, joined at one end by a straight segment perpendicular to them, the current running out along one wire and back along the other; $O$ is the midpoint of the joining segment.

(a) The semicircle of radius R closed by its diameter, current I clockwise; O is the midpoint of the diameter, on the wire. — (a) The semicircle of radius $R$ closed by its diameter, current $I$ clockwise; $O$ is the midpoint of the diameter, on the wire.

(b) Two long parallel wires a distance l apart joined by a short segment, current I out along one and back along the other; O is the midpoint of the segment. — (b) Two long parallel wires a distance $l$ apart joined by a short segment, current $I$ out along one and back along the other; $O$ is the midpoint of the segment.

Given

$I = 8.0$ A
(a) $R = 10$ cm
(b) $l = 20$ cm

Find

the force per unit length on the wire at

O

, magnitude and direction

Solution

Step 1The force on the wire at O comes from the rest of the wire

A short piece $dl$ of the wire at $O$ feels the Ampere force $I\,d\vect l\times\vect B$ , where $\vect B$ is the field at $O$ of every other part of the circuit; the piece exerts no net force on itself. Per unit length, with $d\vect l$ perpendicular to $\vect B$ in both shapes below,

F_1 = I B \tag{1}

A straight piece whose line passes through $O$ gives nothing at $O$ : each of its elements $d\vect l$ is parallel to the vector $\vect r$ from it to $O$ , and the Biot-Savart cross product $d\vect l\times\vect r$ vanishes. So in (a) the diameter, and in (b) the joining segment, contribute no field at $O$ .

Step 2(a) The field of the semicircle at its center

Every element of the arc is at the distance $R$ from $O$ and perpendicular to the radius, so all the Biot-Savart contributions add, over the arc length $\pi R$ :

B = \frac{\mu_0 I}{4\pi R^2}\int_{\text{arc}} dl = \frac{\mu_0 I}{4\pi R^2}\,\pi R

B = \frac{\mu_0 I}{4R}

(2)

For the clockwise current, the right-hand rule puts $\vect B$ into the page at $O$ .

(a) B at O from the arc is into the page (the cross); the force on the wire at O points away from the arc. — (a) $\vect B$ at $O$ from the arc is into the page (the cross); the force on the wire at $O$ points away from the arc.

(b) B at O from the two long wires is into the page; the force on the segment at O points away from the long wires. — (b) $\vect B$ at $O$ from the two long wires is into the page; the force on the segment at $O$ points away from the long wires.

Step 3(a) The force per unit length

At $O$ the current runs left along the diameter. With $\vect i$ to the right, $\vect j$ up the page and $\vect k$ out of it, $d\vect l$ is along $-\vect i$ and $\vect B$ along $-\vect k$ , so the force is along

(-\vect i)\times(-\vect k) = \vect i\times\vect k = -\vect j

downward, away from the arc. Its magnitude is $I$ times (2), by (1); numerically,

F_1 = \frac{(4\pi\times 10^{-7}\text{ H/m})(8.0\text{ A})^2}{4\,(0.10\text{ m})}

F_1 = \frac{\mu_0 I^2}{4R} \approx 0.20\text{ mN/m}

(3)

The force lies in the plane of the loop, perpendicular to the diameter, pointing away from the arc: the arc’s current runs opposite to the diameter’s, and opposite currents repel.

Step 4(b) The field of the two long wires

Each long wire starts at an end of the segment and runs off to infinity; $O$ lies on the perpendicular to it through its end, at the distance $b = l/2$ . A straight wire seen from a point at the distance $b$ from its line gives, by the Biot-Savart law integrated along it,

B = \frac{\mu_0 I}{4\pi b}\,(\sin\alpha_1 + \sin\alpha_2)

with $\alpha_1$ and $\alpha_2$ the angles its ends make with the perpendicular, here $0$ (the near end, at the foot) and $90^\circ$ (the end at infinity):

B_{\text{one}} = \frac{\mu_0 I}{4\pi b}\,(0 + 1) = \frac{\mu_0 I}{4\pi (l/2)} = \frac{\mu_0 I}{2\pi l}

The upper wire carries its current to the right with $O$ below it, the lower wire to the left with $O$ above it; by the right-hand rule both give $\vect B$ into the page at $O$ , so the two add:

B = 2B_{\text{one}} = \frac{\mu_0 I}{\pi l}

(4)

Step 5(b) The force per unit length

At $O$ the current runs up the segment, from the lower wire to the upper one, so $d\vect l$ is along $\vect j$ and $\vect B$ along $-\vect k$ :

\vect j\times(-\vect k) = -\vect i

to the left, away from the long wires. Its magnitude is $I$ times (4), by (1); numerically,

F_1 = \frac{(4\pi\times 10^{-7}\text{ H/m})(8.0\text{ A})^2}{\pi\,(0.20\text{ m})}

F_1 = \frac{\mu_0 I^2}{\pi l} \approx 0.13\text{ mN/m}

(5)

The force lies in the plane of the wires, along the long wires and away from them: the bend is pushed outward.

Check

Both results have the form $\mu_0 I^2$ over a length, which is a force per unit length in N/m. Both fields also follow by symmetry, without integrating. (a) A full circle gives $\mu_0 I/(2R)$ at its center, and its two halves contribute equally there, so the semicircle gives $\mu_0 I/(4R)$ , as in (2). (b) Two infinite antiparallel wires $l$ apart give $2\mu_0 I/(\pi l)$ midway between them (each $\mu_0 I/(2\pi\cdot l/2)$ , adding); cutting each at the line through $O$ leaves two halves that contribute equally, so the half-infinite pair gives half of that, $\mu_0 I/(\pi l)$ , as in (4).

Remark

The force on a current loop from its own field always tends to enlarge it: in (a) the diameter is pushed away from the arc, in (b) the closing segment away from the wires, so both bends try to open out. This is the same outward magnetic pressure that bursts the winding of a coil carrying a large current; the loop’s self-inductance grows as it expands, and at fixed current a circuit is always pushed toward larger inductance.