Worked examples · Sheet 3

Forces between parallel wires

Theory used: Force between two parallel wires Field of an infinite straight wire

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

The two conductors of a power cable are straight and parallel, at distance $d = 5.0 mm$ , and carry the current $I = 25 A$ in opposite senses. Compute the force per unit length between the conductors and the force on a segment of length $ℓ = 1.5 m$ , stating whether it is attractive or repulsive.

\frac{F}{ℓ} = \frac{μ _{0} I ^{2}}{2 π d}, F = \frac{μ _{0} I ^{2}}{2 π d} ℓ

\frac{F}{ℓ} F = \frac{2 \cdot 1 0 ^{- 7} \cdot 2 5 ^{2}}{5.0 \cdot 1 0 ^{- 3}} N/m = \frac{1.25 \cdot 1 0 ^{- 4}}{5.0 \cdot 1 0 ^{- 3}} N/m = 2.5 \cdot 1 0^{- 2} N/m = 2.5 \cdot 1 0^{- 2} \cdot 1.5 N \approx 3.8 \cdot 1 0^{- 2} N

The distance between the conductors is much smaller than their length, so the conductors can be treated as infinite wires and the theorem on the force between two parallel wires applies, with $I_{1} = I_{2} = I$ . The currents are in opposite directions, therefore the force is repulsive: the two conductors tend to move apart. The force on a segment of $1.5 m$ is about $4 \cdot 1 0^{- 2} N$ , small under normal conditions; since it depends on the square of the current, it becomes significant in short circuits, when the current increases many times over.

Exercise 2

Three infinite straight wires, parallel and coplanar, are equally spaced, with distance $d = 4.0 cm$ between adjacent wires. They carry the currents $I_{1} = 10 A$ , $I_{2} = 15 A$ and $I_{3} = 30 A$ in the same sense, with wire $2$ in the middle. Compute the force per unit length on each wire.

Three wires with currents in the same direction: each wire is attracted by the other two, and the resultant force on each points toward the side where the attractions are larger

f_{ij} = \frac{μ _{0} I _{i} I _{j}}{2 π d _{ij}}

f_{12} f_{23} f_{13} = \frac{2 \cdot 1 0 ^{- 7} \cdot 10 \cdot 15}{0.040} N/m = 7.5 \cdot 1 0^{- 4} N/m = \frac{2 \cdot 1 0 ^{- 7} \cdot 15 \cdot 30}{0.040} N/m = 2.25 \cdot 1 0^{- 3} N/m = \frac{2 \cdot 1 0 ^{- 7} \cdot 10 \cdot 30}{0.080} N/m = 7.5 \cdot 1 0^{- 4} N/m

\frac{F _{1}}{ℓ} \frac{F _{2}}{ℓ} \frac{F _{3}}{ℓ} = f_{12} + f_{13} = 1.5 \cdot 1 0^{- 3} N/m toward wire 2 = f_{23} - f_{12} = 1.5 \cdot 1 0^{- 3} N/m toward wire 3 = f_{23} + f_{13} = 3.0 \cdot 1 0^{- 3} N/m toward wire 2

By the superposition principle the force on each wire is the sum of the forces due to the other two, and each pair of wires interacts according to the theorem on the force between two parallel wires; $f_{ij}$ denotes the magnitude of the force per unit length between wires $i$ and $j$ , which for wires $1$ and $3$ are at distance $2 d$ . The currents are all in the same direction, so each force is attractive.

Wire $1$ is attracted to the right by both the others, and the two forces add up. Wire $2$ is attracted to the left by wire $1$ and to the right by wire $3$ ; the attraction of wire $3$ is greater, therefore the resultant points toward wire $3$ . Wire $3$ is attracted to the left by both. As a check, the internal forces of a system have zero resultant by Newton’s third law: counting the forces to the right as positive we have $1.5 \cdot 1 0^{- 3} + 1.5 \cdot 1 0^{- 3} - 3.0 \cdot 1 0^{- 3} = 0$ .

Exercise 3

A long straight wire, lying on a horizontal table, carries the current $I_{1} = 80 A$ . Above it, parallel to it, there is a second wire, free to move vertically along smooth guides, with mass per unit length $\frac{m}{ℓ} = 6.0 \cdot 1 0^{- 3} kg/m$ , carrying the current $I_{2} = 40 A$ in the sense opposite to $I_{1}$ . Compute the height $h$ at which the second wire remains in equilibrium and determine whether the equilibrium is stable for vertical displacements.

\frac{μ _{0} I _{1} I _{2}}{2 π h} ℓ = m g ⟹ h = \frac{μ _{0} I _{1} I _{2}}{2 π \frac{m}{ℓ} g}

h = \frac{2 \cdot 1 0 ^{- 7} \cdot 80 \cdot 40}{6.0 \cdot 1 0 ^{- 3} \cdot 9.81} m = \frac{6.4 \cdot 1 0 ^{- 4}}{5.886 \cdot 1 0 ^{- 2}} m \approx 1.1 \cdot 1 0^{- 2} m

On the upper wire act the weight, pointing downward, and the magnetic force due to the lower wire. The currents are in opposite directions, so by the theorem on the force between two parallel wires the magnetic force is repulsive, that is, it points upward. The smooth guides exert only horizontal forces, therefore at equilibrium the magnetic force on a segment of length $ℓ$ equals the weight of that segment; the length cancels and the mass per unit length remains. The wire floats about $1.1 cm$ above the lower wire.

If the wire moves down slightly, the distance $h$ decreases and the repulsive force, inversely proportional to $h$ , exceeds the weight; the resultant then points upward and brings the wire back toward the equilibrium position. If the wire moves up, the magnetic force becomes smaller than the weight and the wire comes back down. The equilibrium is therefore stable for vertical displacements.

Exercise 4

A rigid rectangular loop lies in the same plane as an infinite straight wire carrying the current $I_{1} = 20 A$ . The sides of the loop parallel to the wire are $a = 10 cm$ long; the nearer side is $c = 2.0 cm$ from the wire and the farther one $c + b = 7.0 cm$ . The loop carries the current $I_{2} = 5.0 A$ , which in the side near the wire has the same sense as $I_{1}$ . Compute the resultant force on the loop.

Rectangular loop beside a wire: the near side, with current in the same direction, is attracted; the far side, with current in the opposite direction, is repelled with a smaller force

F_{1} = \frac{μ _{0} I _{1} I _{2} a}{2 π c}, F_{2} = \frac{μ _{0} I _{1} I _{2} a}{2 π ( c + b )}, F = F_{1} - F_{2} = \frac{μ _{0} I _{1} I _{2} a}{2 π} (\frac{1}{c} - \frac{1}{c + b})

F_{1} F_{2} F = \frac{2 \cdot 1 0 ^{- 7} \cdot 20 \cdot 5.0 \cdot 0.10}{0.020} N = 1.0 \cdot 1 0^{- 4} N = \frac{2 \cdot 1 0 ^{- 7} \cdot 20 \cdot 5.0 \cdot 0.10}{0.070} N \approx 2.9 \cdot 1 0^{- 5} N \approx (10.0 - 2.9) \cdot 1 0^{- 5} N \approx 7.1 \cdot 1 0^{- 5} N

The force on the loop is the sum of the forces on the four sides. The two sides parallel to the wire are each at a constant distance from the wire, so on each of them the field of the wire is uniform and the theorem on the force between two parallel wires applies, with the length $a$ of the side. In the near side the current is in the same direction as $I_{1}$ and the force $F_{1}$ is attractive; in the far side the current, which travels the loop in a single sense, is in the opposite direction, and the force $F_{2}$ is repulsive.

The two sides perpendicular to the wire carry opposite currents and are in the same field, which varies only with the distance from the wire; therefore on each element of one of them acts a force opposite to that on the corresponding element of the other, and their contributions cancel. The resultant force is therefore $F_{1} - F_{2} \approx 7.1 \cdot 1 0^{- 5} N$ , pointing toward the wire, because the attracted side is closer.