Now consider Ampère’s experimental arrangement of two parallel wires of length l, distance d apart. FIGURE 29.45a shows the currents $I_1$ and $I_2$ in the same direction; FIGURE 29.45b shows the currents in opposite directions. We will assume that the wires are sufficiently long to allow us to use the earlier result for the magnetic field of a long, straight wire: $B={\mu }_{0}I\text{/}2\pi r$.

FIGURE 29.45 Magnetic forces between parallel current-carrying wires.

As Figure 29.45a shows, the current $I_2$ in the lower wire creates a magnetic field $\vec{B}{}_2$ at the position of the upper wire. $\vec{B}{}_2$ points out of the figure, perpendicular to current $I_1$. It is field $\vec{B}{}_2$, due to the lower wire, that exerts a magnetic force on the upper wire. Using the right-hand rule, you can see that the force on the upper wire is downward, thus attracting it toward the lower wire. The field of the lower current is not a uniform field, but it is the same at all points along the upper wire because the two wires are parallel. Consequently, we can use the field of a long, straight wire, with $r=d$, to determine the magnetic force exerted by the lower wire on the upper wire:

As an exercise, you should convince yourself that the current in the upper wire exerts an upward-directed magnetic force on the lower wire with exactly the same magnitude. You should also convince yourself, using the right-hand rule, that the forces are repulsive and tend to push the wires apart if the two currents are in opposite directions.

Thus two parallel wires exert equal but opposite forces on each other, as required by Newton’s third law. Parallel wires carrying currents in the same direction attract each other; parallel wires carrying currents in opposite directions repel each other.

EXAMPLE 29.14 A current balance

Two stiff, 50-cm-long, parallel wires are connected at the ends by metal springs. Each spring has an unstretched length of 5.0 cm and a spring constant of $0.025\text{N}\text{/}\text{m}$. The wires push each other apart when a current travels around the loop. How much current is required to stretch the springs to lengths of 6.0 cm?

MODEL Two parallel wires carrying currents in opposite directions exert repulsive magnetic forces on each other.

VISUALIZE FIGURE 29.46 shows the “circuit.” The springs are conductors, allowing a current to travel around the loop. In equilibrium, the repulsive magnetic forces between the wires are balanced by the restoring forces $F_{\text{Sp}}=k\text{Δ}y$ of the springs.

FIGURE 29.46 The current-carrying wires of Example 29.14.

FIGURE 29.46 Full Alternative Text

SOLVE Figure 29.46 shows the forces on the lower wire. The net force is zero, hence the magnetic force is $F_{\text{B}}=2F_{\text{Sp}}$. The force between the wires is given by Equation 29.27 with $I_1=I_2=I\text{:}$

$$F_{\text{B}}={\displaystyle \frac{{\mu }_{0}l{I}^2}{2\pi d}}=2F_{\text{Sp}}=2k\text{Δ}y$$

where k is the spring constant and $\text{Δ}y=1.0\text{cm}$ is the amount by which each spring stretches. Solving for the current, we find

$$I=\sqrt{{\displaystyle \frac{4\pi kd\text{Δ}y}{{\mu }_{0}l}}}=17\text{A}$$

REVIEW Devices in which a magnetic force balances a mechanical force are called current balances. They can be used to make very accurate current measurements.