Forces Between Parallel Conductors

Revisiting the right-hand grip rule before we add a second wire

Two current-carrying wires exert forces on each other. To explain why, begin with the field made by one wire. For a straight wire viewed end-on, point your right thumb with conventional current; your curled fingers show the circular magnetic-field direction.

The magnetic field strength at a distance $r$ from a long straight wire carrying current $I$ is:

The field decreases with distance, strongest near the wire, falling off as $1/r$. This equation is the key step in deriving the force between two parallel wires.

Right-hand grip rule: thumb = current direction, fingers curl = $\vec{B}$ direction. For an effectively long straight wire, $B=\mu_0I/(2\pi r)$, so $B\propto1/r$. Use $\mu_0\approx4\pi\times10^{-7}$ T m A$^{-1}$; it is measured, not exact, in the current SI.

Pause, copy the highlighted grip rule and field formula into your book before moving on.

Each wire sits in the magnetic field produced by the other

We just saw that a current-carrying wire creates a circular magnetic field $B = \mu_0 I/2\pi r$ around itself. That raises a question: if a second current-carrying wire sits in that field, does it feel a force? This card answers it → yes, via the motor effect; same-direction currents attract, opposite-direction repel.

Consider two effectively long, straight, parallel wires separated by perpendicular distance $d$, carrying currents $I_1$ and $I_2$. Wire 1 produces $B_1=\mu_0I_1/(2\pi d)$ at Wire 2. Since the field is perpendicular to Wire 2, $F_2=B_1I_2l$. Dividing by the common interacting length gives the pair-force model below.

Force per unit length: $F/l = \mu_0 I_1 I_2 / (2\pi d)$ (N m$^{-1}$). Same-direction currents attract; opposite-direction currents repel. $F/l \propto I_1 I_2$ and $\propto 1/d$. Reversing both currents leaves magnitude and direction unchanged.

Add the highlighted parallel-wire force formula and direction rule to your notes before the check below.

The wire-force standard applied before 20 May 2019

We just saw how to calculate the force per unit length between ideal parallel wires. That raises a question: how does that familiar $2\times10^{-7}$ N m$^{-1}$ result relate to today's ampere? This card answers it by separating the pre-2019 mechanical definition from the current fixed-charge definition.

Before 20 May 2019, the SI defined the ampere through the idealised force between parallel conductors. The historical wording was:

For school calculations, $\mu_0\approx4\pi\times10^{-7}$ N A$^{-2}$ remains an excellent approximation. Substituting $I_1=I_2=1$ A and $d=1$ m therefore gives $F/l\approx2\times10^{-7}$ N m$^{-1}$. In the current SI, write approximately, not exactly.

Historical pre-2019 definition: two ideal parallel wires carrying 1 A at 1 m separation produced $2\times10^{-7}$ N/m. Current SI: the ampere follows from the exact elementary charge $e=1.602\,176\,634\times10^{-19}$ C, while $\mu_0$ is measured. Using its approximate value still reproduces $F/l\approx2\times10^{-7}$ N/m for the ideal wire example.

Pause, write the highlighted ampere definition and verification into your book before moving on.

Turn the force law into a graph whose gradient has physical meaning

The ideal equation assumes conductors are much longer than their separation, straight, parallel, thin and in vacuum. Real ends bend the field, wires heat and expand, current can drift, the supports add restoring forces, and the wires may move so $d$ changes. A non-zero intercept suggests a zero offset or background force; scatter larger than the uncertainty bars suggests uncontrolled variation.

Linear test: plot $y=F/l$ against $x=I_1I_2/d$. The model $y=[\mu_0/(2\pi)]x$ predicts a straight line through the origin. The gradient estimates $\mu_0/(2\pi)$; compare its uncertainty interval with the accepted measured value rather than demanding exact agreement.