How Do You Calculate the Force Between Two Parallel Current-Carrying Wires?

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Multiply by the parallel length $L$ if the question asks for the total force: $F=\frac{\mu_0 I_1 I_2L}{2\pi r}$.

The Reasoning

This relationship belongs to D.3 Motion in electromagnetic fields. Here, $I_1$ and $I_2$ are the currents, $r$ is the perpendicular distance between the centres of the wires, and $\mu_0=4\pi\times10^{-7}\ \text{T m A}^{-1}$ is the permeability of free space.

Wire 1 produces a magnetic field at wire 2:

A wire of length $L$, carrying current $I_2$ perpendicular to this field, experiences $F=B_1I_2L$. Substituting for $B_1$ gives the required equation.

The direction of the magnetic field around each wire is found using the right-hand grip rule. The field from wire 1 acts on moving charge carriers in wire 2, producing a force perpendicular to both the current and field. Applying the hand rule to both wires shows that equal-direction currents pull the wires together, whereas opposite-direction currents push them apart. The forces form a Newton's third-law pair: they are equal in magnitude and opposite in direction.

Worked Example

Two parallel wires carry currents of $3.0\ \text{A}$ and $5.0\ \text{A}$. Their separation is $0.20\ \text{m}$. For a parallel length of $2.0\ \text{m}$:

As a useful intermediate check, the force per unit length is $1.5\times10^{-5}\ \text{N m}^{-1}$. Multiplying by the parallel length of $2.0\ \text{m}$ produces the stated total force.

If the currents have the same direction, this force is attractive; if they have opposite directions, it is repulsive.

A common misconception is to report (F/L) as the total force. Always check whether the question asks for force per unit length, measured in $\text{N m}^{-1}$, or force, measured in $\text{N}$.

Exam Technique

For a Calculate question, show the equation, substitution, and final value. Use the centre-to-centre separation, convert all quantities to SI units, and state attraction or repulsion when the current directions are provided.