Two long, straight wires are parallel and 10 centimeters apart. One carries a current of 2.0 amps, the other a current of 5.0 amps. What is the magnitude of the force per unit length between the wires?
We can call the distance separating the wires, 10 centimeters, 𝑑. And we can call the current in the first wire, 2.0 amps, 𝐼 sub one and the current in the second wire, 5.0 amps, 𝐼 sub two. We want to solve for the force per unit length between the wires. We’ll call that 𝐹 over 𝐿. In a situation with two long, straight wires that are parallel to one another, with the wires both carrying current and separated from one another by a distance 𝑑, there’s a mathematical relationship that describes the force, it’s a magnetic force, between these parallel wires.
That magnetic force per unit length, 𝐹 sub 𝐵 divided by 𝐿, is equal to the permeability of free space times the product of the two currents all over two 𝜋 times the distance between the current-carrying wires 𝑑. 𝜇 naught, which is a constant, we’ll treat as exactly 1.26 times 10 to the negative sixth teslas meters per ampere. So the force between the wires per unit length is equal to 𝜇 naught times 𝐼 one times 𝐼 two over two 𝜋 𝑑. We know each of these values. And so we’re ready to plug in and solve for 𝐹 over 𝐿.
When we do plug in, we’re careful to use units of meters for our distance 𝑑. When we enter all these values on our calculator, we find 𝐹 over 𝐿 is 2.0 times 10 to the negative fifth newtons per meter. That’s the magnitude of the force per unit length between these current-carrying wires.