Two long parallel wires are separated by 6.0 mm. The current in one of the wires is twice the other current. If the magnitude of the force on a 3.0-m length of one of the wires is equal to 8.0 μN, what is the greater of the two currents?
A) 0.20 A
B) 0.40 A
C) 40 mA
D) 20 mA
E) 0.63 A

Question

Quizplus · Accepted Answer

The force between two parallel currents is given by the formula $F = \frac{{\mu_0 I_1 I_2 L}}{{2\pi d}}$, where $F$ is the force, $\mu_0$ is the permeability of free space ($4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}$), $I_1$ and $I_2$ are the currents in the wires, $L$ is the length of the wires, and $d$ is the distance between the wires. Given that one current is twice the other ($I_2 = 2I_1$), the force can be rewritten as $F = \frac{{\mu_0 I_1 (2I_1) L}}{{2\pi d}}$. Solving for $I_1$ (the smaller current) and then doubling it to find the greater current ($I_2$) leads to the answer. Given $F = 8.0 \times 10^{-6}$ N, $L = 3.0$ m, and $d = 6.0 \times 10^{-3}$ m, solving the equation for $I_1$ and then finding $I_2 = 2I_1$ yields $I_2 \approx 0.40$ A.

Two Long Parallel Wires Are Separated by 6