A force per unit length of 48 $\mu$ N/m is produced when two wires separated by 1.5 mm both carry the same current. The current the wires carry is
A) 0.36 A.
B) 0.60 A.
C) 0.77 A.
D) 1.0 A.

Question

Quizplus · Accepted Answer

The force per unit length between two parallel wires carrying currents $I_1$ and $I_2$ is given by the formula $\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$, where $d$ is the distance between the wires, $L$ is the length of the wires, and $\mu_0$ is the permeability of free space ($4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}$). Given that the force per unit length is $48 \, \text{N/m}$, the distance $d = 1.5 \, \text{mm} = 1.5 \times 10^{-3} \, \text{m}$, and that both wires carry the same current ($I_1 = I_2 = I$), we can rearrange the formula to solve for $I$:$$48 = \frac{(4\pi \times 10^{-7}) I^2}{2\pi (1.5 \times 10^{-3})}$$Solving for $I$ gives $I \approx 0.60 \, \text{A}$.

A Force Per Unit Length of 48 μ\muμ N/m Is Produced When Two Wires Separated by 1

A Force Per Unit Length of 48 $\mu$ N/m Is Produced When Two Wires Separated by 1