Question

A straight, how wire carries a current of 20 A. Another wire carrying equal current is placed parallel to it. If the force acting on a length of 10 cm of the second wire is 2.0 × 10⁻⁵ N, what is the separation between them?

Answer 1

Given:
Magnitude of current in both wires, i₁ = i₂ = 20 A
Force acting on 0.1 m of the second wire, F = 2.0 × 10⁻⁵ N
∴ Force per unit length = $\frac{2\times {10}^{-5}}{0.1}$ = 2.0 × 10⁻⁴ N/m
Now,
Let the separation between the two wires be d.
Thus, the force per unit length is given by
$$\begin{gathered} \frac{F}{l}=\frac{{\mathrm{\mu }}_0{i}_1{i}_2}{2\mathrm{\pi }d} \\ \Rightarrow 2.0\times {10}^{-4}=\frac{2\times {10}^{-7}\times 20\times 20}{d} \\ \Rightarrow d=0.4\mathrm{m}=40\mathrm{cm} \end{gathered}$$