Answer:
[tex]\dfrac{F}{l}=2\times 10^{-7}\ N/m[/tex]
Explanation:
Given,
current in the wire I₁ = 1 A
I₂ = 1 A
distance between them, r = 1 m
using Force per unit length formula
[tex]\dfrac{F}{l}=\dfrac{\mu_0I_1I_2}{2\pi r}[/tex]
[tex]\mu_0 = magnetic\ permeability\ of\ free\ space = 4\pi\times 10^{-7}[/tex]
[tex]\dfrac{F}{l}=\dfrac{4\pi \times10^{-7}\times 1\times 1}{2\pi\times 1}[/tex]
[tex]\dfrac{F}{l}=2\times 10^{-7}\ N/m[/tex]
Hence, the magnitude of force per unit length is equal to [tex]\dfrac{F}{l}=2\times 10^{-7}\ N/m[/tex]