Respuesta :
Answer:
[tex]d=1.68\times 10^{-3}\ m[/tex]
Explanation:
It is given that,
Length of the wire, l = 2.4 m
Force per unit length of the wire, [tex]\dfrac{F}{l}=0.127\ N/m[/tex]
Current in top wire, [tex]I_1=29\ A[/tex]
Current in bottom wire, [tex]I_2=37\ A[/tex]
The permeability of free space is, [tex]\mu_o=1.25664\times 10^{-6}\ N/m^2[/tex]
Let d is the distance of separation between the wires so that the top wire will be held in place by magnetic repulsion. The magnetic force per unit length is given by :
[tex]\dfrac{F}{L}=\dfrac{\mu_o I_1I_2}{2\pi d}[/tex]
[tex]d=\dfrac{\mu_o I_1I_2}{2\pi F/L}[/tex]
[tex]d=\dfrac{1.25664\times 10^{-6}\times 29\times 37}{2\pi \times 0.127}[/tex]
[tex]d=1.68\times 10^{-3}\ m[/tex]
So, the distance of separation between the wires is [tex]1.68\times 10^{-3}\ m[/tex]. Hence, this is the required solution.
