Answer:
The magnitude of force per unit length of one wire on the other is [tex]2.7945\times 10^{-5}\ N[/tex] and the direction is away from one another
The magnitude of force per unit length of one wire on the other is [tex]2.7945\times 10^{-5}\ N[/tex] and the direction is towards each other.
Explanation:
[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi\times 10^{-7}\ N/A^2[/tex]
[tex]i_1[/tex] = Current in first wire = 2.9 A
[tex]i_2[/tex] = Current in second wire = 5.3 A
r = Gap between the wires = 11 cm
Force per unit length
[tex]F_{12}=F_{21}=\frac{\mu_0 i_1i_2}{2\pi r}\\ =\frac{4\pi\times 10^{-7}\times 2.9\times 5.3}{2\pi 0.11}\\ =2.7945\times 10^{-5}\ N[/tex]
The magnitude of force per unit length of one wire on the other is [tex]2.7945\times 10^{-5}\ N[/tex] and the direction is away from one another
[tex]F_{12}=F_{21}=\frac{\mu_0 i_1i_2}{2\pi r}\\ =\frac{4\pi\times 10^{-7}\times 2.9\times 5.3}{2\pi 0.11}\\ =2.7945\times 10^{-5}\ N[/tex]
The magnitude of force per unit length of one wire on the other is [tex]2.7945\times 10^{-5}\ N[/tex] and the direction is towards each other.
