Answer:

F/L =  8*10^-4 N/m

Explanation:

To calculate the magnitude of the force per meter in the central wire, you take into account the contribution to the force of the others two wires:

[tex]F_N=F_{1,2}+F_{2,3}[/tex]   (1)

F1,2 : force between first and second wire

F2,3 : force between second and third wire

The force per meter between two wires of the same length is given by:

[tex]\frac{F}{L}=\frac{\mu_oI_1I_2}{2\pi r}[/tex]

μo: magnetic permeability of vacuum =  4pi*10^-7 T/A

r: distance between wires

Then, you have in the equation (1):

[tex]\frac{F_N}{L}=\frac{\mu_oI_1I_2}{2\pi r}+\frac{\mu_oI_2I_3}{2\pi r}\\\\\frac{F_N}{L}=\frac{\mu_oI_1}{2\pi r}[I_2+I_3][/tex]

But

I1 = I2 = I3 = 10A

r = 10cm = 0.1m

You replace the values of the currents and the distance r and you obtain:

[tex]\frac{F_N}{L}=\frac{\mu_oI^2}{\pi r}\\\\\frac{F_N}{L}=\frac{(4\pi*10^{-7}T/A)(20A)^2}{2\pi (0.1m)}=8*10^{-4}\frac{N}{m}[/tex]

hence, the net force per meter is 8*10^-4 N/m