Answer:
The magnitude of the force between the two wires is [tex]17855.9*10^{-5} N[/tex]
Explanation:
If there are two parallel rectilinear conductors through which two electric currents of the same direction I1 and I2 circulate,both conductors will generate a magnetic field on each other, giving rise to a force between them.
To calculate the value of this force, first, according to the law of Biot and Savart, the magnetic field produced by conductor 1 over 2 is obtained, which will be given by the equation:
B1=[tex]\frac{u_{o}*I1 }{2*\pi*a }[/tex] equation 1
B1: Magnetic field produced by conductor 1
[tex]u_{o}[/tex] = free space permeability
a= distance between wires
I1= current carrying wire 1
This magnetic field exerts on a segment L of the conductor 2 through which a current of intensity I2 circulates, a force equal to:
F1-2= I2*L*B1 Equation2
We replaced B1 of the equation 1 in the equation 2:
F1-2=[tex]I2*L*\frac{u_{o}*I1 }{2*\pi *a}[/tex]
F1-2= [tex]\frac{u_{o}*I1*I2*L }{2*\pi *a}[/tex]
If we calculate the force exerted by conductor 2 on conductor 1 we would arrive at exactly the same value:
F2-1= F1-2
For this problem, the magnitude of the force between the two parallel cables that conduct current in the same direction is:
F1-2=F2-1=F
[tex]u_{o} =4*\pi *10^{-7}[/tex]Wb/A.m
I1=I1=35A
L=43M
a =5.9 cm=[tex]5.9*10^{-2}[/tex] m
[tex]F=\frac{4*\pi *10^{-7}*35*35*43 }{2*\pi *5.9*10^{-2} }[/tex]
[tex]F=17855.9*10^{-5} N[/tex]
Answer: The magnitude of the force between the two wires is [tex]17855.9*10^{-5} N[/tex]
