Question

What is the magnitude of the force between two parallel wires 43.0 m long and 5.9 cm apart, each carrying 35.0 A in the same direction?

Answer 1

Answer:

The magnitude of the force between the two wires is $17855.9*10^{-5} N$

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=$\frac{u_{o}*I1 }{2*\pi*a }$ equation 1

B1: Magnetic field produced by conductor 1

 $u_{o}$  = 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=$I2*L*\frac{u_{o}*I1 }{2*\pi *a}$

F1-2= $\frac{u_{o}*I1*I2*L }{2*\pi *a}$

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

$u_{o} =4*\pi *10^{-7}$Wb/A.m

I1=I1=35A

L=43M

a  =5.9 cm=$5.9*10^{-2}$ m

$F=\frac{4*\pi *10^{-7}*35*35*43 }{2*\pi *5.9*10^{-2} }$

$F=17855.9*10^{-5} N$

Answer: The magnitude of the force between the two wires is $17855.9*10^{-5} N$