Question

Two long, parallel, current-carrying wires lie in an xy-plane. The first wire lies on the line y = 0.340 m and carries a current of 27.5 A in the +x direction. The second wire lies along the x-axis. The wires exert attractive forces on each other, and the force per unit length on each wire is 295 µN/m. What is the y-value (in m) of the line in the xy-plane where the total magnetic field is zero?

Answer 1

Answer:

The y-value of the line in the xy-plane where the total magnetic field is zero  $U = 0.1355 \ m$

Explanation:

From the question we are told that

    The distance of wire one from two along the y-axis is    y = 0.340 m

   The current on the first wire is  $I_1 = (27.5i) A$

    The force per unit length on each wire is  $Z = 295 \mu N/m = 295*10^{-6} N/m$

Generally the force per unit length is mathematically represented as

         $Z = \frac{F}{l} = \frac{\mu_o I_1I_2}{2\pi y}$

=>      $\frac{\mu_o I_1I_2}{2\pi y} = 295$

Where  $\mu_o$ is the permeability of free space with a constant value of  $\mu_o = 4\pi *10^{-7} \ N/A2$

substituting values

       $\frac{ 4\pi *10^{-7} 27.5 * I_2}{2\pi * 0.340} = 295 *10^{-6}$

=>    $I_2 = 18.23 \ A$

Let U  denote the  line in the xy-plane where the total magnetic field is zero

So  

      So the force per unit length of  wire 2  from  line  U is equal to the force per unit length of wire 1  from  line  (y - U)      

   So  

         $\frac{\mu_o I_2 }{2 \pi U} = \frac{\mu_o I_1 }{2 \pi(y - U) }$

substituting values

          $\frac{ 18.23 }{ U} = \frac{ 27.5 }{(0.34 - U) }$

         $6.198 -18.23U = 27.5U$

          $6.198=45.73U$

          $U = 0.1355 \ m$