Question

Two parallel wires separated by a distance of 0.6 m each carry current in the same direction. One wire is carrying a current of 4 A and the other is carrying a current of 9 A. The magnitude of the magnetic field exactly in between the two wires is ____ T.

Answer 1

Answer:

The value is  $B = 3.33 *10^{-6} \ T$

Explanation:

From the question we are told that

  The distance of separation is  $d = 0.6 \ m$

  The current on the one wire is $I_1 = 9 \ A$

  The current on the second wire is $I_2 = 4 \ A$

Generally the magnitude of the field exerted between the current carrying wire is

        $B = B_1 - B_2$

Here $B_1$ is the magnetic field due to the first wire which is mathematically represented as

         $B_1 = \frac{\mu_o * I_1 }{2 \pi * d_1}$

Here $d_1$ is the distance to the half way point of the separation and the value is  

    $d_1 = 0.3 \ m$

$B_2$ is the magnetic field due to the first wire which is mathematically represented as

         $B_2 = \frac{\mu_o * I_2 }{2 \pi * d_2}$

Here $d_2$ is the distance to the half way point of the separation and the value is  

    $d_2 = 0.3 \ m$  

This means that $d_1 = d_2 = a = 0.3$

So

     $B = \frac{\mu_o * I_1 }{2 \pi * d_1} - \frac{\mu_o * I_2 }{2 \pi * d_2}$

=>  $B = \frac{\mu_o * (I_1 - I_2)}{2 \pi *0.3 }$

=>  $B = \frac{ 4\pi * 10^{-7} * (9- 4)}{2 * 3.142 *0.3 }$

=>  $B = 3.33 *10^{-6} \ T$