Question

Two long parallel wires carry currents of 10 A in opposite directions. They are separated by 40 cm. What is the magnetic field in the plane of the wires at a point that is 20 cm from one wire and 60 cm from the other

Answer 1

Answer:

The value is  $B = 6.67 *10^{-6} \ T$

Explanation:

From the question we are told that  

     The current on first  wire is $I_1 = 10 \ A$

     The current on the second wire is $I_2= -10 \ A$

     The distance of separation is $d = 40 \ cm = 0.4 \ m$

     The distance from the first wire considered is a =  20 cm  =  0.2 m

      The distance from the first wire considered is b = 60 cm  =  0.60 m

Generally the magnetic field in the plane of the wires is mathematically represented as

         $B = B_1 + B_2$

The reason for the positive sign is because the current are moving in opposite sides

Here  

      $B_1 = \frac{\mu_ o * I_1 }{2 * \pi * a }$

and

      $B_2 = \frac{\mu_ o * I_2}{2 * \pi * b }$

So

      $B = \frac{\mu_ o * I_1 }{2 * \pi * a } + \frac{\mu_ o * I_2}{2 * \pi * b }$

=>  $B = \frac{\mu_ o * 10 }{2 * \pi * a } + \frac{\mu_ o * (-10)}{2 * \pi * b }$

Here $\mu_o$ is the permeability of free space with value $\mu_o = 4\pi * 10^{-7} N/A^2$

So

=>  $B = \frac{ 4\pi * 10^{-7}* 10 }{2 *3.142 } [\frac{1}{0.2} - \frac{1}{0.6} ]$

=>  $B = 6.67 *10^{-6} \ T$