Answer:
The value is  [tex]B = 6.67 *10^{-6} \ T[/tex]
Explanation:
From the question we are told that Â
   The current on first  wire is [tex]I_1 = 10 \ A[/tex]
   The current on the second wire is [tex]I_2= -10 \ A[/tex]
   The distance of separation is [tex]d = 40 \ cm = 0.4 \ m[/tex]
   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
     [tex]B = B_1 + B_2[/tex]
The reason for the positive sign is because the current are moving in opposite sides
Here Â
   [tex]B_1 = \frac{\mu_ o * I_1 }{2 * \pi * a }[/tex]
and
   [tex]B_2 = \frac{\mu_ o * I_2}{2 * \pi * b }[/tex]
So
   [tex]B = \frac{\mu_ o * I_1 }{2 * \pi * a } + \frac{\mu_ o * I_2}{2 * \pi * b }[/tex]
=> Â [tex]B = \frac{\mu_ o * 10 }{2 * \pi * a } + \frac{\mu_ o * (-10)}{2 * \pi * b }[/tex]
Here [tex]\mu_o[/tex] is the permeability of free space with value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
So
=> Â [tex]B = \frac{ 4\pi * 10^{-7}* 10 }{2 *3.142 } [\frac{1}{0.2} - \frac{1}{0.6} ][/tex]
=> Â [tex]B = 6.67 *10^{-6} \ T[/tex]
