Answer:
[tex]B_{net}[/tex] =[tex](2\times 10^{-5}T/A).(I-25A)[/tex]
Explanation:
Given :
the distance between the two wires is = 2 cm
= 0.02 m
Current in one of the wire, I = 25 A
Now we know that, when the direction of the current is same that is in one direction , the magnetic field at the mid point in between the two wires will oppose each other.
Therefore we know that
[tex]B_{net}[/tex] = [tex]\frac{\mu _{o}\times I_{1}}{2\pi r_{1}}-\frac{\mu _{o}\times I_{2}}{2\pi r_{2}}[/tex]
=[tex]\frac{(4\pi \times 10^{-7}T.m/A)}{2\pi (1\times 10^{-2}m)}\times (I-25A)[/tex]
= [tex](2\times 10^{-5}T/A).(I-25A)[/tex]
