Given:
Current in the equal wires: i= 10 A
Distance between the two wires: a = 2 cm = 0.02 m
Distance between the wire and the point : d = 2 cm = 0.02 m

From the diagram l₁ and l₂ are the wires coming out of the plane.
These two wires and point P form an equilateral triangle of side 0.02 m.
By right hand rule, Magnetic field at P due to l₁ is shown by arrow B₁ tangent to the field’s path and Magnetic field at P due to l₂ is shown by B₂ tangent to the field’s path.
By geometry we can calculate the angle between B₁ and B₂ which turns out to be 60° .
Formula used:
By Ampere’s Law for a current carrying wire is
Where,

B is the magnitude of magnetic field,

μ₀ is the permeability of free space, μ₀= 4π × 10^-7 T mA^-1

d is the distance between the current carrying wire and the required point.
Magnetic field due to wire l₁:

∴ B₁ = 1 × 10^-4 T
Magnetic Field due to l₂ :
∴ B₂ = 1 × 10^-4 T
B₁=B₂ as same magnitude of current is flowing through both the wires and point P is located at same distance from each wires.
Now, angle between and : θ = 60°
Resultant of two vectors is given as
B_R is the resultant magnetic field at P.

∴ B_R = 1.73 × 10^-4 T
Hence, resultant magnetic field at a point 2 cm away from due to two current carrying wires is 1.73 × 10^-4 T.