Two long parallel wires are at a distance 2d apart. They carry steady equal current flowing out of the plane of the paper. The variation of the magnetic field along the line XX' is given by

We know, the magnetic field due to a long current carrying wire is given as B = μ₀I/(2πr), where μ₀ is the permeability of free space, I is the current, and r is the distance from the wire.

Let the two wires be at x = -d and x = d. The magnetic field due to the wire at x = -d at a point x on the line XX' is given by:
B₁ = μ₀I/(2π(x+d))
The direction of this field will be upwards (using the right-hand rule).

The magnetic field due to the wire at x = d at a point x on the line XX' is given by:
B₂ = μ₀I/(2π(d-x))
The direction of this field will be downwards.

The net magnetic field B at a point x on XX' is given by the vector sum of B₁ and B₂.
B = B₁ - B₂ = μ₀I/(2π(x+d)) - μ₀I/(2π(d-x))
B = (μ₀I/(2π)) * [(1/(x+d)) - (1/(d-x))]
B = (μ₀I/(2π)) * [(d-x - (x+d))/((x+d)(d-x))]
B = (μ₀I/(2π)) * [(-2x)/(d² - x²)]
B = (-μ₀Ix/(π(d² - x²)))

At x = 0, B = 0.
At x = d, B tends to -∞.
At x = -d, B tends to +∞.
For x > 0, B is negative (downwards).
For x < 0, B is positive (upwards).
The graph will show that the magnetic field is zero at the midpoint (x=0), increases to infinity as it approaches either wire, and is negative between the wires and positive outside the wires. The graph will be asymptotic to x = d and x = -d.