Two long wires carrying same currents in opposite directions are placed at separation D as shown.Predict variation of magnetic field as one moves from the point O and A

Explanation

Between the wires:
Magnetic field at `P:`
Due to `①`,`B_(1)=(mu_(0)i)/(2pix)`, upward
Due to `②`,`B_(2)=(mu_(0)i)/(2pi(d-x))`, upward
`B_(P)=B_(1)+B_(2)=(mu_(0)i)/(2pi) (1/x+1/(d-x))`
`=(mu_(0)i)/(2pi). d/(x(d-x))`, upward
`B_(P)` is minimum, if `x(d-x)` is maximum
`x(d-x)` is maximum if `x=d-x implies x=d//2`
[The product of two parts is maximum, if parts are equal]
As `x=d/2,B` is minimum, `(B_(P))_(min)=(2mu_(0)i)/(pid)`, upward
As `x to 0,B to oo`
`x to d,B to oo`
For `x=0 ,x=d//2`, magnetic field decreases reaching to minimum value at `x=d//2`.
From `x=d//2` to `x=d` , magnetic field increases and tends to infinite at `x=d`.
Left of `O`: At `x` distance from `O`.

At `P: B_(1)=(mu_(0)i)/(2pix)`, downward, `B_(2)=(mu_(0)i)/(2pi(x+d))`, upward
`B_(P)=(mu_(0)i)/(2pi)(1/x-1/(x+d))`, downward
`x to 0,B_(P) to oo`
`x to oo,B_(P) to 0`
Right of `A:`

At `P: B_(1)=(mu_(0)i)/(2pi(d+x))`, upward, `B_(1)=(mu_(0)i)/(2pix)`, downward
`B_(P)=(mu_(0)i)/(2pi)(1/x-1/(d+x))`, downward
`x to 0, B_(P) to oo`
`x to oo, B_(P) to 0`
`B` versus `x` graph: