Given that,
Current =i
Angle =α
Magnetic field =B
Now, magnetic field at a point a from current carrying conductor making angle θ1 and θ2 with the ends of the wire is
B=μ0I4πa(sinθ1+sinθ2)
Now, a=xsinθ2
θ1=θ2
θ2=900
Now, the magnetic field
Blower=μ0I4π⎡⎢ ⎢ ⎢ ⎢⎣1+sin(θ2)xsin(θ2)⎤⎥ ⎥ ⎥ ⎥⎦
And Bupper is also same
No, if the current was flowing anticlockwise both these field would point upward outside the page
Now, net magnetic field
B=Blower+Bupper
B=2×μ0I4π⎡⎢ ⎢ ⎢ ⎢⎣1+sin(θ2)xsin(θ2)⎤⎥ ⎥ ⎥ ⎥⎦
B=μ0I2πx⎡⎢ ⎢ ⎢ ⎢⎣1+sin(θ2)sinθ2⎤⎥ ⎥ ⎥ ⎥⎦
B=μ0I2πx⎡⎢ ⎢ ⎢ ⎢ ⎢⎣{sin(θ4)+cos(θ4)}22sin(θ4)cos(θ4)⎤⎥ ⎥ ⎥ ⎥ ⎥⎦
Now, dividing numerator and denominator by cos2θ4
B=μ0I4πx⎡⎢ ⎢ ⎢ ⎢ ⎢⎣(1+tan(θ4))2tanθ4⎤⎥ ⎥ ⎥ ⎥ ⎥⎦
B=μ0I4πx[(1+tan(θ4))2]cot(θ4)
Now, the magnetic field is
B=Kcot(θ4)
Where, K=μ0I4πx[{1+tan(θ4)}2]
Hence, the magnetic field is B=Kcot(θ4)