Answer
given,
force per unit length = 350 µN/m
current, I = 22.5 A
y = y = 0.420 m
[tex]\dfrac{F}{L}= \dfrac{KI_1I_2}{d}[/tex]
[tex]I_2 = \dfrac{F}{L}\dfrac{d}{KI_1}[/tex]
[tex]I_2 = 350\times 10^{-6}\times \dfrac{0.42}{2 \times 10^{-7}\times 22.5}[/tex]
I₂ = 32.67 A
distance where the magnetic field is zero
[tex]\dfrac{4\pi \times 10^{-7}\times 32.67}{2\pi y_1}=\dfrac{4\pi \times 10^{-7}\times 22.5}{2\pi (0.42-y_1)}[/tex]
[tex]y_1 = 0.248\ m[/tex]
there the distance at which the magnetic field is zero in the two wire is at 0.248 m.