1. B1 into the plane and B2 out of the plane.
2. B1 is the field of the ring and B2 the field of due to straight conductor.
B = B1 – B2 = 0
B1 = B2
\(\frac{\mu_0}{2}\frac{1}{a}=\frac{\mu_0}{2x}\frac{1}{d}\)
\(\frac{a}{2a}\) = \(\frac{2}{2\times \pi\times 20\times10^-2}\)
2a = 40π × 10-2m = 1,256m
Diameter d = 1.256 m.