20 A, perpendicular out of the page
40 A, perpendicular out of the page
40 A, perpendicular into the page
20 A, perpendicular into the page
Magnetic field at 'O' will be due to 'PS' and 'QN' only
i.e B₀ = Bps + Bqn → Both inwards
Let current in each wire = i
∴B₀ = μ₀i/(4πd) + μ₀i/(4πd)
∴i = 20A