The magnetic field at center ‘O’ of the arc in figure is
μoI4π×r2+π
μoI2πrπ4+2−1
μ04π×Ir2−π
μ04π×Ir2+π4
Magnetic field due to straight wires and quarter circle is given by
Bcircular=14μ0I2r=μ0I8r Bstraight=2μ0I4π(rcos450)sin900−sin450 =2μ0I2πr1−12 Hence, Bnet = μ0I8r+2μ0I2πr1−12 = μoI2πrπ4+2−1