In figure, ABCDEFA was a square loop of side /, but is folded in two equal parts so that half of it lies in the x-z plane and the other half lies in the y-z plane. The origin O is center of the frame . The loop carries current ‘i’. The magnetic field at the center is

Due to FABC, the magnetic field at O is along y-axis and due to CDEF, the magnetic field is along x axis. Hence, the field will be of the form A[i^+j^] Calculating field due to fulBC: Due to AB: B→AB=μ0i4πl2sin45°+sin45°j^=2μ0i2πlj^ Due to BC: B→AB=μ0i4πl2sin0°+sin45°j^=μ0i22πlj^ B→FA=μ0i22πlj^  Hence, B→FABC=μ0iπl122+122+22i^ ⇒  E→FABC=2μ0iπl(j^)  Similarly due to CDEF:B→CDEF=2μ0lπli^ ⇒  B→net =2μ0iπl(i^+j^)