In the figure shown ABCDEFA was a square loop of side , but is folded in two equal parts so that half of it lies in XZ plane and the other half lies in the YZ plane. The origin O is centre of the frame also. The loop carries current. The magnetic field at the centre is:
√2μ0iπl(^i+^j)
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 FABC :
due to BC:
→BBC=μ0i4πl(l2)(sin00+sin450)^j=μ0i2√2πl^j
Similarly due to FA:
→BFA=μ0i2√2πl^i→BPABC=μ0iπl[12√2+12√2+√22]^i⇒→BFABC=√2μ0iπl(^j)
Hence, similarly due to CDEF:
→BCDEF=√2μ0iπl(^i)⇒→Bnet=√2μ0iπl(^i+^j)