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Question

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:




A

μ0i22πl(^i^j)

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B

μ0i4πl(^i+^j)

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C

2μ0iπl(^i+^j)

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D

μ0i2πl(^i+^j)

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Solution

The correct option is C

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=μ0i22πl^j
Similarly due to FA:
BFA=μ0i22πl^iBPABC=μ0iπl[122+122+22]^iBFABC=2μ0iπl(^j)
Hence, similarly due to CDEF:
BCDEF=2μ0iπl(^i)Bnet=2μ0iπl(^i+^j)


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