In Fig ABCDEFA was a square loop of side l, 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 centre of the frame also. The loop carries current 'i'. The magnetic field at the centre is

Options

A. `(mu_0i)/(2sqrt2pil) (hati-hatj)`
B. `(mu_0i)/(4pil) (-hati-hatj)`
C. `(sqrt2mu_0i)/(pil) (hati+hatj)`
D. `(mu_0i)/(sqrt2pil) (hati+hatj)`

Answer

C

Explanation

(c) Due to FABC, the magnetic field at O is alpng y-axis and
due to CDEF, the magnetic field is along x-axis.

Hence, the field will be of the form `A[hati+hatj]`
Calculating field due to FABC:
Due to AB:
`vecB_(AB)=(mu_0i)/(4pi(l/2)) (sin45^@+sin45^@)hatj=sqrt2(mu_0i)/(2pil) hatj`
Due to BC:
`vecB_(AB)=(mu_0i)/(4pi(l/2)) (sin0^@+sin45^@)hatj=(mu_0i)/(2sqrt2pil) hatj`
`vecB_(FA)=(mu_0i)/(2sqrt2pil) hatj`
Hence, `vecB_(FABC)=(mu_0i)/(pil)[1/(2sqrt2)+1/(2sqrt2)+(sqrt2)/2] hati`
`implies vecE_(FABC)=(sqrt2mu_0i)/(pil)(hatj)`
Similarly due to CDEF: `vecB_(CDEF)=(sqrt2mu_0i)/(pil) hati`
`implies vecB_("net")=(sqrt2mu_0i)/(pil) (hati+hatj)`