A square loop `ABCD` of side `'a'` carrying current `I` is folded about an axis passing through its centre (`y`-axis). The two halves are inclined at angle `30^(@)` with `xy` plane as shown. Find the magnetic field at origin. The origin coincides with the centre of loop.

Options

A. `(mu_(0)Isqrt(3))/(sqrt(2)pia)`
B. `(mu_(0)Isqrt(3))/(2sqrt(2)pia)`
C. `(3mu_(0)I)/(2sqrt(2)pia)`
D. `(sqrt(6) mu_(0)I)/(pia)`

Answer

D

Explanation

`B_("net")=2(B_(half))cos theta`

Where `B_("half")=1/2[(4mu_(0))/(4pi) I/((a//2))] (cos 45^(@)+cos45^(@))`
`=(mu_(0)Isqrt(2))/(pia)`
So, `B=2((mu_(0)Isqrt(2))/(pia)) (sqrt(3))/2=(sqrt(6)mu_(0)I)/(pi a)`