A current loop `ABCD` is held fixed on the plane of the paper as shown in figure. The arcs `BC( radius = b) and DA ( radius = a)` of the loop are joined by two straight wires `AB and CD` at the origin `O` is 30^(@)`. Another straight thin wire with steady current `I_(1)` flowing out of the plane of the paper is kept at the origin .

Due to the process of the current `I_(1)` at the origin:

Options

A. 0
B. `(mu_0 (b-a) i)/(24 ab)`
C. `(mu_0 I)/(4pi) [ (b-a)/(ab) ]`
D. `(mu_0 I)/(4pi ) [ (2(b-a) + pi/3 (a + b) ]`

Answer

C

Explanation

Net magnetic field due to loop ABCD at O is B ` = B_(AB) + B_(BC) + B_(CD) + B_(DA)`
`B = 0 + (mu_0 I)/(4pi a) xx pi/6 + 0 - (mu_0 I)/(4pi b) xx pi/6`