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 .

The magnitude of the magnetic field (B) due to the loop `ABCD` at the origin (o) is :

Options

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

Answer

B

Explanation

Net magnetic field due to loop ABCD at O is
`B=B_(AB)+B_(BC)+B_(CD)+B_(OA)=0+(mu_(0)I)/(4pia)xx(pi)/6+0-(mu_(0)I)/(pib)xx(pi)/6=(mu_(0)I)/(24a)=(mu_(0)I)/(24b)=(mu_(0)I)/(24ab)(b-a)`