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. `(mu_0I)/(4pi)[(b-a)/(ab)]`
B. `(mu_0I)/(4pi)[(2(b-a)+pi/3(a+b)]`
C. `(mu_0I(b-a))/(24ab)`
D. `zero`

Answer

C

Explanation

The point O lies along the lines AB as well as CD. `therefore` They do not produce any magnetic induction at Q. The field produced by the wire DA at O is out of the plane of paper (+ve) and that produced by the wire BC at O is into the plane of the paper (negative). `therefore` The resultant feild at O due to the loop ABCD is `B=B_(AB)+B_(BC)+B_(CD)+B_(DA)` Since Bc and AD, subtend on angle `theta'=pi/6 at O,therefore B=0-(mu_0I)/(4piB) xx pi/6 +0+(mu_0I)/(4pia) xx pi/6 =(mu_0I)/(24)[1/a-1/b]=(mu_0I(b-a))/(24ab)`.