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

Answer

A

Explanation

The magnetic field at `O` due to current in `DA` is
`B_(1) = ( mu_(0))/( 4 pi) (I)/(a) xx ( pi)/( 6)` ( directed vertically upwards)
The magnetic field at `O` due to current in `BC` is
`B_(2) = ( mu_(0))/( 4 pi) (I)/(b) xx ( pi)/( 6)` ( directed vertically downwards)
The magnetic field due to current `AB and CD` at `O` is zero .
Therefore the net magnetic field is
`B = B_(1) - B_(2)` ( directed vertically upwards)
( mu_(0))/( 4 pi) (I)/(a) xx ( pi)/( 6) - ( mu_(0))/( 4 pi) (I)/(b) xx ( pi)/( 6) = (mu_(0)I)/( 24) ((1)/(a) - (1)/(b)) = (mu_(0)I)/( 24ab) (b-a)`