Correct Answer - A
According to Biot savart Law, the magnetic induction at a point to a current carrying elemnet i `delta` l is given by
`deltaB=(mu_(0))/(4pi)(ideltalxxr)/(r^(3))or B=(mu_(0))/(4pi)(ideltalsintheta)/(r^(2))`
Directed normal to plane containing `deltal` and
r theta being angle between `delta`l and r
r `theta` being angle between `delta` l and r
Magnetic field due to linear portions of wire
If we consider anuy element of lenth `deltal` of either linear portion ab or de, the angle between `delta` l and r is 0 or `pi` therefore.
`sintheta=0`
Hence, magnetic induction due to both linear portins of wire is
`(mu_(0))/(4pi)(ideltalsintheta)/(r)=0`
Field due to semi circular are
Now angle between a current element `delta` l of semi circular arc and the radius vector of the elment to point c is `pi//2`
Therefoe the magnitude of magnetic induction B at O due to this element is
`deltaB=(mu_(0))/(4pi)(ideltalsinpi//2)/(r^(2))=(mu_(0)ideltal)/(4pir^(2))`
Hence magnetic indution due to whole semi circular loop is `B=Sigma(mu_(0))/(4pi)(ideltal)/(r^(2))=(mu_(0))/(4pir^(2))=(mu_(0))/(4pi)(i)/(i^(2))Sigmadeltai=(mu_(0)i)/(4piR^(2))(pir)=(mu_(0)i)/(4r)`
