Correct Answer - B
(b)
Let us compute the magnetic field due to any one segment:
`B=(mu_0i)/(4pi(d sin alpha)) (cos0^@+cos(180-alpha))`
`=(mu_0I)/(4pi(d sin alpha))(1-cos alpha)=(mu_0I)/(4pid) tan (alpha)/2`
Resultant field will be
`B_(n et)=2B=(mu_0I)/(2pid)tan (alpha)/2 implies K=(mu_0I)/(2pid)`