A long straight wire, carrying current I, is bent at its midpoint to form an angle of 45o. The magnetic field at point P which is at distance R from the point of bending must be

For segment BC, perpendicular distance isr=Rcos⁡45∘B and C ends are on the same side of the perpendicular PN.Therefore, α becomes negative and β is positive.Hence, α=−45∘ and β=90∘UsingB=μ0I4πr(Sin⁡α+Sin⁡β)⇒B=(2−1)μ0I4πR(into the plane of paper)