Let the given circular $A B C$ part of wire subtends an angle $\theta$ at its centre. Then, magnetic field due to this circular part is
$B'=B_{c} \times \frac{\theta}{2 \pi}=\frac{\mu_{0}}{4 \pi} \times \frac{2 \pi i}{e} \times \frac{\theta}{2 \pi} $
$\Rightarrow \, B'=\frac{\mu_{0}}{4 \pi} \cdot \frac{i}{r} \theta $
Given, $i=40 \, A, \, r=3.14 \,cm =3.14 \times 10^{-2}\, m$
$\theta =360^{\circ}-90^{\circ}=270^{\circ}=\frac{3 \pi}{2} \,rad $
$\therefore \, B'=\frac{10^{-7} \times 40}{3.14 \times 10^{-2}} \times \frac{3 \pi}{2}=60 \times 10^{-5} $
$B'=6 \times 10^{-4}\, T$
