Explanation

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