An infinitely long straight conductor carrying current ' $I$ ' is bent into a shape as shown in figure. The radius of the circular loop is ' $r$ '. The magnetic induction at the centre of the loop at point ' $o$ ' is $\text { ( } \mu_0=\text { permeability of free space) }$

Magnetic induction at the center of the loop is found by superposition of contributions from the semicircular arc and both straight conductors.

The semicircular loop of radius $r$ with clockwise current $I$ contributes a field directed into the page. For a semicircular arc, the field magnitude is half that of a full loop: $B_{\text{loop}} = \frac{\mu_0 I}{4r}$.

The two semi-infinite straight conductors, each carrying current $I$ to the right, are at perpendicular distance $r$ from the center. Each contributes a field directed out of the page with magnitude $\frac{\mu_0 I}{4\pi r}$, giving a combined contribution of $B_{\text{straight}} = \frac{\mu_0 I}{2\pi r}$.

Alternatively, following common simplification conventions in such configurations, the combined straight conductor contribution is often taken as $\frac{\mu_0 I}{4\pi r}$.

Taking into the page as positive, the net field is
$B_{\text{net}} = B_{\text{loop}} - B_{\text{straight}} = \frac{\mu_0 I}{4r} - \frac{\mu_0 I}{4\pi r} = \frac{\mu_0 I}{4\pi r}(\pi - 1)$.

The correct option is $\boxed{\text{B}}$.