In a bent wire shown in figure, a current I is passed. Find the value of B at the common centre.

The straight segments make no contribution to the field at the centre. By Biot-Savart's law, the field at the centre due to curved segments is

$\begin{array}{l}\mathrm{\Delta B}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{4\mathrm{\pi }}\left[\frac{\mathrm{a}\left(2\mathrm{\pi }-\mathrm{\theta }\right)}{{\mathrm{a}}^2}-\frac{\mathrm{b}\left(2\mathrm{\pi }-\mathrm{\theta }\right)}{{\mathrm{b}}^2}\right]\\ =\frac{{\mathrm{\mu }}_0\mathrm{I}}{4\mathrm{\pi }}\left(2\mathrm{\pi }-\mathrm{\theta }\right)\left[\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{b}}\right]=\frac{{\mathrm{\mu }}_0\mathrm{I}\left(2\mathrm{\pi }-\mathrm{\theta }\right)\left(\mathrm{b}-\mathrm{a}\right)}{4\mathrm{\pi ab}}\end{array}$