A thin uniform wire of length L and mass M is bent to form a semicircle. What is the gravitational field at the centre?

$\begin{array}{l}\overrightarrow{\mathrm{dF}}=+\frac{\mathrm{GM}}{\mathrm{L}}\frac{\left(\mathrm{R}\mathrm{d\theta }\right)}{{\mathrm{R}}^2}\overrightarrow{\mathrm{OA}}\\ =+\frac{\mathrm{GM}}{\mathrm{LR}}[\widehat{\mathrm{i}}\mathrm{cos\theta }\mathrm{d\theta }+\widehat{\mathrm{j}}\mathrm{sin\theta }\mathrm{d\theta }]\\ \therefore \overrightarrow{\mathrm{F}}=+\frac{\mathrm{GM}}{\mathrm{LR}}\underset{0}{\overset{\mathrm{\pi }}{\int }}[\widehat{\mathrm{i}}\mathrm{cos\theta d\theta }+\widehat{\mathrm{j}}\mathrm{sin\theta d\theta }]\\ =+\frac{\mathrm{GM}}{\mathrm{LR}}{\left[\widehat{\mathrm{i}}\mathrm{sin\theta }\mathrm{cos\theta }\widehat{\mathrm{j}}\right]}_0^{\mathrm{\pi }}=\frac{\mathrm{GM}}{\mathrm{LR}}[0+2\widehat{\mathrm{j}}]\\ \\ \mathrm{Now},\mathrm{\pi R}=\mathrm{L},\\ \therefore \mathrm{R}=\frac{\mathrm{L}}{\mathrm{\pi }}\\ \therefore \overrightarrow{\mathrm{F}}=\frac{\mathrm{GM}}{\mathrm{L}.\frac{\mathrm{L}}{\mathrm{\pi }}}2\widehat{\mathrm{j}}=\frac{2\mathrm{\pi GM}}{{\mathrm{L}}^2}\widehat{\mathrm{J}}\end{array}$