Gravitational field at the centre of a semicircle formed by a thin wire AB of mass M and length $l$ is

Consider a small element of mass dm

$\mathrm{dm}=\frac{\mathrm{M}}{\mathrm{L}}\cdot \mathrm{dx}=\frac{\mathrm{M}}{\mathrm{L}}[\mathrm{R}\cdot \mathrm{d}\theta ];\mathrm{L}=\pi \mathrm{R}\Rightarrow \mathrm{R}=\frac{\mathrm{L}}{\pi }$

The gravity field at 0 due to the element

$\mathrm{dE}=\frac{\mathrm{Gdm}}{{\mathrm{R}}^2}=\frac{\mathrm{G}}{{\mathrm{R}}^2}\cdot \frac{\mathrm{M}}{\mathrm{L}}\cdot \mathrm{Rd}\theta =\frac{\mathrm{GM}}{\mathrm{LR}}\mathrm{d}\theta$

$\mathrm{dE}=\frac{\mathrm{GM}\pi }{{\mathrm{L}}^2}\cdot \mathrm{d}\theta$

${\mathrm{E}}_{\mathrm{x}}={\int }_0^{\pi } {\mathrm{dEx}}_{\mathrm{x}}={\int }_0^{\pi } \mathrm{dEcos}\theta ={\int }_0^{\pi } \frac{\mathrm{GM}\pi }{{\mathrm{L}}^2}\cos \theta \mathrm{d}\theta =0$

${\mathrm{E}}_{\mathrm{y}}={\int }_0^{\pi } {\mathrm{dEy}}_{\mathrm{y}}={\int }_0^{\pi } \mathrm{dEsin}\theta$

$={\int }_0^{\pi } \frac{\mathrm{GM}\pi }{{\mathrm{L}}^2}\sin \theta \mathrm{d}\theta =\frac{2\pi \mathrm{GM}}{{\mathrm{L}}^2}$

$\therefore {\mathrm{E}}_{\mathrm{y}}=\frac{2\pi \mathrm{GM}}{{\mathrm{L}}^2}\text{ along }\mathrm{y}\text{-axis }$