A very long conducting wire is bent in a semi-circular shape from $A$ to $B$ as shown in figure. The magnetic field at point $P$ for steady current configuration is given by :
Answer: (C) $\dfrac{\mu_0 i}{4R}\left[1 - \dfrac{2}{\pi}\right]$ pointed away from page
P is the centre of the semicircle AB of radius $R$. The wire has three parts.
Semicircle: the current goes from A (top) round the left side to B (bottom), which is anticlockwise as seen, so its field at P points out of the page:
$$B_1 = \frac{\mu_0 i}{4R}$$
Upper straight wire: semi-infinite, ending at A, directly above P at distance $R$. The current flows to the right and P is below the wire, so the field is into the page:
$$B_2 = \frac{\mu_0 i}{4\pi R}$$
Lower straight wire: semi-infinite, ending at B. The current flows to the left and P is above the wire, so the field is also into the page:
$$B_3 = \frac{\mu_0 i}{4\pi R}$$
Net field (out of the page positive):
$$B = \frac{\mu_0 i}{4R} - \frac{2\mu_0 i}{4\pi R} = \frac{\mu_0 i}{4R}\left[1 - \frac{2}{\pi}\right]$$
pointing away from (out of) the page.
Solution by Sreeraj P, M.Sc Physics