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: 4. \(\frac{\mu_0 i}{4R}\left[1-\frac{2}{\pi}\right]\) pointed away from page

Add the fields of the three parts at P: two semi-infinite straight wires and one semicircle.

• Each straight wire ends at A or B, and P is at perpendicular distance \(R\) from it, so \(B=\dfrac{\mu_0i}{4\pi R}\). By the right-hand rule both point into the page, giving \(\dfrac{\mu_0 i}{2\pi R}\) in total.
• The semicircle: \(B=\dfrac{\mu_0 i}{4R}\). The current goes from A round to B (anticlockwise as seen), which points out of the page.

Net \(=\dfrac{\mu_0i}{4R}-\dfrac{\mu_0i}{2\pi R}=\dfrac{\mu_0 i}{4R}\left[1-\dfrac2\pi\right]\), pointing away from the page.