An infinitely long wire is bent in the form of a semicircle at the end as shown in the figure. It carries current I along abcdo. If radius of the semicircle be R, then the magnetic field at ${}^{\text{'}}\mathrm{O}{ }^{\text{'}}$ which is the centre of the circular part is

Given data:

A wire with infinite length has been shaped into a semicircle at one end, as depicted in the figure. The wire is conducting current I along the path of abcdo.

Formula used: $B_{semicircle}=\frac{{\mu }_{0}I}{4R}$

Detailed Solution:

As we know $B_{semicircle}=\frac{{\mu }_{0}I}{4R}$

Therefore, B total is 

$$\begin{gathered} =\frac{{\mu }_{0}I}{4\pi R}+\frac{{\mu }_{0}I}{4R} \\ =\frac{{\mu }_{0}I}{4R}\left[\frac{1}{\pi }+1\right] \\ =\frac{{\mu }_{0}I}{4\pi R}\left[\pi +1\right] \end{gathered}$$