A very long conducting wire is bent in a semi- circular shape from...

Let's consider the following diagram:

The formula to calculate the magnetic field $\left(\overset{\to}{B}1\right)$ due to wire 1 is given by

$\overset{\to}{B}1 = \dfrac{{\mu}_{0}i}{4\piR} ⊗ . . . \left(1\right)$

The formula to calculate the magnetic field \left(\overset{\to}{B}2\right)$due to wire 2 is given by</p><p>$\overset{\to}{B}2 = \dfrac{{\mu}_{0}i}{4R} ⊙ . . . \left(2\right)$</p><p>And, the formula to calculate the magnetic field \left(\overset{\to}{B}3\right)$ due to wire 3 is given by

$\overset{\to}{B}3 = \dfrac{{\mu}_{0}i}{4\piR} ⊗ . . . \left(3\right)$

Hence, the net magnetic field at P can be written as

$\begin{cases}\overset{\to}{B} & = & \overset{\to}{B}1 + \overset{\to}{B}2 + \overset{\to}{B}3 \\ = & \dfrac{{\mu}_{0}i}{4\piR} ⊗ + \dfrac{{\mu}_{0}i}{4R} ⊙ + \dfrac{{\mu}_{0}i}{4\piR} ⊗ \\ = & \dfrac{{\mu}_{0}i}{2\piR} ⊗ + \dfrac{{\mu}_{0}i}{4R} ⊙ \\ = & - \dfrac{{\mu}_{0}i}{2\piR} ⊙ + \dfrac{{\mu}_{0}i}{4R} ⊙ \\ = & \dfrac{{\mu}_{0}i}{4R}\left[1 - \dfrac{2}{\pi}\right] ⊙\end{cases}$