Answer
$B = 1.0~mT$
Work Step by Step
We can write the expression for the magnetic field produced by a current in a straight wire:
$B = \frac{\mu_0~i}{2~\pi~R}$
Then the expression for the magnetic field produced at point $a$ due to the a current in the bottom wire is:
$B = \frac{\mu_0~i}{4~\pi~R}$
We can write the expression for the magnetic field produced by an arc of current:
$B = \frac{\mu_0~i~\phi}{4~\pi~R}$
To find the magnitude of the magnetic field at point $a$, we can double the magnetic field due to the bottom wire and add the magnetic field due to half a circle of current:
$B = 2\times \frac{\mu_0~i}{4~\pi~R}+ \frac{\mu_0~i~\phi}{4~\pi~R}$
$B = \frac{\mu_0~i}{2~\pi~R}+ \frac{\mu_0~i~\pi}{4~\pi~R}$
$B = \frac{\mu_0~i}{2~\pi~R}+ \frac{\mu_0~i}{4~R}$
$B = \frac{(4\pi\times 10^{-7}~H/m)~(10~A)}{(2\pi)~(5.0\times 10^{-3}~m)}+ \frac{(4\pi\times 10^{-7}~H/m)~(10~A)}{(4)~(5.0\times 10^{-3}~m)}$
$B = 1.0~mT$
