Half down and half up!

By Ampere's Law, the magnetic field around a cable with current(I) flowing along its axis is $B=\frac{μ_0i}{2πr}$ , where "r" is the distance between the axis of the cable and the point where we are calculating field.

Considering the half cylinder as a collection of infinitesimal cables, let's first calculate the field created by a half hollow cylinder(negligible thickness).

An element in this object will create a field $dB=\frac{μ_0i}{2πr}$ however, this field is radial, so if θ is the angle between the vertical and the element's position, the vertical component of the field will be $dB=\frac{μ_0i}{2πr}senθ$ .It is intuitive that horizontal components of field will cancel along the half-cylinder. Making $i=Kdl$ , where K is current density and integrating, we get: $\int dB=\int \frac{μ_0Kdl}{2πr}senθ= \frac{μ_0K}{2πr} \int^{180}_0 senθdl$ ,but $dl=rdθ$ and $K= \frac{i'}{\pi r}$ , thus:

$B=\frac{μ_0i'}{\pi ^2 r}$

Making $i'=Jda$ (where $J=\frac{I}{\frac{\pi r^2}{2}}$ is current density), with $da=\pi rdr$ and integrating again to expand this to a solid half-cylinder:

$B=\int^R_0 \frac{μ_0J\pi rdr}{\pi ^2r} \rightarrow B=\frac{2μ_0I}{\pi ^2 r}$

Finally, the field in the proposed situation will be the superposition of $B_1$ and $B_2$ :

$B=B_1+B_2 \rightarrow B=\frac{2μ_0I_1}{\pi ^2 r} + \frac{2μ_0I_2}{\pi ^2 r}$

Plugging the given values:

$B=48μT$