Magnetic fields from current flows and resistor circuit

Question:
The wire splits into two ways, which are bent circularly with radius of $a$. The upper half wire has resistance $2R$ and the lower part has $R$. Find the magnetic fields at the center of the circle in terms of the total current $I$.

Answer:
The magnetic field created by current $I_1$ is denoted as $B_1$. From Biot-Savart law,
\begin{eqnarray}
\vec{B_1} &=& \oint \frac{\mu_0}{4\pi}\frac{I d\vec{l}\times \vec{r}}{r^2}  \\
    &=& \frac{\mu_0}{4\pi}\frac{I_1}{a^2}\int_{0\rightarrow \pi a}dl (-\vec{z})  \\
    &=& -\frac{\mu_0}{4\pi}\frac{I_1}{a^2}\pi a\vec{z} \\
    &=& -\frac{\mu_0 I_1}{4a}\vec{z}
\end{eqnarray}
Likewise, we can obtain the magnetic field created by the lower part.
\begin{equation}
\vec{B_2} = \frac{\mu_0 I_2}{4a}\vec{z}
\end{equation}
This is a parallel connection, and we can use
\[
\frac{I_1}{I_2}=\frac{R}{2R} \quad \rightarrow \quad I_1=\frac{1}{2}I_2
\]
The current is conserved.
\[
I = I_1 + I_2 \quad \rightarrow \quad I_2 = I - I_1
\]
Therefore,
\[
I_1=\frac{1}{2}(I - I_1) \quad \rightarrow \quad I_1 = \frac{I}{3}
\]
Hence, the other current will be
\[
I_2 = \frac{2I}{3}
\]
The magnetic field is expressed as
\[
\vec{B}=\vec{B_1}+\vec{B_2}=\frac{\mu_0}{4a}(I_2-I_2)\vec{z}=\frac{\mu_0}{4a}\left(\frac{2I}{3}-\frac{I}{3}\right)\vec{z}=\frac{\mu_0 I}{12a}\vec{z}
\]
The magnetic field is directed toward us.

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