Magnetic field at origin due to infinite wires and semicircular turn

[PhysicsRock]

Homework Statement

An infinite wire is bent to resemble a U. The U-part is a semicircle with radius $R = 5.14 \, \text{cm}$. Calculate the magnetic field at point $P$, which is the center of curvature of the semicircular part.

Relevant Equations

Biot-Savart law $\vec{B}(\vec{r}) = \frac{\mu_0 I}{4 \pi} \int \frac{d\vec{s}^\prime \times (\vec{r} - \vec{r}^\prime)}{\vert \vec{r} - \vec{r}^\prime \vert^3}$.

Right now, I am trying to calculate the field due to the left straight wire. For clearance, I have oriented the contraption such that the straight wires go from $z = 0$ to $z = \infty$ and pass through $x = \pm R$, i.e. the semicircle is below the $x$-axis. The current starts at $z = \infty$ on the left wire and flows towards $z = 0$. That makes the point $P$ the coordinate origin. Thus, Biot-Savart tells us that

$$
\vec{B}(\vec{p}) = \vec{B}(0) = -\frac{\mu_0 I}{4 \pi} \int \frac{d\vec{s}^\prime \times \vec{r}^\prime}{(r^\prime)^3} = -\frac{\mu_0 I}{4 \pi} \int \frac{d\vec{s}^\prime \times \hat{r}^\prime}{(r^\prime)^2}
$$

Since the vector product $\cdot \times \cdot$ is distributive, I can calculate the magnetic field of each individual contributor separately. For the left wire (i.e the one passing through $x=-R$), I use

$$
d\vec{s}^\prime = dz \cdot \hat{z}
$$

and

$$
r^\prime = \sqrt{R^2 + z^2}, \hat{r}^\prime = \begin{pmatrix} -\cos(\alpha) \\ 0 \\ \sin(\alpha) \\ \end{pmatrix}.
$$

Here, I introduced the angle $\alpha$ as the angle between the $x$-axis and the "arm" that's sliding along the wire in the integration process. Then, the cross product is

$$
d\vec{s}^\prime \times \hat{r}^\prime = dz \begin{pmatrix} 0 \\ 0 \\ 1 \\ \end{pmatrix} \times \begin{pmatrix} -\cos(\alpha) \\ 0 \\ \sin(\alpha) \\ \end{pmatrix} = dz \begin{pmatrix} 0 \\ -\cos(\alpha) \\ 0 \\ \end{pmatrix}
$$

This is where I start wondering already. It doesn't seem right that the field vector is pointing in the $y$ direction, as by the right hand rule it should be a closed circle, parallel to the $x$-$y$-plane. What did I do wrong here?

 

[BvU]

Use some imagination....
How is the field from your configuration related to the field from two infinite wires plus one circle?

$\$

 

[PhysicsRock]

Use some imagination....
How is the field from your configuration related to the field from two infinite wires plus one circle?

$\$

As mentioned above, it should be a superposition of the three individual fields, due to $\vec{a} \times (\vec{b} + \vec{c} + \vec{d}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c} + \vec{a} \times \vec{d}$. What I derived above only applies to the leftmost wire. Correct me if I'm wrong, but typically the magnetic field of a wire consists of circular lines around the wire. That doesn't add up with the field vector pointing in the $y$-direction here though.

 

[haruspex]

It doesn't seem right that the field vector is pointing in the y direction, as by the right hand rule it should be a closed circle, parallel to the x-y-plane.

A circle in the xy plane has a tangent in the y direction at two points. Does one of those happen to be point P?

As @BvU hints, there is a much easier way. If there were a second instance of the U, same plane, rotated 180 degrees, would it exert the same field at P? Is there a way to decompose the sum of those two Us into two much simpler circuits?