Magnetic field at origin due to infinite wires and semicircular turn

AI Thread Summary
The discussion focuses on calculating the magnetic field at the origin due to an infinite wire and a semicircular turn. The user applies the Biot-Savart law to determine the magnetic field contribution from the left wire, but expresses confusion about the direction of the resulting field vector, which appears to point in the y-direction instead of conforming to the expected circular pattern. The conversation suggests that the field should be a superposition of contributions from multiple sources, including two infinite wires and a semicircle. Participants propose considering symmetry and alternative configurations to simplify the calculation. The need for clarity in the magnetic field's direction and the relationship between the different components is emphasized throughout the thread.
PhysicsRock
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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 seperately. 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?
 
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Use some imagination....
How is the field from your configuration related to the field from two infinite wires plus one circle?

##\ ##
 
BvU said:
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.
 
PhysicsRock said:
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?
 
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