Magnetic field at origin due to infinite wires and semicircular turn

• PhysicsRock
In summary, the conversation discusses the calculation of the magnetic field due to a left straight wire using Biot-Savart's law. The magnetic field is a superposition of the fields from the individual contributors, with the current flowing towards the origin at point P. The field vector pointing in the y direction is due to the angle between the x-axis and the "arm" sliding along the wire in the integration process. The field from this configuration is related to the field from two infinite wires and one circle, and can be decomposed into simpler circuits by considering a second instance of the configuration rotated 180 degrees.
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 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?

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?

nasu

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