- #1

PhysicsRock

- 116

- 18

- 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?

$$

\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?