# Biot-Savart Law: infinity wire

Aroldo
Hey!

1. Homework Statement

One must simply calculate the magnetic field at a distance s to the wire, which carries a steady current $$I$$

## Homework Equations

Should I write the point vector as:
$$\mathbf{r} = s\hat{s} + \phi \hat{\phi} + z \hat{z}$$
or
$$\mathbf{r} = s\hat{s} + z \hat{z}$$ ?

## The Attempt at a Solution

I am not solving it as the author does. I am trying to use spherical coordinates, therefore I am writing:
$$\mathbf{B}(\mathbf{r}) = \frac{\mu_0}{4\pi}\int_{-\infty}^\infty{\frac{d\mathbf{l'} \times (\mathbf{r} - \mathbf{r'})}{|(\mathbf{r} - \mathbf{r'})|^{3/2}}}$$


Where:
$$d\mathbf{l'} = dz \hat{z}$$
$$\mathbf{r} - \mathbf{r'} = s \hat{s} + z \hat{z} - z'\hat{z} = s \hat{s} + (z-z')\hat{z}$$

$$\mathbf{B}(\mathbf{r}) = \frac{\mu_0}{4\pi} \frac{2 I}{s} \hat{\phi}$$

But, if I consider the vector as:
$$\mathbf{r} = s\hat{s} + \phi \hat{\phi} + z \hat{z}$$

(it seems to me more general) The answer has a component in the s-direction, which is incorrect.

Please, what is wrong in my reasoning?

Homework Helper
Gold Member
But, if I consider the vector as:
$$\mathbf{r} = s\hat{s} + \phi \hat{\phi} + z \hat{z}$$

(it seems to me more general) The answer has a component in the s-direction, which is incorrect.

If you draw a figure that attempts to show how the three terms ## s\hat{s} + \phi \hat{\phi} + z \hat{z}## combine to give ##\mathbf{r}##, you'll see why the ## \phi \hat{\phi}## term should not be included. It is important to understand the meaning of the unit vector ##\hat{s}##.

Note that in spherical coordinates ##(r, \theta, \phi)##, the position vector is just ##\mathbf{r} = r \hat{r}##. It is not ##\mathbf{r} = r \hat{r} + \theta \hat{\theta}+\phi \hat{\phi}##

Aroldo
Aroldo
If you draw a figure that attempts to show how the three terms ## s\hat{s} + \phi \hat{\phi} + z \hat{z}## combine to give ##\mathbf{r}##, you'll see why the ## \phi \hat{\phi}## term should not be included. It is important to understand the meaning of the unit vector ##\hat{s}##.

Note that in spherical coordinates ##(r, \theta, \phi)##, the position vector is just ##\mathbf{r} = r \hat{r}##. It is not ##\mathbf{r} = r \hat{r} + \theta \hat{\theta}+\phi \hat{\phi}##
Thank you a lot!