# Question regarding Biot-Savart law

1. Oct 24, 2013

### cwbullivant

In my E&M class yesterday, in going over an example for the Biot-Savart law, I couldn't quite understand the initial setup of a problem (this may be more of a math question, but given the source, I figured it ought to be posted here). The attached picture is a crude MS Paint reproduction of the graph shown in class (letters are color coded to correspond to the line or figure they represent); I don't have a working scanner at the moment, and even if I did, this is a lot more legible than my handwriting.

Ok, the question:

In the diagram, we defined $$\vec{ds} , \vec{r}, |r| , and \sin{\theta}$$

ds, r, and sin θ were quite straightforward.

But in defining $$\vec{r}$$, we used:
$$\vec{r} = y\hat{y} - z\hat{z}$$ (using y and z as unit vectors instead of j and k)

Given that the vector r points from (y, z) = (0, -z) to (y, 0), I'd have expected it to be defined as:

$$\vec{r} = y\hat{y} + z\hat{z}$$

As that's how every math class I've taken would have defined it; the calculation appeared to work out (dB pointing into the page) using the definition put on the board, but I can't understand why it was chosen.

2. Oct 25, 2013

### vanhees71

I also don't understand, why often the Biot-Savart Law and other formulae like it are presented in such an awkward manner. It's just the solution of the Poisson equation for the vector potential in Coulomb gauge for static magnetic fields, which reads (in Heaviside-Lorentz units)
$$\Delta \vec{A}=-\vec{j}.$$
The solution is then simple to state by analogy to the solution for the scalar potential via Coulomb's Law, i.e.,
$$\vec{A}=\frac{1}{4 \pi} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \frac{\vec{j}(\vec{x}')}{|\vec{x}-\vec{x}'|}.$$
To get the magnetic field, you have to take the curl of this expression
$$\vec{B}=\vec{\nabla} \times \vec{A}.$$
Differentiating under the integral you need to calculate an expression of the form
$$\vec{\nabla} \times \vec{a} f(\vec{x})=-\vec{a} \times \vec{\nabla} f(\vec{x}),$$
because $\vec{a}$ is a vector independent of $\vec{x}$. Then we use
$$\vec{\nabla} \frac{1}{|\vec{x}-\vec{x}'|}=-\frac{\vec{x}-\vec{x}'}{|\vec{x}-\vec{x}'|^3}.$$
This finally gives
$$\vec{B}(\vec{x})=\frac{1}{4 \pi} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' \vec{j}(\vec{x}') \times \frac{\vec{x}-\vec{x}'}{|\vec{x}-\vec{x}'|^3}.$$

For a current along a wire this simplifies to
$$\vec{B}(\vec{x})=\frac{I}{4 \pi} \int_{C} \mathrm{d} \vec{x}' \times \frac{\vec{x}-\vec{x}'}{|\vec{x}-\vec{x}'|^3}.$$
Notet that this formula only holds for a closed loop $C$. The reason for this is that the Biot-Savart Law for a static field can only be the solution of the magnetotstatic Maxwell equations if $\vec{\nabla} \cdot \vec{j}=0$, because otherwise the constraint $\vec{\nabla} \cdot \vec{B}=0$ (no magnetic monopoles!) is violated.

Unfortunately the English Wikipedia gives the Biot-Savart Law in the usual awkward notation. For those who understand German, here it's given in the more precise notation (including a derivation):

http://de.wikipedia.org/wiki/Biot-Savart-Gesetz

3. Oct 25, 2013

### cwbullivant

I'm afraid I'm not familiar with the Posson equation, gauges, or Heaviside-Lorentz units.

The law itself I understood when carrying out the calculation. I simply couldn't why the r-vector used in the equation was defined with a -z instead of a +z.

4. Oct 25, 2013

### mikeph

The contribution from a wire element at (0,-z) will be the same as that from a wire element at (0,+z).

My guess is he's done it for the semi-infinite wire and multiplied by two. The reflective symmetry of the problem along the z=0 plane should mean you get the same answer upon swapping z for -z.