# Biot–Savart Law

1. Apr 7, 2013

### Astrum

I'm not sure I get how:$$\frac{\mu_{0}}{4\pi}\int_{C}\frac{I(d\vec{l}\times\vec{r})}{r^{3}}$$ = $$\frac{\mu_{0}}{4\pi}\int_{C}\frac{I(d\vec{l}\times\hat{r})}{r^{2}}$$

the r is a displacement vector in the first, and in the second it's a unit vector, but why is this so?

Last edited: Apr 7, 2013
2. Apr 7, 2013

### HomogenousCow

The second expression is correct, the first one is not.

3. Apr 7, 2013

### Astrum

Sorry, meant to put a cubed there. Fixed

4. Apr 7, 2013

### qbert

There's nothing really to get: $\widehat{r} = \frac{ \vec{r} }{r}$.

5. Apr 7, 2013