# Step in a proof that $\textrm{div} \ \mathbf{B} = 0$ from Biot-Savart's law

Notation: The magnetic field $\mathbf{B}$ generated by a point charge $e$ moving with velocity $\mathbf{v}$ is given by Biot-Savart's law

$$\mathbf{B} = \frac{\mu_0 e\ \mathbf{v} \wedge \mathbf{r}}{4\pi r^3}$$ where $\mathbf{r}$ is the vector from the charge to the point at which the field is measured, $r = \left| \mathbf{r} \right|$, and $\wedge$ denotes vector product.

Question: According to my book: Since $\mathbf{r}\, / \, r^3 = - \,\textrm{grad} \left(1\, /\, r \right)$, we have $$\textrm{div} \left( \mathbf{v} \wedge \frac{\mathbf{r}}{r}\right) = \mathbf{v} \wedge \textrm{curl} \left( \textrm{grad} \frac{1}{r}\right) = 0.$$

What I have is $$\textrm{div} \left( \mathbf{v} \wedge \frac{\mathbf{r}}{r}\right) = \mathbf{v} \, . \, \textrm{curl} \left( \textrm{grad} \frac{1}{r}\right) - \left( \textrm{grad} \frac{1}{r} \right) \, . \, \textrm{curl} \ \mathbf{v},$$ so I think there is a typo in the book ($\, \wedge$ should be $\, . \,)$. However, I still don't know how to go from my equation to the correct one. Is it because $\textrm{curl} \ \mathbf{v} = 0$? If so, why? I understand that the curl of a gradient vanishes identically.

• v1: The right hand side of your last equation must be wrong because the right hand side is a vector while the left hand side is a scalar. Mar 12, 2014 at 23:52
• I think there is a typo in the book. I explained this in my edit. Mar 12, 2014 at 23:54
• Hi fctaylor25. If you haven't already done so, please take a minute to read the definition of when to use the homework tag, and the Phys.SE policy for homework-like problems. Mar 13, 2014 at 0:32
• @Qmechanic I re-added the tag. Thanks for the link, I didn't realize this question should be labelled homework, even though I ran into it during self-study. Mar 13, 2014 at 2:23
• I feel it is important to say that the formula you have for the magnetic field of a point charge is incorrect in general. Here is the correct formula. Your formula is correct in the limit $v \ll c$ Mar 13, 2014 at 4:11

$\textrm{curl} \ \mathbf{v} = 0$, since $\mathbf{v}$ is not a function of the position at which the field is measured, i.e., it is not a function of the variable with respect to which we are differentiating.