Prove Biot-Savart law, assuming that $\vec{A}=\frac I c\int \frac{d\vec{L}}{r}$

Question

Prove Biot-Savart law, assuming that $$\vec{A}=\frac I c\int \frac{d\vec{L}}{r}$$ $$\vec{B}=\nabla\times \vec{A}$$ Any hint on what to do next?

Answer 1

For staters, try putting $\vec A$ into $\vec B$.

You need to find an identity that helps you put:

$\nabla \times ({I\over c}\int {d\vec L \over r})$

into something more useful. (Most people don't know this off the top of their heads, but your book's appendix or other math tables should be quite helpful!)

Some more algebra should follow, but you should be able to get the Biot-Savart Law from this.

Answer 2

If what are you trying to write is:

$$ \overrightarrow{A}(r) = \frac{I}{c} \oint \frac{d \overrightarrow{L}(r')}{|r - r'|} $$ Where $ d \overrightarrow{L} $ is the line element of the circuit, then you could use this identity: $$ \nabla \wedge (f \overrightarrow{A}) = f \; \nabla \wedge \overrightarrow{A} + \nabla f \wedge \overrightarrow{A} $$ to get: $$ \overrightarrow{B}(r) = \frac{I}{c} \nabla_r \wedge \oint \frac{d \overrightarrow{L}(r')}{|r - r'|} = \frac{I}{c} \oint \nabla_r \left( \frac{1}{|r-r'|}\right) \wedge d \overrightarrow{L}(r') $$ Now you use that $$ \nabla_r \left( \frac{1}{|r-r'|}\right) = -\frac{\overrightarrow{r} - \overrightarrow{r'}}{|r-r'|^3} $$ To finally get Biot-Savart law: $$ \overrightarrow{B}(r) = \frac{I}{c} \; \oint \frac{d \overrightarrow{L}(r') \wedge (\overrightarrow{r} - \overrightarrow{r'})}{|r-r'|^3} $$