Can we derive Biot Savart law from Coulomb's law?

Can we derive Biot Savart law from Coulomb's law by setting a current source at an instant in place of one of the charges? Let's say $dl$ is the length of the conductor having $dQ$ charge which applies some force at the other charge at some instant of time. Taking derivates on both sides we get some current term. But how to prove it?

By Coulomb's law

$dE = k \frac{dq}{r^2}$

$\dfrac{dE}{dt} = k\dfrac{I}{r^2}$

$d\vec{l}\times \dfrac{d\vec{E}}{dt} = k I d\vec{l} \times \dfrac{ \vec{r}}{r^3}$

Is it correct? If yes, how should I proceed ahead?

• No, you can't. At this stage of Electromagnetism the electric and magnetic fields are still disjoint entities. Coulomb and Biot-Savart's laws are both experimentally obtained. Jun 26 '16 at 14:59
• But moving one of the charges like I did above by placing a current source, a similar expression is obtained only the constant is a problem. Jun 26 '16 at 15:05
• But you just have an electric field in your expression, no magnetic field. Jun 26 '16 at 15:09
• But since there's something visible in that equation so isn't there any way to do it even if we take electromagnetic waves and speed of light relation with permeability into consideration . Jun 26 '16 at 15:26
• Use $dE/dt=\nabla \times B$ and see? But caution that you would not be "deriving BS law from Coulomb". You would be deriving BS law from Coulomb+Faraday, which is essentially to prove that Maxwell Eqns. are self-consistent. Jun 26 '16 at 16:02