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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?

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    $\begingroup$ 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. $\endgroup$
    – QuantumBrick
    Commented Jun 26, 2016 at 14:59
  • $\begingroup$ 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. $\endgroup$
    – Ananyo Bhattacharya
    Commented Jun 26, 2016 at 15:05
  • $\begingroup$ But you just have an electric field in your expression, no magnetic field. $\endgroup$
    – Jahan Claes
    Commented Jun 26, 2016 at 15:09
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    $\begingroup$ 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 . $\endgroup$
    – Ananyo Bhattacharya
    Commented Jun 26, 2016 at 15:26
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    $\begingroup$ 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. $\endgroup$
    – pathintegral
    Commented Jun 26, 2016 at 16:02

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I am sorry that people always say no rather than I don't know. The comment of Ananyo is correct, like v^B produces E, v^E produces B, and you have on the left hand dl/dt which is v. So you are correct only needing some refinement in reasoning and what each variable means. A further indication that you are correct is that if you take the Coulomb static force expression and put it in the retarded potential expression (with finite speed of propagation) you can recover all of Maxwell equations including magnetism and radiation. So you are doing the same at least symbolically. And as Einstein equations suggest, Magnetic fields are Electric fields viewed from a moving frame.

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you are basically trying to undergo a transition from a law which is valid for static charges(or non relativistic speeds) to one which is valid for steady currents. That is why, simple differentiation is erroneous and does not include any magnetic field term in dE/dt.

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