# Farady's law and div B = 0

I'm reading a book on electromagnetism and I am a bit confused about some things in Maxwells equations. This is what I don't like about many physics books: they are very wordy, but at the end you don't know what is an experimental fact, what is a "theorem", what is an assumption and so on,...

Anyway the questions are:

1. Is Faraday's law a consequence of Coulomb's and Biot-Savart's (or Ampere's) law?

2. Is divB = 0 consequence of Biot-Savart's law i.e. if we found the magnetic monopole, would that mean that Biot-Savart's law is not true? (I think so, but I am not 100% sure)

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1. No.
2. No.

More precisely, Biot-Savart relates the curl of $B$ to currents, which is independent of specifying the divergence.

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I disagree slightly with genneth on question number 2. In its most stripped-down form, as I see it, the Biot-Savart law states that all magnetic fields created by static currents are of the form $$\mathbf{B} = \frac{\mu_0}{4\pi} \int_C \frac{I d\mathbf{l} \times \mathbf{r}}{|\mathbf{r}|^3}.$$ From this you can recover Ampere's law and the condition that $\textrm{div}\mathbf{B}=0$ for fields caused by static currents: the fist by taking the circulation of $\mathbf{B}$ and doing the appropriate integrals, and the second by taking the divergence of $\mathbf{B}$ and noticing that the integrand vanishes for each $d\mathbf{l}$, so $\nabla\cdot\mathbf{B}=0$ for every $\mathbf{r}$.

As Steve B points out, though, since the Biot-Savart law only holds for static currents, this proof of $\nabla\cdot\mathbf{B}=0$ only holds in that case. In a general dynamic situation you need make an independent assumption: in the general case you need to independently postulate all four of Maxwell's equations; the actual fields generated by charges and currents are derived from the Lienard-Wiechert potentials that solve them.

Regarding your first question, the answer is indeed no: both Coulomb's and Biot-Savart's laws are valid, strictly speaking, only for static phenomena. Faraday's law is the first dynamical correction to them. If you want to see that the mathematics is not enough to get Faraday's law, then consider that similarly to $$\textrm{Biot-Savart law}\Rightarrow\nabla\cdot\mathbf{B}=0$$ above, you can prove $$\textrm{Coulomb's law}\Rightarrow\nabla\times\mathbf{E}=0,$$ which contradicts Faraday's law.

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