Everyday magnetism as a consequence of perfect balance of electrical effects

Consider two charges moving in space, both at the same speed and parallel to each other. Because they are moving, they will behave like two currents and will have a magnetic field associated with them. An observer who was riding along with the two charges, however, would see both charges as stationary and would say that there is no magnetic field.

[few lines later]

What we are saying, then, is that magnetism is really a relativistic effect. In the case of the two charges we just considered, travelling parallel to each other, we would expect to have to make relativistic corrections to their motion, with terms of order $$v^2/c^2$$. These corrections must correspond to the magnetic force.

[few lines later]

So $$v^2/c^2$$ is about $$10^{-25}$$. Surely a negligible "correction". But no! Although the magnetic force is, in this case, $$10^{-25}$$ of the "normal" electrical forces between the moving electrons, remember that the "normal" electrical forces have disappeared because of the almost perfect balancing out — because the wires have the same number of protons as electrons. The balance is much more precise than one part in $$10^{-25}$$, and the small relativistic term which we call the magnetic force is the only term left. It becomes the dominant term.

Chapter 1, Volume 2, Feynman Lectures on Physics

1. How can I derive the above mathematically starting from Maxwell's equations (and maybe Lorentz force law)?

Somewhere earlier, the book states the following:

$$c^2(\text{circulation of }\vec{B}\text{ around C}) = \frac{d}{dt}(\text{flux of }\vec{E}\text{ through S}) + \frac{\text{flux of electric current through S }}{\epsilon_0}\tag{1.9}$$

The constant $$c^2$$ that appears in the Eq. (1.9) is the square of the velocity of light. It appears because magnetism is, in reality, a relativistic effect of electricity.

Chapter 1, Volume 2, Feynman Lectures on Physics

1. Since magnetism is entirely an artefact of electrical effects and special relativity, would it be possible to derive two (or at least one) of Maxwell's equations from the rest?

OFF-TOPIC:

What other common phenomenon is due to the effects of special relativity at not-so-high speeds making a significant observable difference?

• – rob Jan 6 at 19:24