Special relativity, electromagnetic fields, charge and Q

Question

Is it true, were Coulomb's constant k to be several orders of magnitude smaller, that there would be no (or increasingly negligible) magnetic fields generated by moving charges? The reason being the charge imbalance in the frame of the moving electrons would produce a smaller charge imbalance force, and therefor in the frame of the positive charges (the conductor) there would be a smaller magnetic field.

How would a smaller k affect magnets then?

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Additional comments to answers below

I see the connection through free space permittivity, though I hadn't thought of it that way. Though I didn't mention it, I had in mind the huge size of k as compared to, say, G, the gravitational constant.

It's fascinating (fascination is often in inverse proportion to understanding) to me that the slow drift velocity is nonetheless relativistic because k is so large.

Answer 1

The reason being the charge imbalance in the frame of the moving electrons would produce a smaller charge imbalance force

Since

$$k = \frac{1}{4\pi \epsilon_0}$$

decreasing $k$ is equivalent to increasing permittivity $\epsilon_0$ of free space. Assuming the permeability $\mu_0$ of free space is held constant, decreasing $k$ decreases the invariant speed $c$

$$c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} = \sqrt{\frac{4 \pi k}{\mu_0}}$$

So, for example, if $k$ were decreased by 4 orders of magnitude

$$k' = \frac{k}{10^4}$$

the speed of light would decrease by 2 orders of magnitude

$$c' = \frac{c}{10^2}$$

and so relativistic effects, e.g., length contraction, would become greater at a given speed.

$$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{1}{\sqrt{1 - \frac{\mu_0}{4\pi k}v^2}}$$

In other words, while the coupling constant $k$ gets smaller, the charge density increase, due to length contraction, gets larger. It can be shown that these changes offset one another.