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This is the common impression.

But, is it just a rule of thumb and correct only in some wavelength region or does it hold universally?

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It is more important in the sense of accelerating charged particles.

In vacuum, it is easy to show that the ratio of the amplitudes of the electric field and magnetic field is $c$ (in SI units).

The Lorentz force on a charged particle is $$\vec{F} = q (\vec{E} + \vec{v} \times \vec{B})$$

If we just consider the magnitudes of the electric versus the magnetic forces acting on a charged particle, then (for a charge in vacuum), we can see that $$ \frac{F_B}{F_E} = \frac{v}{c}$$

Thus unless the charges are moving relativistically then it is the electric part of the Lorentz force that is dominant.

In a conducting material, the balance between electric and magnetic fields changes. In a very good conductor then then $$\frac{E}{B} = \left( \frac{\omega}{\mu_0 \mu_r \sigma}\right)^{1/2} = c\left( \frac{\omega \epsilon_0}{\sigma}\right)^{1/2},$$ with a a phase difference of $\pi/4$ between the E- and B-fields. In general this ratio will be $\ll c$ in a good conductor where $\sigma \gg \omega \epsilon_0$.

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  • $\begingroup$ This is a great answer and one of the only answers on the internet I've managed to find that speaks about the $\pi/4$ phase-difference between the E and B fields in a good conductor. Do you know if there's any physical mechanism or reason behind this phase difference? I have posted a question about this over here physics.stackexchange.com/questions/630956/… but I am yet to get an answer $\endgroup$ Apr 21, 2021 at 16:37

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