# Why are magnetic fields so much weaker than electric?

In EM radiation, the magnetic field is $3*10^8$ times smaller than the electric field, but is it valid to say it's "weaker". These fields have different units, so I don't think you can compare them, but even so it seems like we only interact with the electric field of EM radiation, not the magnetic field. Why is this?

-
Note that in the Gaussian units system, $\mathbf E$ and $\mathbf B$ share the same unit. – Kyle Kanos Jul 29 '14 at 2:27
Note that although the fields may, in some sense, have the same magnitude, the magnetic force is weaker by a factor of $v/c$. – Javier Jul 29 '14 at 2:49
Keep in mind the weakness of the electric dipole field too. The magnetic field we observe from magnets is a dipole field, so will certainly be weaker than the direct field of a pole, which is what is at play in atoms etc. It is the electric field potential that plays the leading role. – anna v Jul 29 '14 at 3:33
Thus when an EM wave impinges on an atom it will often see the poles of the electrons, whereas the magnetic can only interact with the magnetic dipoles of the atoms etc.which are much weaker. That is why the electric is more important, even though the energy carried is the same, as Johannes states. – anna v Jul 29 '14 at 3:43

As you already indicated, physical units need to be considered. When working in SI units, the ratio of electric field strength over magnetic field strength in EM radiation equals 299 792 458 m/s, the speed of light $c$.

However, the numerical value for $c$ depends on the units used. When working in units in which the speed of light $c=1$, one would conclude that both fields are equal in magnitude.

A better way to look at this is to consider the energy carried by an electromagnetic wave. It turns out that the energy associated with the electric field is equal to the energy associated with the magnetic field. So in terms of energies electric and magnetic fields are equals.

-
Thats perfect, I was also wondering wether they had the same energy. Thanks! – aPhysicist Jul 29 '14 at 2:43
Looking at it in terms of energies can be misleading, sometimes by quite a bit. If you set a charged particle to oscillate driven by the electric and magnetic fields of a plane wave, then the electric component of the Lorentz force will be vastly bigger than the magnetic part. – Emilio Pisanty Jul 29 '14 at 19:54
Of course @Emilio is right, but interestingly the size of the difference in the forces is a function of the mass of the particle, not it's charge. If we were to imagine a very low mass test particle--say a charged neutrino (!)--then the difference gets to be much smaller – dmckee Jul 30 '14 at 1:09

It's a quirk of units: notice that the conversion between them is dimensionful, and has the value $3\times 10^8\,\mathrm{m/s}$, which is the speed of light. In the CGS system both fields have the same units, and field-squared is an energy density.

In SI units, the energy density for a configuration of fields is given by \begin{align} \frac{dU}{dV} &= \frac12\left( \epsilon_0 E^2 + \frac1{\mu_0} B^2 \right)\\ &= \frac{\epsilon_0}2 \left( E^2 + c^2 B^2 \right) \end{align} which tells you that, given the field strength ratio $c$ you have calculated for electromagnetic waves, the energy of an EM wave is shared equally between its electric and magnetic components.

We do tend to usually think about the interaction between EM waves and materials in terms of the electric field component, but that's a bias because it's relatively easy to liberate free electric charges, and magnetic interactions only occur at second order. Spend some time around folks thinking about plasma physics, or wrap your head around the behavior of the sun's magnetic field reversal cycle, for a very different perspective.

-