In electrodynamics, is the electric field always perpendicular to the magnetic field? if so, is there a simple way to prove this?

Question

Using the Maxwell equations and making no assumptions about the sources (charge and current densities), can one argue that $\vec{E}\perp \vec{B}$ at all times? Assuming that the EM field stems from a single source or set thereof.

I know that it is fairly easy to show that this is the case when the sources both vanish (i.e. when $\rho = 0$ and $\vec{J} = 0 $).

$$\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_{0}}$$ $$\nabla \cdot \vec{B} = 0$$ $$\nabla \times \vec{E} = -\frac{\partial\vec{B}}{\partial t}$$ $$\nabla \times \vec{B} = \mu_{0}\vec{J}+\frac{1}{c^2}\frac{\partial\vec{E}}{\partial t}$$

That is, if $\vec{E}$ and $\vec{B}$ both obey all of Maxwell's equations without exception,then does there always exist some vector $\vec{K}(\vec{x},t)$ such that

$$\vec{B}=\vec{K} \times \vec{E}?$$

If not, is there some way to specify a set of conditions under which this would be the case? One such condition would undoubtedly be the case of where there are no sources.

Or, at least, is there some explicit way to prove the counter point?

Answer 1

No, this is not true. A simple example is a charged capacitor in the earth's magnetic field--unless you hold it just right, E and B are usually not orthogonal.

Even if there are no sources around, you can get E non-perpendicular to B (and even parallel to B) by superimposing multiple light waves traveling in different directions with different polarizations and wavelengths.

Answer 2

The electric field is definitely not always perpendicular to the magnetic field. You can visualize this very easily by a counter-example:

Take a homogeneous magnetic field $\vec B$, e.g., due to a permanent magnet or inside a long wire coil. Then insert an electric point charge into this field. You can easily recognize that the electric $\vec E$ and the magnetic field $\vec B$ can have any angle between $0$ and $180$ degrees.

PS: The Coulomb field of the point charge is, of course, a solution to the Maxwell equation $$\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_{0}}$$

Additional counter-example with a "single source":

Take a circular ring of positive charge rotating so that a stationary annullar electric current flows. On the axis on one side of the ring away from its center, you have always an electric field pointing away from the ring current. And you have also a magnetic field parallel or antiparallel to the electric field depending on which side of the ring you are looking at.

This disproves your claim that $\vec{E}\perp \vec{B}$ at all times for a "single source".

Answer 3

It depends what you mean by "a single source or set thereof". For example, may we take, as a set of sources, charges moving round a circular track at constant speed, equally spaced from one another? If so, consider a point, P, some way from the centre of the track and on its axis (line through the circle centre at right angles to the plane of the circle). At such a point the electric field and magnetic field due to the charges on the track are both directed along the axis and so parallel or antiparallel to each other.

This result is easily obtainable using mainly symmetry. It could be made quantitative using Coulomb's law and the Biot-Savart law and these, in turn, can (with care and time!) be deduced from Maxwell's equations.

Later addition The set-up in the first paragraph could be achieved in practice by spinning a disc (made of an insulator) with a charged edge.