# Why are electric and magnetic fields perpendicular in an electromagnetic wave?

I want to know why electric and magnetic fields are perpendicular in an electromagnetic wave and what would happen if they weren't.

This can be explained by Maxwell's equations. For which one needs to know the basics of vector calculus, gradient, divergence and curl.

When there are no charges:

\begin{aligned} c\nabla\times \boldsymbol E(\boldsymbol r,t) + \partial_t \boldsymbol B(\boldsymbol r,t) = 0\\ \nabla\cdot \boldsymbol B(\boldsymbol r,t) = 0 \\ c\nabla\times \boldsymbol B(\boldsymbol r,t) - \partial_t \boldsymbol E(\boldsymbol r,t) = 0\\ \nabla\cdot \boldsymbol E(\boldsymbol r, t) = 0 \end{aligned}

Where, $\boldsymbol E$ and $\boldsymbol B$ are the electric field strength and the magnetic induction, respectively and $c$ is the speed of light in free space.

The spatial and time periodicity of the radiation be utilized to write Maxwell's equations in Fourier transformed form:

\begin{aligned} c\boldsymbol q\times \boldsymbol E(\boldsymbol q,\omega)− \omega B(\boldsymbol q,\omega) = 0 \\ \boldsymbol q \cdot \boldsymbol B(\boldsymbol q, \omega) = 0 \\ c\boldsymbol q\times \boldsymbol B(\boldsymbol q,\omega)+\omega \boldsymbol E(\boldsymbol q,\omega) = 0\\ \boldsymbol q \cdot \boldsymbol E(\boldsymbol q, \omega) = 0 \end{aligned} where $\boldsymbol q$ is a wave vector.

From the third equation we get $$\omega \boldsymbol B(\boldsymbol q, \omega)=c\boldsymbol q\times \boldsymbol E(\boldsymbol q, \omega)$$

Now we take the scalar product with $\omega\boldsymbol E(\boldsymbol q, \omega)\cdot \boldsymbol B(\boldsymbol q, \omega) = c\omega\boldsymbol E(\boldsymbol q, \omega)\cdot \boldsymbol q\times \boldsymbol E(\boldsymbol q, \omega)$ but from the first equation we know that $i\boldsymbol q \cdot \boldsymbol E(\boldsymbol q, \omega) = 0$

Therefore, $$\boldsymbol E(\boldsymbol q, \omega)\cdot \boldsymbol B(\boldsymbol q, \omega) = 0$$

For the scalar product between two vectors to be zero either one of them is the zero vector or they are perpendicular to each other.

Therefore, the electric and magnetic fields are perpendicular.

Also, for a propagating EM wave, the $\boldsymbol E$ and $\boldsymbol B$ fields are always perpendicular in a homogenous, linear, anisotropic medium. This type of media includes many things like air, water, glass (without stress or tempering). However, in inhomogenous, non-linear, or isotropic media, the $\boldsymbol E$ and $\boldsymbol B$ fields may not be perpendicular, e.g. in a crystal (which is isotropic).

Although this explains the math of your question, I would also add that its just God's design that these two energies are perpendicular to each other :)

• I think that you meant "isotropic" where you said "anisotropic", and "anisotropic" where you said "isotropic". – user93237 Mar 5 '16 at 19:35
• Oh Ya! Pardon my mistake. The words got flipped – user118008 Mar 5 '16 at 19:40