Is $B=E/c$ true for all electromagnetic waves (in vacuum)? The relation is true (and rather easy to prove using Maxwell's equations) for a plane wave. Since any wave can be written as a sum of plane waves, I figured it should hold for arbitrary waves, but a more careful analysis of the problem has me stumped.
I can write $\vec{E}$ and $\vec{B}$ as
$$
\vec{E} = \sum_j\vec{E}_j,\\
\vec{B} = \sum_j\vec{B}_j,
$$
where all the $\vec{E}_j$ and $\vec{B}_j$ are plane waves. But these are vector sums, so I can't just factor a $1/c$ out of one the sums and call it a day; $\vec{E}=\sum_j\vec{E}_j$ and $\vec{B}=\sum_j\vec{B}_j$ do not imply $E=\sum_j E_j$ and $B=\sum_j B_j$ (which, in fact, will not hold unless all of the $\vec{E}_j$ are (anti)parallel to each other and all of the $\vec{B}_j$ are (anti)parallel to each other, in which case $\vec{E}$ is itself a plane wave and I'm back to square one).
I tried doing the very simple case $\vec{E}=\vec{E}_1+\vec{E}_2$ (and likewise for the magnetic field) (with $B_1=E_1/c$, $B_2=E_2/c$, $|\vec{k}_1|=|\vec{k}_2|=:k$) and using Maxwell's equations. After a lot of algebra, I arrived at
$$
B=\frac{E}{c}\\\Leftrightarrow\\\frac{2}{\omega^2}\left((\vec{k}_1\cdot\vec{k}_2)(\vec{E}_1\cdot\vec{E}_2)-(\vec{k}_1\cdot\vec{E}_2)(\vec{k}_2\cdot\vec{E}_1)\right)+k_{1x}\,\!^2E_{1x}\,\!^2+k_{1y}\,\!^2E_{1y}\,\!^2+k_{1z}\,\!^2E_{1z}\,\!^2+k_{2x}\,\!^2E_{2x}\,\!^2+k_{2y}\,\!^2E_{2y}\,\!^2+k_{2z}\,\!^2E_{2z}\,\!^2=\frac{2}{c^2}\,(\vec{E}_1\cdot\vec{E}_2)
$$
or, equivalently,
$$
B=\frac{E}{c}\\\Leftrightarrow\\2(\vec{B}_1\cdot\vec{B}_2)+k_{1x}\,\!^2E_{1x}\,\!^2+k_{1y}\,\!^2E_{1y}\,\!^2+k_{1z}\,\!^2E_{1z}\,\!^2+k_{2x}\,\!^2E_{2x}\,\!^2+k_{2y}\,\!^2E_{2y}\,\!^2+k_{2z}\,\!^2E_{2z}\,\!^2=\frac{2}{c^2}\,(\vec{E}_1\cdot\vec{E}_2),
$$
at which point I don't know how to proceed (the above doesn't seem to simplify into anything reasonable and is starting to make me think the relationship $B=E/c$ might not necessarily hold for arbitrary waves).
Bit lost, seem to recall this result just being thrown out there when I was an undergraduate (was proven only for plane waves), not sure how to prove (or disprove) it for arbitrary waves. Any help would be appreciated.
 A: No! In fact all conclusions you can think of for plane waves are not true In general! . The E and B fields do NOT have to be inphase. And B doesn't equal E/C for all waves. That is why you weren't able to prove it!
This is because we are drawing conclusions based on MONOCHROMATIC PLANE WAVES, It is well known that a solution to the homogenous wave equation added with another solution to the homogenous  wave equation gives me another solution.
In the case of a standing wave solution to the homogenous wave equation. The E and B fields are $\pi /2 $ out of phase spatially AND in time
This obviously also means that E/B doesn't equal C at all times.
Most people assume the conclusions about plane monochromatic waves apply to all solutions of the homogenous wave equation But they do not!
A brilliant answer -
Question about intensity of EM waves
EDIT:
originally I wrote
Mathematically
$E_{1}\cdot B_{1} =0$
Where $E_{1},B_{1} $ are a plane monochromatic wave
let's say I want to add another plane wave to this solution
adding $E_{2},B_{2} $ to the $ E_{1},B_{1} $ wave, which is also a perfectly valid solution
Obviously
$E_{2} \cdot B_{2} =0$
However this doesn't imply
$(E_{1} + E_{2}) \cdot (B_{1} + B_{2}) = 0$
Meaning E and B solutions to the homogenous wave equation aren't necessarily perpendicular.
Correction:
I forgot to also apply the thinking that for each monochromatic wave  the E and B fields are related by C, Thus the dotproduct DOES imply they are always perpendicular! ( atleast for the quick    maths that I did for the summation of 2 plane waves..)
My other points still stand though.
A: The vectorial problem you stumped on is justified since the correct expression is:
$$c=\frac{|E|}{|B|}$$
The above known expression uses only the scalar amplitude values of $E$ and $B$ since $c$ is not a velocity thus is not a vector but the scalar, speed value of light in a vacuum.
