# Calculation of magnetic field from electric field

by maggas
Tags: calculation, electric, field, magnetic
 P: 2 1. The problem statement, all variables and given/known data In this tute on EM waves, we were given the Electric Field $$\textbf{E}=\textbf{E}_0\text{exp}(i(\textbf{k}\cdot\textbf{x} - \omega t))$$ which after a fair bit of algebra yields the magnetic field $$\textbf{B}=(\hat{\textbf{k}}\times\textbf{E})/c$$ Similarly the inverse problem I had to solve, given the Magnetic Field $$\textbf{B}=\textbf{B}_0\text{exp}(i(\textbf{k}\cdot\textbf{x} - \omega t))$$ yields $$\textbf{E}=c\textbf{B}\times\hat{\textbf{k}}$$ The tute also gives a hint that this can be solved in a few lines, without heavy algebra, using Lagrange's formula. 2. Relevant equations Maxwell's Equations Lagrange's formula: $$\textbf{a}\times(\textbf{b}\times\textbf{c}) = (\textbf{a}\cdot\textbf{c})\textbf{b} - (\textbf{a}\cdot\textbf{b})\textbf{c}$$ 3. The attempt at a solution I can only solve the question the long winded way, and would like to know how it can be solved using this identity rather than equating many equations to solve coefficients! EDIT: Forgot to mention only using simplified Maxwell's Equations, i.e. Gauss' = 0 and Ampere's has no J term
 P: 2 I've managed to solve this myself, thanks for the help $$\textbf{B}=(\hat{\textbf{k}}\times\textbf{E})/c$$ $$c\textbf{B}=(\hat{\textbf{k}}\times\textbf{E})$$ Then using Lagrange Triple Product $$\hat{\textbf{k}}\times(\hat{\textbf{k}}\times\textbf{E}) = (\hat{\textbf{k}}\cdot\textbf{E})\hat{\textbf{k}} - (\hat{\textbf{k}}\cdot\hat{\textbf{k}})\textbf{E}$$ $$\hat{\textbf{k}}\times c\textbf{B} = - \textbf{E}$$ Therefore $$\textbf{E} = c\textbf{B} \times \hat{\textbf{k}}$$ //as required