• Tags: wave-equation, k, omega
• Difficulty: Moderate

A plane EM wave propagating in the +z direction: E = $E_0$ sin(kz - omegat) x-hat, B = $B_0$ sin(kz - omegat) y-hat. Wave number k = 2pi/lambda, angular frequency omega = 2pif, phase velocity v = $\frac{omega}{k}$ = flambda = c. The relation $E_0$/$B_0$ = $\frac{omega}{k}$ = c confirms the wave speed. To find direction of propagation: it's along E x B. Given E and B directions, use the right-hand rule. If E is along x and B is along y, propagation is along z. If E is along y and B along -x, propagation is along y x (-x) = ... one must be careful with signs. JEE often gives E or B and asks for the other: use $B_0$ = $\frac{E_0}{c}$ and determine direction from E x B = c*(direction of propagation). The wave carries energy in the direction of S = $\frac{1}{mu_0}$(E x B).