# Speed of light (Maxwell)

Imagine an electromagnetic wave with E(vector) field in the y-direction. Show that Equation

$\nabla \times E =- dB/dt$

applied to the harmonic wave B(vector) and E(vector)

$E=E_0*\cos(kx-\omega t)$

$F=B_0*\cos(kx-\omega t)$

yelds the fact that

$E_0=c*B_0$

Answer

If the propagation of the wave is in the x direction and E=Ey in the y direction, by the rule of right screw (i= j x k , k = i x j, j= k x i) this means that B=Bz in the z direction.

$dE/dx = -k*E_0*sin(kx-\omega*t)$

$dB/dt= \omega*B_0*sin(kx-\omega*t)$

Now $dE/dx = -dB/dt$ means

$k E_0 =\omega*B_0$

$E_0 = (\omega/k) *B_0$

But we know that $omega = 2*\pi/T$ and $k = 2*\pi/x$ , therefore

$E_0 = (x/T)*B_0$

$E_0 = c*B_0$