Question

In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of $2.0\times {10}^{10}Hz$ and amplitude $48V{m}^{-1}$.

(a) What is the wavelength of the wave?

(b) What is the amplitude of the oscillating magnetic field?

(c) Show that the average energy density of the E field equals the average energy density of the B field. [c = 3 $\times {10}^{8}m{s}^{-1}$]

Answer 1

$\left(a\right)$ The wavelength is given by $\lambda =\frac{c}{v}=\frac{3\times {10}^8}{2\times {10}^{10}}=1.5\times {10}^{-2}m$

$\left(b\right)$ $B_o=E_o/c=1.6\times {10}^{-7}T$

$\left(c\right)$ Energy density due to the electric field, $E_E=\frac{1}{2}{ϵ}_o{E}^2$

Energy density due to the magnetic field, $E_B=\frac{1}{2}{B}^2/{\mu }_o$

also, $c=\frac{1}{\sqrt{{ϵ}_0{\mu }_o}}$

on solving above equations, $E_E=E_B$