• # question_answer In a plane EM wave, the electric field oscillates sinusoidally at a frequency of $2.0\times {{10}^{10}}Hz$ and amplitude $48\,V{{m}^{-1}}.$ (i) What is the wavelength of a wave? (ii) What is the amplitude of the oscillating magnetic field? (iii) Show that the average density of E field equals to the average energy density of the B field.

(i) Wavelength, $\lambda =\frac{c}{v}=\frac{3\times {{10}^{8}}}{2.0\times {{10}^{10}}}=1.5\times {{10}^{-2}}m$ (ii) ${{B}_{0}}=\frac{{{E}_{0}}}{c}=\frac{48}{3\times {{10}^{8}}}=1.6\times {{10}^{-7}}T$ (iii) Average energy density of E field ${{u}_{E}}=\frac{1}{2}{{\varepsilon }_{0}}E_{0}^{2}$ Average energy density of B field ${{u}_{B}}=\frac{1}{2{{\mu }_{0}}}B_{0}^{2}$ But ${{E}_{0}}=c{{B}_{0}}$ and ${{c}^{2}}=\frac{1}{{{\mu }_{0}}{{\varepsilon }_{0}}}$ $\therefore \,\,{{u}_{E}}=\frac{1}{2}{{\varepsilon }_{0}}E_{0}^{2}=\frac{1}{2}{{\varepsilon }_{0}}{{(c{{B}_{0}})}^{2}}$ $=\frac{1}{2}{{\varepsilon }_{0}}.\frac{1}{{{\mu }_{0}}{{\varepsilon }_{0}}}.{{B}^{2}}=\frac{1}{2{{\mu }_{0}}}B_{0}^{2}={{u}_{B}}$