• # question_answer In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of $2\times {{10}^{10}}$ Hz and amplitude 48 V/m. (i) What is the wavelength of the wave? (ii) What is the amplitude of the oscillating magnetic field? (iii) Show that the average energy density of the E field equals the average energy density of the B field.

Given, frequency of oscillation, $f=2\times {{10}^{10}}$ Hz Electric field amplitude,    ${{E}_{0}}=48V/m$ Since, we know that,       $c=3\times {{10}^{8}}m/s$ (i) Wavelength of the waves,             $\lambda =\frac{c}{f}=\frac{3\times {{10}^{8}}}{2\times {{10}^{10}}}=1.5\times {{10}^{-2}}m$ (ii) Using the formula,$c=\frac{{{E}_{0}}}{{{B}_{0}}}$ The amplitude of the oscillating magnetic field,                         ${{B}_{0}}=\frac{{{E}_{0}}}{c}=\frac{48}{3\times {{10}^{8}}}$                         $=1.6\times {{10}^{-7}}\,\,T$ (iii) The average energy density of electric field, ${{u}_{E}}=\frac{1}{2}{{\varepsilon }_{0}}E_{0}^{2}$                     ?(i) We know that,    $\frac{{{E}_{0}}}{{{B}_{0}}}=c$ Putting in Eq. (i), we get $\therefore$      ${{u}_{E}}=\frac{1}{2}{{\varepsilon }_{0}}\cdot {{c}^{2}}B_{0}^{2}$                                ?(ii) Speed of electromagnetic waves,             $c=\frac{1}{\sqrt{{{\mu }_{0}}{{\varepsilon }_{0}}}}$                         ?(iii) Putting Eq. (iii) in Eq. (ii), we get             ${{u}_{E}}=\frac{1}{2}{{\varepsilon }_{0}}B_{0}^{2}\cdot \frac{1}{{{\mu }_{0}}{{\varepsilon }_{0}}}$             $=\frac{1}{2}\cdot \frac{B_{0}^{2}}{{{\mu }_{0}}}={{u}_{B}}$ Thus, the average energy density of the E field equals the average energy density of B field.