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.

 

Answer:

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.