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.

Answer:

(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 \[{{c}^{2}}=\frac{1}{{{\mu }_{0}}{{\varepsilon }_{0}}}\] \[\therefore \,\,{{u}_{E}}=\frac{1}{2}{{\varepsilon }_{0}}E_{0}^{2}=\frac{1}{2}{{\varepsilon }_{0}}{{(c{{B}_{0}})}^{2}}\] \[=\frac{1}{2}{{\varepsilon }_{0}}.\frac{1}{{{\mu }_{0}}{{\varepsilon }_{0}}}.{{B}^{2}}=\frac{1}{2{{\mu }_{0}}}B_{0}^{2}={{u}_{B}}\]