A) the wavelength of the wave is \[1.5\times {{10}^{-5}}\text{ }m\]
B) the amplitude of the oscillating magnetic field is \[16\times {{10}^{-3}}T\]
C) the average energy density of the E is equal to the average energy density of the B. \[[c=3\times {{10}^{8}}\,m\,{{s}^{-1}}.]\]
D) None of these
Correct Answer: C
Solution :
[c] Here c\[=3\times {{10}^{8}}m/s\], \[v=2\times {{10}^{10}}\,Hz\], \[{{E}_{0}}=48V/m\] \[\lambda =?\], \[{{B}_{0}}=?\], \[\overrightarrow{{{u}_{E}}}=?\], \[\overrightarrow{{{u}_{B}}}=?\] \[\because \,\,c=v\lambda \] \[\Rightarrow \,\lambda =\frac{c}{v}=\frac{3\times {{10}^{8}}}{2\times {{10}^{10}}}=1.5\times {{10}^{-2}}m\] \[\frac{{{E}_{0}}}{{{B}_{0}}}=c\Rightarrow {{B}_{0}}=\frac{48}{3\times {{10}^{8}}}=16\times {{10}^{-8}}T\]. Energy density due to electric field and magnetic field \[{{u}_{E}}=\frac{1}{2}{{\varepsilon }_{o}}{{E}^{2}}\,\And \,{{u}_{B}}=\frac{1}{2}{{\mu }_{o}}{{B}^{2}}\] We have \[E=cB\] and \[{{C}^{2}}=\frac{1}{{{\mu }_{o}}{{\varepsilon }_{o}}}\] so, \[{{U}_{E}}=\frac{1}{2}{{\varepsilon }_{o}}{{(cB)}^{2}}=\frac{1}{2}{{\varepsilon }_{o}}\left( \frac{1}{{{\mu }_{o}}{{\varepsilon }_{o}}} \right){{B}^{2}}=\frac{1}{2{{\mu }_{o}}}{{B}^{2}}={{U}_{B.}}\]
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