Electromagnetic Waves

JEE Main level, single correct, medium

50 of 165

Question 50 of 165

The ratio of average electric energy density and total average energy density of electromagnetic wave is:

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Answer: (D) 12\dfrac{1}{2}

Average electric energy density is stated as uE=12ϵ0E2{u}_{E}=\dfrac{1}{2}{\epsilon }_{0}{E}^{2}
Total average energy density U=12ϵ0E2+12μ0B2U=\dfrac{1}{2}{\epsilon }_{0}{E}^{2}+\dfrac{1}{2{\mu }_{0}}{B}^{2}
=12ϵ0E2+12μ0E2c2=\dfrac{1}{2}{\epsilon }_{0}{E}^{2}+\dfrac{1}{2{\mu }_{0}}\dfrac{{E}^{2}}{{c}^{2}} (As B=EcB=\dfrac{E}{c})
=12ϵ0E2+12μ0E2ϵ0μ0=12ϵ0E2+12E2ϵ0=ϵ0E2=\dfrac{1}{2}{\epsilon }_{0}{E}^{2}+\dfrac{1}{2{\mu }_{0}}{E}^{2}{\epsilon }_{0}{\mu }_{0} =\dfrac{1}{2}{\epsilon }_{0}{E}^{2}+\dfrac{1}{2}{E}^{2}{\epsilon }_{0} ={\epsilon }_{0}{E}^{2}
Hence, the ratio, uEU=12\dfrac{{u}_{E}}{U}=\dfrac{1}{2}.

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