Electromagnetic Waves

JEE Main level, single correct, medium

37 of 165

Question 37 of 165

In a plane EM wave, the electric field oscillates sinusoidally at a frequency of 5×1010Hz5\times {10}^{10}Hz and an amplitude of 50Vm–150V{m}^{–1}. The total average energy density of the electromagnetic field of the wave is : [Use ϵ0=8.85×10–12C2N−1m−2{\epsilon }_{0}=8.85\times {10}^{–12}{C}^{2}{N}^{-1}{m}^{-2}]

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Answer: (A) 1.106×10–8Jm–31.106\times {10}^{–8}J{m}^{–3}

Total average energy density of the electromagnetic field is given by, UE=12ϵ0(E0)2{U}_{E}=\dfrac{1}{2}{\epsilon }_{0}{\left({E}_{0}\right)}^{2}.
⇒UE=12×8.85×10−12×(50)2\Rightarrow {U}_{E}=\dfrac{1}{2}\times 8.85\times {10}^{-12}\times {\left(50\right)}^{2}
=1.106×10–8Jm–3=1.106\times {10}^{–8}J{m}^{–3}

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