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Question: The rms value of the electric field of the light coming from the sum is 720 N \(c^{- 1}\).The averag...

The rms value of the electric field of the light coming from the sum is 720 N c−1c^{- 1}.The average total energy density of the electromagnetic wave is :

A

3.3×10−3Jm−33.3 \times 10^{- 3}Jm^{- 3}

B

4.58×10−6Jm−34.58 \times 10^{- 6}Jm^{- 3}

C

6.37×10−9Jm−36.37 \times 10^{- 9}Jm^{- 3}

D

81.35×10−12Jm−381.35 \times 10^{- 12}Jm^{- 3}

Answer

4.58×10−6Jm−34.58 \times 10^{- 6}Jm^{- 3}

Explanation

Solution

: Total average energy density of electromagnetic wave is

υav=12ε0Erms2+12μ0BrmsB2\upsilon_{av} = \frac{1}{2}\varepsilon_{0}E_{rms}^{2} + \frac{1}{2\mu_{0}}B_{rms}^{B^{2}}

=12ε0Erms2+12μ0(Erms2c2)(∵Brms=Ermsc)= \frac{1}{2}\varepsilon_{0}E_{rms}^{2} + \frac{1}{2\mu_{0}}\left( \frac{E_{rms}^{2}}{c^{2}} \right)\left( \because B_{rms} = \frac{E_{rms}}{c} \right)

=12ε0Erms2+12μ0Erms2ε0μ0(∵c=1μ0ε0)= \frac{1}{2}\varepsilon_{0}E_{rms}^{2} + \frac{1}{2\mu_{0}}E_{rms}^{2}\varepsilon_{0}\mu_{0}\left( \because c = \frac{1}{\sqrt{\mu_{0}\varepsilon_{0}}} \right)

=12ε0Erms2+12ε0Erms2=ε0Erms2= \frac{1}{2}\varepsilon_{0}E_{rms}^{2} + \frac{1}{2}\varepsilon_{0}E_{rms}^{2} = \varepsilon_{0}E_{rms}^{2}

=8.85×10−12×(720)2= 8.85 \times 10^{- 12} \times (720)^{2}

=4.58×10−6Jm−3= 4.58 \times 10^{- 6}Jm^{- 3}