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Show that average value of radiant flux...

Show that average value of radiant flux density 'S' over a single period 'T' is given by `S=1/(2cmu_0)E_0^2`.

Text Solution

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We know that radiant flux density S (called Poynting vector) is given by
`vecS=1/(mu_0) (vecExxvecB)=c^2in_0(vecExxvecB) [ :' c=1/(sqrt(mu_0in_0))]`
Let the e.m. wave be propagating along x-axis. The electric field vector of e.m. wave be along y-axis and
magnetic field vecotr be along z-axis. Therefore,
`vecE=vecE_0cos (kx-omegat) and vecB=vecB_0 cos (kx-omegat) :. vecExxvecB=(vecE_0xxvecB_0)cos^2(kx-omegat)`
`vecS=c^2 epsilon_0(vecExxvecB)=c^2 epsilon_0 (vecE_0xxvecB_0)cos^2 (kx-omegat)`
Average value of the magnitude of radiant flux density over complete cycle is
`s_(av)=c^2in_0|vecE_0xxvecB_0| 1/T int_0^T cos^2(kx-omegat)dt`
`=c^2in_0E_0B_0xx1/TxxT/2 [ :' int_0^T cos^2 (kx-omegat)dt=T/2]`
`s_(av) =(c^2)/2in_0E_0((E_0)/c) [As c=E_0//B_0]`
`=c/2 in_0E_0^2=c/2xx1/(c^2mu_0) E_0^2 [c=1/(sqrt(mu_0 in_r)) or in_0=1/(c^2mu_0)]`
`=(E_0^2)/(2mu_0c)`
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