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Question
Show that average value of radiant flux density ‘S’ over a single period ‘T’ is given by S = `1/(2cmu_0) E_0^2`.
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Solution
Radiant flux density S = `1/mu_0 (vecE xx vecB)`
c = `1/sqrt(mu_0 ε_0)`
or c2 = `1/(mu_0ε_0)`
`1/mu_0 = ε_0c^2`
∴ S = `ε_0c^2 (vecE xx vecB)` ......(I)
Let electromagnetic waves be propagated along X-axis so its electric and magnetic field vectors are along Y and Z axis.
∴ `vecE = E_0 cos(kx - ωt)hatj`
`vecB = B_0 cos(kx - ωt)hatk`
`vecE xx vecB = (E_0B_0) cos^2 (kx - ωt)(hatj xx hatk)`
Put E × B in I
∴ S = `ε_0c^2 E_0B_0 cos^2 (kx - ωt)hati`
So average value of the magnitude of radiant flux density over complete cycle is
`S_(av) = c^2 ε_0E_0B_0 [1/T] int_0^T cos^2 (kx - ωt)dt hati`
= `(c^2ε_0E_0B_0)/T [T/2] = c^2/2 ε_0E_0 (E_0/c)` .....`[because c = E_0/B_0 or B_0 = E_0/c_0]`
= `(cε_0E_0^2)/2 [c = 1/sqrt(mu_0ε_0) or ε_0 = 1/(mu_0c^2)]`
= `c/2 * 1/(mu_0c^2) E_0^2`
`S_(av) = E_0^2/(2 mu_0c)` Hence proved.
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