Mumbai University > Electronics Engineering > Sem 5 > Electromagnetic Engineering

Marks: 10 Marks

Year: May 2016

Poynting theorem generally provides the direction of EM wave and the power associated with it. That can be obtained using cross product of $\bar{E}$ and $\bar{H}$ hence,

$\bar{P}= \bar{E} × \bar{H}=|\bar{E} | a_x ×|\bar{H} | a_y \\ \bar{P}= |\bar{E} ||\bar{H} | (a_x × a_y ) \\ \bar{P}=|P| \overline{a_z} w/m^2 \\ |\bar{E} | a_x$

Where $\bar{P}$ is called pointing vector which shows the direction of EM wave and the power associated with it

$\bar{P}= \bar{E} × \bar{H}$

Taking divergence on both sides

$∇.\bar{P}= ∇.(\bar{E} × \bar{H}) \\ ∇.\bar{P}= \bar{H} (∇×\bar{E} )- \bar{E} (∇×\bar{H}) \\ ∇.\bar{P}= \bar{H} \bigg(-μ \dfrac{d\bar{H}}{dt}\bigg)- \bar{E} \bigg(σ\bar{E} +E \dfrac{d\bar{E}}{dt}\bigg) \\ ∇.\bar{P}= - μ H \dfrac{d\bar{H}}{dt}- \bar{E} σ \bar{E}-E \dfrac{dE }{dt} \\ ∇.\bar{P}=\dfrac{- μ}{2} \dfrac{d}{dt} |\bar{H} |- σ |\bar{E} |^2-\dfrac{E}{2} \dfrac{d}{dt} |\bar{E} |^2$

In integrated format-

$∮_s\bar{P} .d\bar{s}= -∮_u \bigg[σ |\bar{E} |^2+ \dfrac{E}{2} \dfrac{d}{dt} |\bar{E} |^2 v+ \dfrac{μ}{2} \dfrac{d}{dt} |\bar{H} |^2 \bigg]$

Average poynting vector:

$|P|= |\bar{E} ||\bar{H}| \\ |E|= E_o \cos⁡(wt.β_2) \\ |\bar{H} |= H_o \cos⁡(wt.β_2 ) \\ =\dfrac{E_o}{η} \cos⁡(wt.β_2 ) \\ |P|=\dfrac{E_o^2}{η} \cos^2⁡(wt.β_2) \\ \text{Instantaneous power} = \dfrac{E_o^2}{η} \bigg[\dfrac{1- \cos⁡ 2 (wt .β_2}{2}\bigg] \\ P_{avg}=\dfrac{1}{T} ∫_0^T|P|dt \\ =\dfrac{E_o^2}{2Tη} ∫_0^T 1- \cos⁡2 (wt.β_2) \\ P_{avg}= \dfrac{ E_o^2}{2η}$