Poynting theorem

Introduction:-

Poynting theorem is used to get an expression for propagation of energy in a medium.

It gives the relation between energy stored in a time-varying magnetic field and the energy stored in time-varying electric field and the instantaneous power flow out of a given region.

EM waves propagates through space from source to destination. In order to find out power in a uniform plane wave it is necessary to develop a power theorem for the EM field known as poynting theorem.

The direction of power flow is perpendicular to $\vec{E}$ and $\overrightarrow{H}$ in the direction of plane containing $\overrightarrow{E}$ and $\overrightarrow{H}$.

i.e, it gives the direction of propagation .

$\overrightarrow{P}&space;=&space;\overrightarrow{E}X\overrightarrow{H}$  Watts/m2  (or)   VA/m2.

Proof:-

from Maxwell’s  equations $\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{H}&space;=&space;\overrightarrow{J}+\overrightarrow{J_{d}}$

$\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{H}&space;=&space;\overrightarrow{J}+\frac{\partial&space;\overrightarrow{D}}{\partial&space;t}$

$\overrightarrow{J}=\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{H}-\frac{\partial&space;\overrightarrow{D}}{\partial&space;t}$

the above equation has units of the form current density Amp/m2. When it gets multiplied by $\overrightarrow{E}$ V/m. The total units  will  have of the form power per unit volume.

$\overrightarrow{J}\rightarrow$ Amp/m2$\overrightarrow{E}\rightarrow$ Volts/m.

$EJ\rightarrow$ Amp. Volt/m3 $\rightarrow$ Watts/m3 $\rightarrow$ Power/volume.

$\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{H}&space;=&space;\overrightarrow{J}+\frac{\partial&space;\overrightarrow{D}}{\partial&space;t}$

$\overrightarrow{E}.(\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{H})&space;=&space;\overrightarrow{E}.\overrightarrow{J}+\overrightarrow{E}.\frac{\partial&space;\overrightarrow{D}}{\partial&space;t}$

$\overrightarrow{E}.(\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{H})&space;=&space;\overrightarrow{E}.\overrightarrow{J}+\overrightarrow{E}.\frac{\partial&space;(\epsilon&space;\overrightarrow{E})}{\partial&space;t}$

by using the vector identity

$\overrightarrow{\bigtriangledown&space;}.(\overrightarrow{E}&space;X&space;\overrightarrow{H})&space;=&space;\overrightarrow{H}.(\overrightarrow{\bigtriangledown&space;}X&space;\overrightarrow{E}&space;)-\overrightarrow{E}.(\overrightarrow{\bigtriangledown&space;}X&space;\overrightarrow{H}&space;)$

then

$\overrightarrow{H}.(\overrightarrow{\bigtriangledown&space;}X&space;\overrightarrow{E}&space;)-\overrightarrow{\bigtriangledown&space;}.(\overrightarrow{E}&space;X&space;\overrightarrow{H})&space;=&space;\overrightarrow{E}.\overrightarrow{J}+\overrightarrow{E}.\frac{\partial&space;(\epsilon&space;\overrightarrow{E})}{\partial&space;t}$.

from the equation of Maxwell’s $\overrightarrow{\bigtriangledown&space;}&space;X&space;\overrightarrow{E}=-\frac{\partial&space;\overrightarrow{B}}{\partial&space;t}&space;=&space;-\frac{\partial&space;(\mu&space;\overrightarrow{&space;H})}{\partial&space;t}$

$\overrightarrow{H}.(-\frac{\partial&space;(\mu&space;\overrightarrow{&space;H})}{\partial&space;t}&space;)-\overrightarrow{\bigtriangledown&space;}.(\overrightarrow{E}&space;X&space;\overrightarrow{H})&space;=&space;\overrightarrow{E}.\overrightarrow{J}+\overrightarrow{E}.\frac{\partial&space;(\epsilon&space;\overrightarrow{E})}{\partial&space;t}$

$\overrightarrow{\bigtriangledown&space;}.(\overrightarrow{E}&space;X&space;\overrightarrow{H})=-\overrightarrow{E}.\overrightarrow{J}-\overrightarrow{E}.\frac{\partial&space;(\epsilon&space;\overrightarrow{E})}{\partial&space;t}-\overrightarrow{H}.(\frac{\partial&space;(\mu&space;\overrightarrow{&space;H})}{\partial&space;t}&space;)------EQN(I)$

by using the vector identity  $\frac{\partial&space;(\overrightarrow{A}.\overrightarrow{B})}{\partial&space;t}=\overrightarrow{A}.\frac{\partial&space;\overrightarrow{B}}{\partial&space;t}+\overrightarrow{B}.\frac{\partial&space;\overrightarrow{A}}{\partial&space;t}$

if $\overrightarrow{A}&space;=&space;\overrightarrow{B}$    $\Rightarrow&space;\frac{\partial&space;(\overrightarrow{A}.\overrightarrow{A})}{\partial&space;t}=2&space;\overrightarrow{A}.\frac{\partial&space;\overrightarrow{A}}{\partial&space;t}$

$\overrightarrow{A}.\frac{\partial&space;\overrightarrow{A}}{\partial&space;t}=\frac{1}{2}\frac{\partial&space;A^{2}}{\partial&space;t}$

from EQN(I)   ,  $\overrightarrow{\bigtriangledown&space;}.(\overrightarrow{E}&space;X&space;\overrightarrow{H})=-\overrightarrow{E}.\sigma\overrightarrow{E&space;}-\frac{1}{2}\epsilon\frac{\partial&space;E^{2}}{\partial&space;t}-\frac{1}{2}\mu&space;\frac{\partial&space;H^{2}}{\partial&space;t}$

$\overrightarrow{\bigtriangledown&space;}.(\overrightarrow{E}&space;X&space;\overrightarrow{H})=-\sigma&space;E^{2}-\frac{1}{2}\frac{\partial&space;}{\partial&space;t}(\epsilon&space;E^{2}+\mu&space;H^{2})$

by integrating the above equation by over  a volume

$\int_{v}\overrightarrow{\bigtriangledown&space;}.(\overrightarrow{E}&space;X&space;\overrightarrow{H})&space;dv=-\int_{v}\sigma&space;E^{2}dv-\int_{v}\frac{1}{2}\frac{\partial&space;}{\partial&space;t}(\epsilon&space;E^{2}+\mu&space;H^{2})dv$

by converting the volume integral to surface integral

$\oint_{s}(\overrightarrow{E}&space;X&space;\overrightarrow{H}).&space;\overrightarrow{ds}=-\int_{v}\sigma&space;E^{2}dv-\int_{v}\frac{1}{2}\frac{\partial&space;}{\partial&space;t}(\epsilon&space;E^{2}+\mu&space;H^{2})dv$.

the above equation gives the statement of Poynting theorem.

Poynting theorem:-

It states that the net power flowing out of a given volume is equal to the time rate of decrease in the energy stored with in that volume V and the ohmic losses.