Poynting Theorem

The poynting vector can be defined as the energy transported per unit area per unit time at a field point. This note provides us an information on poynting theorem.

Summary

The poynting vector can be defined as the energy transported per unit area per unit time at a field point. This note provides us an information on poynting theorem.

Things to Remember

The poynting vector can be defined as the energy transported per unit area per unit time at a field point. 

 Poynting vector is denoted by \(\vec P\) and can be represented as,

\begin{align*} \vec P &= \frac {1}{\mu_0} (\vec E \times \vec B) \\ \text {or,}\: \vec P &= (\vec E \times \vec H) \end{align*}

Pointing theorem is represented by \(\int \vec P. d\vec s = \frac {d}{dt} \int \frac 12 (\vec E.\vec D + \vec H.\vec B)dv - \int (\vec E. \vec J)dv, \)

MCQs

No MCQs found.

Subjective Questions

Q1:

  1. What is xDSL? What are the difference between ADSL and SDSL?
  2. Explain about Digital Audio/Video Broadcasting.

Type: Short Difficulty: Easy

Q2:

  1. What is xDSL? What are the difference between ADSL and SDSL?
  2. Explain about Digital Audio/Video Broadcasting.

Type: Short Difficulty: Easy

Videos

No videos found.

Poynting Theorem

Poynting Theorem

Poynting vector

Energy is transported through the space by means of electromagnetic waves. The pointing vector can be defined as the energy transported per unit area per unit time at a field point. Pointing vector is denoted by \(\vec P\) and can be represented as,

\begin{align*} \vec P &= \frac {1}{\mu_0} (\vec E \times \vec B) \\ \text {or,}\: \vec P &= (\vec E \times \vec H) \end{align*}

Energy of a charged particle in an electromagnetic field

There are four electromagnetic equations. They are:

  1. \(\nabla . \vec {D} = \rho\)
  2. \(\nabla . \vec B = 0\)
  3. \(\nabla \times \vec E = -\frac {\partial \vec {B}}{\partial t}\)
  4. \(\nabla \times \vec H = \vec j + \frac {\partial \vec {B}}{\partial t}\)

Now, from last two equations it becomes,

\begin{align*} \vec H.(\nabla \times \vec E) &= - \vec H. \frac {\partial \vec B}{\partial t}, \dots (1) \\ \vec E. (\nabla \times \vec H) &= \vec E. \left (\vec J + \frac {\partial \vec D}{\partial t}\right ),\dots (2) \\ \end{align*}

From equations (1) and (2), is can be written as,

\begin{align*} \vec H.(\nabla \times \vec E) - \vec E .(\nabla \times \vec H) &= - \vec H. \frac {\partial \vec B}{\partial t} - \vec E. \left (\vec J + \frac {\partial D}{\partial t}\right ) \\ &= - \left (\vec H. \frac {\partial \vec B}{\partial t}+ \vec E . \frac {\partial \vec D}{\partial t}\right ) - \vec E. \vec J, \dots (3) \end{align*}

By using vector property,

$$ \nabla .(\vec E \times \vec H) = \vec H.(\nabla \times \vec E) - \vec E.(\nabla \times \vec H), \dots (4) $$

Using equation (3) and (4), it becomes,

\begin{align*} \nabla . (\vec E \times \vec H) = - \left (\vec H. \frac {\partial \vec B}{\partial t} + \vec E. \frac {\partial D}{\partial t}\right ) - \vec E. \vec J, \dots (5) \end{align*}

For an isotropic medium, \(\vec D = \epsilon \vec E\) and \(\vec B = \mu \vec H\), then

\begin{align*} \vec E. \frac {\partial \vec D}{\partial t} &= \vec E. \frac {\partial }{\partial t}(\epsilon \vec E) \\ &= \frac 12 \epsilon \frac {\partial }{\partial t}(\vec E. \vec E) \\ & = \frac 12 \frac {\partial }{\partial t}(\vec E. \epsilon \vec E) \\ &= \frac 12 \frac {\partial }{\partial t} (\vec E. \vec D) \\ &= \frac {\partial }{\partial t} \left (\frac {\vec E.\vec D}{2} \right ) \\ \end{align*}

\begin{align*} \text {Similarly,} \\ \vec H. \frac {\partial \vec B}{\partial t} &= \vec .\:\frac {\partial }{\partial t}(\mu \vec H) = \frac {\partial }{\partial t} \left (\frac {\vec H. \vec B}{2}\right) \\ \text {Now equation}\: (5)\: \text {becomes,}\\ \nabla . (\vec E \times \vec H) &= -\left (\frac {\partial }{\partial t} \frac {\vec H.\vec B}{2} + \frac {\partial }{\partial t} \frac {\vec E. \vec D}{2}\right ) - \vec E. \vec J \\ &= \frac {\partial }{\partial t}\left [\frac 12 (\vec E. \vec D + \vec H . \vec B)\right ] -\vec E.\vec J, \dots (6)\end{align*}

By integrating equation (6), it becomes,

\begin{align*} \nabla .(\vec E \times \vec H)dV = -\int \left [\frac {\partial }{\partial t}\left( \frac 12 (\vec E. \vec D + \vec H . \vec B) \right )\right ] - \int (\vec E. \vec J)dv, \dots (7)\end{align*}

On re-arranging the above equation, it becomes

\begin{align*} \int (\vec E.\vec J)dv + \int \frac 12\frac {\partial }{\partial t}(\vec E. \vec D + \vec H.\vec B)dv + \int (\vec E \times \vec H)\: ds = 0 \end{align*}

Here, the 1st term \(\int (\vec E. \vec J)dv = \frac {dw}{dt}\) is the rate of work done by the field on charges.

And the 2nd term,

\begin{align*} \int \frac 12\frac {\partial }{\partial t}(\vec E. \vec D + \vec H.\vec B)dv &= \frac {d}{d t}\int \frac 12(vec E. \vec D + \vec H.\vec B)dv\\ &= \frac {d}{dt}\int \left (\frac 12 \epsilon E^2 + \frac 12\mu H^2\right )dv\end{align*}

Since, \(\vec D = \epsilon \vec E\) and \(\vec B = \mu \vec H\).

This 2nd term represents the rate of increase of energy stored in electric and magnetic fields in the volume V.

Since pointing vector is represented as

\begin{align*} \vec P = (\vec E \times \vec H), \\ \text {equation}\: (7)\: \text {takes a form}, \\ \int \vec P. d\vec s = \frac {d}{dt} \int \frac 12 (\vec E.\vec D + \vec H.\vec B)dv - \int (\vec E. \vec J)dv, \dots (8)\end{align*}

Expression represented by equation (8) is known as poynting theorem.

Bibliography

P.B. Adhikari, Bhoj Raj Gautam, Lekha Nath Adhikari. A Textbook of Physics. kathmandu: Sukunda Pustak Bhawan, 2011.

Jha, V. K.; 'Lecture title'; Maxwell's Equation; St. Xavier's College, Kathmandu; 2016.

Lesson

Maxwell's Electromagnetic Equations

Subject

Physics

Grade

Bachelor of Science

Recent Notes

No recent notes.

Related Notes

No related notes.