Poynting Theorem

Last Updated : 04 Jan, 2024

The Poynting Theorem which was named After the British Physicist John Henry Poynting is a concept in electromagnetism that describes the energy inflow in an electromagnetic field. It Establishes a connection between the electromagnetic fields and the rate of energy transfer in a given region of space. Mathematically, it’s expressed as the cross-product of the electric field( E) and the magnetic field( H). It represents the power per unit area, or intensity, of the electromagnetic field. In this article, we will be Going Through The Poynting Theorem and its Mathematical Representations and Derive Some Equations.

Table of Content

What is the Poynting Theorem?
Mathematical Representation
Proof of Poynting Theorem
Derivation of Average power
Derivation of Average Power Density
Advantages and Disadvantages
Application
Solve Example

What is the Poynting Theorem?

Poynting Theorem Explains How Electromagnetic Energy Travels Through Space. It Relates changes in Energy Density to the Poynting Vector, which indicates the direction and Amount of Energy transferred in the Electromagnetic fields.

Given below are some of the basic prerequisites of the Poynting Theorem:

Volume charge density: Volume charge density is represented by ( $\rho$ ). Volume charge density is the measure of the amount of electric charge per unit volume in a given region.
Volume current density: (J): It is a measure of the amount of electric current per unit volume in a given region.
Magnetic field (H): A magnetic field is the area surrounding a magnet or current-carrying conductor where it exerts magnetic forces on nearby objects.
Electric field (E): It is the Region surrounding charge particles where a force is exerted on other charged particles, influencing their movement and behavior.

Mathematical Representation

When electromagnetic waves travel in space, they transport energy. Energy density is linked to the electric and magnetic fields.

The Poynting vector represents the rate of energy propagation per unit area perpendicular to the incident energy flow.

$P = \mathbf{E} \times \mathbf{H}$

$P=EH\sin\theta$

The electric field (E) and magnetic field (H) vectors are oriented perpendicular to each other and to the direction of wave propagation.

where P is rate of energy transfer and poynting vector

$P = \frac{\mathbf{E} \times \mathbf{B}}{\mu}$

Proof of Poynting Theorem

Statement: At any point in a uniform plane wave, the cross product of electric field E and magnetic field H gives a measure of the rate of energy flow per unit area at that point.

$P = \mathbf{E} \times \mathbf{H}$

using the Maxwell equation We can Write as:

$\nabla \times \mathbf{H}= \mathbf{J} + \frac{\partial \mathbf{D}}{\partial t}$ -(1)

$\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$ -(2)

taking a dot product with E

$\mathbf{E} \cdot (\nabla \times \mathbf{H}) = \mathbf{E} \cdot \mathbf{J} + \mathbf{E} \cdot \frac{\partial \mathbf{D}}{\partial t}$ -(3)

using identity, we can Write

$\nabla \cdot (\mathbf{E} \times \mathbf{H}) = \mathbf{H} \cdot (\nabla \times \mathbf{E}) - \mathbf{E} \cdot (\nabla \times \mathbf{H})$

$\mathbf{E} \cdot (\nabla \times \mathbf{H}) = \mathbf{H} \cdot (\nabla \times \mathbf{E}) - \nabla \cdot (\mathbf{E} \times \mathbf{H})$

put the value of Equation 3

$\mathbf{H} \cdot (\nabla \times \mathbf{E}) - \nabla \cdot (\mathbf{E} \times \mathbf{H}) = \mathbf{E} \cdot \mathbf{J} + \mathbf{E} \cdot \frac{\partial \mathbf{D}}{\partial t}$ -(4)

$\nabla \cdot (\mathbf{E} \times \mathbf{H}) = \mathbf{H} \cdot (\nabla \times \mathbf{E}) - \mathbf{E} \cdot \mathbf{J} - \mathbf{E} \cdot \frac{\partial \mathbf{D}}{\partial t}$ -(5)

From Equation 2, take dot product with H

$\nabla \cdot (\nabla \times \mathbf{E}) = -(\mathbf{H} \cdot \frac{\partial \mathbf{H}}{\partial t})\kappa$

….where $\varkappa$ = permeability

put the value in Equation 5

$\nabla \cdot (\mathbf{E} \cdot \mathbf{H}) = -(\mathbf{H} \cdot \frac{\partial \mathbf{H}}{\partial t})\kappa - \mathbf{E} \cdot \mathbf{J} - (\mathbf{E} \cdot \frac{\partial \mathbf{E}}{\partial t})\epsilon$ -(6) Where $D=\epsilon E$

Lets us consider the term

$\frac{\partial}{\partial t}(\mathbf{H} \cdot \mathbf{H}) = \mathbf{H} \cdot \frac{\partial \mathbf{H}}{\partial t} + \mathbf{H} \cdot \frac{\partial \mathbf{H}}{\partial t} = 2(\mathbf{H} \cdot \frac{\partial \mathbf{H}}{\partial t}) = \frac{1}{2} \frac{\partial}{\partial t}(\mathbf{H}^2)$

Similarly, we can Write as

$\mathbf{E} \cdot \frac{\partial \mathbf{E}}{\partial t} = \frac{1}{2} \frac{\partial}{\partial t}(\mathbf{E}^2)$

put these values in Equation 6

$\nabla \cdot (\mathbf{E} \times \mathbf{H}) = -\frac{\kappa}{2}\frac{\partial \mathbf{H}^2}{\partial t} - \mathbf{E} \cdot \mathbf{J} - \frac{\epsilon}{2}\frac{\partial \mathbf{E}^2}{\partial t}$

Taking volume integral

$\int_{v} \nabla \cdot (\mathbf{E} \times \mathbf{H}) \,dv = -\frac{\mu}{2} \int_{v} \frac{\partial}{\partial t}\left(\frac{\mathbf{H}^2}{2}\right) \,dv - \int_{v} (\mathbf{E} \cdot \mathbf{J}) \,dv - \frac{\epsilon}{2} \int_{v} \frac{\partial}{\partial t}(\mathbf{E}^2) \,dv$

$\oint \nabla \cdot (\mathbf{E} \times \mathbf{H}) \,dv = -\int_{v} (\mathbf{E} \cdot \mathbf{J}) \,dv - \frac{\partial}{\partial t} \int_{v} (\mu \mathbf{H}^2/2 + \epsilon \mathbf{E}^2/2) \,dv$

(rate of energy flow) and $\int_{v} (\mathbf{E} \cdot \mathbf{J}) \,dv$ = power loss, $\frac{\partial}{\partial t} \int_{v} \left( \frac{\epsilon \mathbf{E}^2}{2} + \frac{\mu\mathbf{H}^2}{2}\right) \,dv$ =rate of decrease in electromagnetic wave energy.

Hence, poynting theorem is proved

In a uniform plane wave, the first term and next term combine, then something energy loss occurs. These energy rates per unit area are called the poynting vector.

$P = \mathbf{E} \times \mathbf{H}$

Note: Take all Taking All quantity in the vector form.

Derivation of Average power by Poynting Theorem

Let us Derive Average Power by Poynting Theorem:

We know

power density vector $P = \mathbf{E} \times \mathbf{H}$

$P=EH\sin\theta$

$\theta=90$

then,

$|P|=|E||H|$ instantaneous poynting vector

So, $|E| = E \cos(wt \cdot \beta_2)$ -(1)

$|\bar{H}| = H \cos(wt \cdot \beta_2)$

= $\frac{E}{\eta} \cos(wt \cdot \beta_2)$ -(2)

Where $\eta$ is constant

In Equation (1) and (2) putting the average poynting vector

$|P| = \frac{E_0^2}{\eta} \cos^2(wt \cdot \beta_2)$

So,

Instantaneous power = $E_0^2 \frac{1 - \cos^2(wt \cdot \beta_2)}{2\eta }$

For average power, we can proceed as

$P_{\text{avg}} = \frac{1}{T} \int_{0}^{T} |P| \, dt$

$=\frac{E_0^2}{2T\eta} \int_{0}^{T} [1 - \cos^2(wt \cdot \beta_2)] \, dt$

$P_{\text{avg}} = \frac{E_0^2}{2\eta}$

$P_{\text{avg}} = \frac{E_0^2}{2\eta}$

where P_avg is the average power over one period of the periodic signal.

E_o is the amplitude of the signal

$\eta$ is a constant.

Derivation of Average Power Density by Poynting Theorem

Let’s say that average power density is denoted as (S_avg)

$S_{\text{avg}} = \frac{1}{V} \int_S \mathbf{P} \cdot d\mathbf{A}$

$S_{\text{avg}} = -\mu\epsilon\omega^2 \int_S \mathbf{E_o \times H_o} \cdot d\mathbf{A}$

We know that

wave velocity $C = \frac{1}{\sqrt{\epsilon\mu} }$

$\frac{\omega^2 \lambda ^2 }{4\pi^2} = \frac{1}{\epsilon\mu}$

$k = 2\pi/\lambda$ = $2\pi f/C$

$S_{\text{avg}} = \Kappa^2 \int_S \mathbf{E_o \times H_o} \cdot d\mathbf{A}$

$S_{\text{avg}} = \frac{4 \pi^2 f^2 }{C^2} |E_o \times H_o |\int_S d\mathbf{A}$

Where integration of A is Equal to S (surface area),

$S_{\text{avg}} = \frac{4 \pi^2 f^2 S}{C^2} |E_o \times H_o |$

where C is wave velocity and f is frequency.

calculating with help of energy density, then we can Write as

$S_{\text{avg}} = {4 \pi f^2}\mu\epsilon S \,Eo \, H_o$

Advantages and Disadvantages of Poynting Theorem

Given Below are List of Advantages and Disadvantages of Poynting Theorem:

Advantages

This theorem is valuable in the design and analysis of antennas and in helping engineering optimize their performance through energy distribution.
This theorem provides a clear and quantitative understanding of how electromagnetic energy flows through space.
This theorem establishes a connection between changes in electromagnetic energy density and the power absorbed by charges.

Disadvantages

This theorem’s mathematical expressions are very complex.
It is most applicable to time-varying electromagnetic fields and may not be directly useful for analyzing static electromagnetic situations.
This theorem is based on classical electrodynamics and may not fully account for quantum effects in certain situations.
This theorem makes assumptions such as idealized conditions and linear materials.

Application of Poynting Theorem

Wave propagation: Understanding the propagation of electromagnetic waves provides information about the direction and intensity of energy flow with the wave.
Energy Conservation: The energy conservation principle in electromagnetic systems ensures that energy is neither created nor destroyed but only converted between different forms.
Power transfer: This is used to determine the power transfer through a given area in an electromagnetic field.
Radiation Pressure: It is used to determine radiation pressure, a force that can be harnessed in technologies such as optical tweezers.

Solve Example on Poynting Theorem

Q1. The Poynting vector of an electromagnetic wave in $S = 120(\sin)^2 (9.0rad/m)z+2.7 10^9rad/s)t)k$ .What is the wavelength?

Solve:

we know that

$\mathbf{E} = E_0 \sin(kz - \omega t) \mathbf{i}$

direction of wave propagation (kz-wt): the wave is propagating in the x direction.

where $k=2\pi \lambda f=w/2\pi$

where w=angular frequency and k=wave number

$P = \frac{\mathbf{E} \times \mathbf{B}}{\mu_0}$ = $\frac{E_0 B_0 \sin^2(kz - \omega t)}{\mu_0}$

so ,Given

k=9

wave length = $\lambda$ =2*pi/9= 0.698 meter

Q2. A radio transmitter emits an electromagnetic wave with an electric field magnitude of 4v/m and a magnetic of 2 UoT. Calcualated the manitude of the poynting vector at a point where the angle E and B is 30 degree.

Solve:

Given :

E=4, B=2,

$\theta$ =30 degree

use the formula

$P=4 \times2\times10^{-6} \sin(30)=8 \times 10^{-6} \sin (30)P=4 \times 10^{-6} W/m^2.$

Q3. A LBM with an electric field magnitude of 10V/m and a magnetic field magnitude of 4 UoT is used an optical experiment .calculate the poynting vector magnitude.

solve:

Given,

E=10,B=4

$\theta$ =90

So, P=E x B

P= $10\times4\times10^{-6}$

P= $40 \times 10^{-6} \frac{W}{m^2}$

Conclusion

Poynting Theoram is a important concept of electromagnetism and it can be defined as the Change in Electromagnetic Energy Density at a Point in a Space that will equal to the Negative of the Divergence of the Poynting Vector Minus the Power Dissipated as the Heat per Unit Volume. In this article, we have tried to cover the poynting theoram as well as its representation and proof. We have also discussed the two derivations related to puynting theorem and also discussed some examples on it. In simple words, Poynting theoram gives us understanding of transportation of electromagnetic energy and how it is converted in a given region of space.

FAQs on Poynting Theorem

1. What happens if the angle theta between E and B is zero or 180 degree ?

Then poynting vector will be zero ,indicating no energy transport.

2. Can the poynting vector be used for used electric and magnetic fields ?

Yes, the poynting vector is applicable to both electric and magnetic fields .the transfer of energy between these fields in an electromagnetic wave.

3. What practical use of the poynting vector ?

Analyzing energy flow in antennas ,transmission lines, and electromagnetic devices.

4. Can Poynting’s Theorem apply to both fields?

Yes, the Poynting Theorem can be Applied in Both Fields.

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Poynting Theorem

What is the Poynting Theorem?

Mathematical Representation

Proof of Poynting Theorem

Derivation of Average power by Poynting Theorem

Derivation of Average Power Density by Poynting Theorem

Advantages and Disadvantages of Poynting Theorem

Advantages

Disadvantages

Application of Poynting Theorem

Solve Example on Poynting Theorem

Conclusion

FAQs on Poynting Theorem

1. What happens if the angle theta between E and B is zero or 180 degree ?

2. Can the poynting vector be used for used electric and magnetic fields ?

3. What practical use of the poynting vector ?

4. Can Poynting’s Theorem apply to both fields?

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