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Energy Density And Intensity Of EM Waves - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Energy Density and Intensity of EM waves is considered one the most difficult concept.

  • 48 Questions around this concept.

Solve by difficulty

An electromagnetic wave of frequency 1x1014 hertz is propagating along the z-axis. The amplitude of the electric field is 4 V/m.If \epsilon _{0} = 8.8 x 10-12 C2/N-m2, then the average energy density of the electric field will be :

For the plane electromagnetic wave given by \mathrm{E}=\mathrm{E}_0 \sin (\omega \mathrm{t}-\mathrm{kx})$ and $\mathrm{B}=\mathrm{B}_0 \sin (\omega t-k x), the ratio of average electric energy density to average magnetic energy density is

The energy density associated with electric field E and magnetic field B of an electromagnetic wave in free space is given by

 ( \varepsilon_0 -  permittivity of freespace , \mu_0 - permeability of free space)

 

Concepts Covered - 1

Energy Density and Intensity of EM waves

Energy Density and Intensity of EM waves-

1. Energy Density-

\begin{array}{l}{\text { The electric and magnetic fields in a plane electromagnetic wave }} {\text { are given by }} \\ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {\qquad \begin{array}{l}{E=E_{0} \sin \omega(t-x / c)} \\ \\ {\text { and, } B=B_{0} \sin \omega(t-x / c)}\end{array}}\end{array}

\begin{array}{l}{\text { In any small volume } d V, \text { the energy of the electric field is }} \\ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {\qquad U_{E}=\frac{1}{2} \varepsilon_{0} E^{2} d V}\end{array}

\begin{array}{l}{\text { And the energy of the magnetic field is } U_{B}=\frac{1}{2 \mu_{0}} B^{2} d V} \\ \\ \\ {\text { Thus, the total energy is } U=\frac{1}{2} \varepsilon_{0} E^{2} d V+\frac{1}{2 \mu_{0}} B^{2} d V} \\ \\ \\ {\text { The energy density is } u=\frac{1}{2} \varepsilon_{0} E^{2}+\frac{1}{2 \mu_o} B^{2}} \\ \\ \\ {u=\frac{1}{2} \varepsilon_{0} E_{0}^{2} \sin ^{2} \omega(t-x / c)+\frac{1}{2 \mu_{0}} B_{0}^{2} \sin ^{2} \omega(t-x / c)}\end{array}

For taking the average over a long time, the sin2 terms have an average value of  {1}/{2}

 \text { So, } u_{\mathrm{av}}=\frac{1}{4} \varepsilon_{0} E_{0}^{2}+\frac{1}{4 \mu_{0}} B_{0}^{2}

\text {Now, as we know, } \quad E_{0}=c B_{0} \text { and } \mu_{0} \varepsilon_{0}=\frac{1}{c^{2}} \text { so that, } \mu_{0}=\frac{1}{\varepsilon_{0} c^{2}} \text { and, }

                                        B_{0}=\frac{E_{0}}{c} \frac{1}{4 \mu_{0}} B_{0}^{2}=\frac{\varepsilon_{0} c^{2}}{4}\left(\frac{E_{0}}{c}\right)^{2}=\frac{1}{4} \varepsilon_{0} E_{0}^{2}

\begin{array}{l}{\text { Thus, the electric energy density is equal to the magnetic density }} {\text { in average, }} \\ \\ {\text { or, } \quad u_{\mathrm{av}}=\frac{1}{4} \varepsilon_{0} E_{0}^{2}+\frac{1}{4} \varepsilon_{0} E_{0}^{2}=\frac{1}{2} \varepsilon_{0} E_{0}^{2}} \\ \\ {\text { Also } u_{\mathrm{av}}=\frac{1}{4 \mu_{0}} B_{0}^{2}+\frac{1}{4 \mu_{0}} B_{0}^{2}=\frac{1}{2 \mu_{0}} B_{0}^{2}}\end{array}

 

Intensity (I): The energy crossing per unit area per unit time, perpendicular to the direction of propagation of EM wave is called intensity.

So,

                \text { i.e. } I=\frac{\text { Total EM energy }}{\text { Surface area } \times \text { Time }}=\frac{\text { Total energy density } \times \text { Volume }}{\text { Surface area } \times \text { Time }}

 

                                             \Rightarrow I=u_{a v} \times c=\frac{1}{2} \varepsilon_{0} E_{0}^{2} c=\frac{1}{2} \frac{B_{0}^{2}}{\mu_{0}} \cdot c \ \ \ \frac{W a t t}{m^{2}}

Momentum: Electromagnetic waves also carries momentum. As we know linear momentum is associated with energy and speed.

So we can write that if wave incident on a completely absorbing surface then momentum delivered will be equal to - 

                                                                                            p=\frac{u}{c}

But if the wave is incident on a totally reflected surface, then the momentum will be equal to - 

                                                                                        -p=\frac{2 u}{c}

 

Poynting vector (\vec{S} ) : It is defined as the rate of flow of energy crossing a unit area in electromagnetic waves. So, 

                                                             \vec{S}=\frac{1}{\mu_{o}}(\vec{E} \times \vec{B})=c^{2} \varepsilon_{0}(\vec{E} \times \vec{B})

Unit of Poynting vector is Watt/m2. Now, as we know that in electromagnetic waves, \vec{E} and \vec{B} are perpendicular to each other. So, 

                                                            |\vec{S}|=\frac{1}{\mu_{0}} E B \sin 90^{\circ}=\frac{E B}{\mu_{0}}=\frac{E^{2}}{\mu C}

The importance of the Poynting vector is that the direction of the Poynting vector \vec{S} at any point gives the wave's direction of travel and direction of energy transport to the point.

The average value of the Poynting vector - 

                                                               \vec{S}=\frac{1}{2 \mu_{0}} E_{0} B_{0}=\frac{1}{2} \varepsilon_{0} E_{0}^{2} c=\frac{c B_{0}^{2}}{2 \mu_{0}}

As we can notice that direction of Poynting vector can be given by the vector product so, The direction of \vec{S}does not oscillate but its magnitude varies between zero and a maximum each quarter of the period.

 

 

Radiation pressure: It is defined as the momentum imparted per second per unit area on which the light falls

So, for the perfectly absorbing body, we can write in terms of the Poynting vector - 

                                                                                         P_{a}=\frac{ S}{c}

And for perfectly reflecting surface - 

                                                                                         P_{r}=\frac{2 S}{c}

 

Wave impedance (Z): As the word, impedance tells that obstruction inflowing of something, similarly here, the medium offers hindrance to the propagation of the wave. Such hindrance is called wave impedance and it is given by - 

                                                                         Z=\sqrt{\frac{\mu}{\varepsilon}}=\sqrt{\frac{\mu_{r}}{\varepsilon_{r}}} \sqrt{\frac{\mu_{0}}{\varepsilon_{0}}}

For vacuum or free space - 

                                                                              Z=\sqrt{\frac{\mu_{0}}{\varepsilon_{0}}}=376.6 \Omega

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Energy Density and Intensity of EM waves

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