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

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Syllabus

Quick Facts

Energy Density and Intensity of EM waves is considered one the most difficult concept.
49 Questions around this concept.

Solve by difficulty

EASY

A radiation of energy 'E' falls normally on a perfectly reflecting surface. The momentum transferred to the surface is (C=Velocity of light)

$\frac{2E}{C}$

$\frac{2E}{C^{2}}$

$\frac{E}{C^{2}}$

$\frac{E}{C}$

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Out of the following options which one can be used to produce a propagating electromagnetic wave?

A charge moving at constant velocity

A stationary charge

A chargeless particle

An accelerating charge

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Concepts Covered - 1

Energy Density and Intensity of EM waves

Energy Density and Intensity of EM waves-

1. Energy Density-

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

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

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

For taking 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 that that the linear momentum is associated with energy and speed, similarly if a portion of EM wave of energy u propagating with speed c, then linear momentum

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 totally reflected surface, then the momentum will be equals 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,

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

Importance of 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 the point.

Average value of 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 vector product so, The direction of $\vec{S}$ does not oscillate but it's magnitude varies between zero and a maximum each quarter of period.

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

So, for the perfectly absorbing body, we can write in terms of 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 tell that obstruction in flowing of something, similarly here ,the medium offers hindrance to the propagation of wave. Such hindrance is called wave impedance and it is given by -

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