Learning Objectives

To enable the learners to understand the following concepts / topics, analyse and appreciate them incorporating the previous knowledge, enhance their application skills in various concepts, develop values and life skills required in furthering their overall personality.

• Maxwell’s equations and production of electromagnetic waves.
• Concept of displacement current.
• Source / origin of electromagnetic waves.
• Nature of EM waves and their expressions.
• Properties of EM waves.
• Intensity, radiation pressure and momentum of EM-waves.
• Poynting vector $(\mathbf{S})$
• Various types of waves is EM- spectrum, their names, frequencies and uses.
• Applications of EM waves.

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Introduction

James Clark Maxwell, in taking stock of our knowledge of electricity and magnetism, summarised basic results in terms of four equations known as Maxwells equations. He also showed that one of these equations known as Ampere’s circuital law needed some modification. This was achieved by introducing the concept of displacement current. Maxwell also shared that the electric and magnetic fields satisfy equation of wave motion and co-exist with one another. He postulated the existence of what is now known as electromagnetic waves. In this unit we will discuss these ideas in a very simple, elementary way.

Maxwells Equations

The four Maxwell’s equations are:

(1) Gauss Law of Electrostatics

According to Gauss law the electric flux $\phi _{\mathrm{E}}$ through any closed surface is $\frac{\mathrm{q}}{\varepsilon _{0}}$; where $\mathrm{q}$ is charge inside the closed surface. Expressed mathematically:

$$ \begin{equation*} \phi _{\mathrm{E}}=\oint _{\mathrm{S}} \mathbf{E} \cdot \mathrm{d} \mathbf{S}=\frac{\mathrm{q}}{\varepsilon _{0}} \quad \qquad \ldots . \ldots (1) \end{equation*} $$

$\mathrm{q}$ is charge in side closed surface $\mathrm{S}$

(2) Gauss Law of Magnetostatics

The net magnetic flux $\phi _{\mathrm{m}}$ through any closed surface is zero. Expressed mathematically:

$$ \begin{equation*} \phi _{\mathrm{m}}=\oint \mathbf{B} \cdot \mathrm{d} \mathbf{S}=0 \quad \qquad \ldots . \ldots (2) \end{equation*} $$

This is basically a restatement of the fact that magnetic monopoles donot exist.

(3) Ampere is Circuital Law

The law states that the line integral of magetic field over any closed surface equals $\mu _{0} \mathrm{I}$, where I is current inside the closed loop considered. Expressed mathematically:

$$ \begin{equation*} \oint \mathbf{B} . \mathrm{d} \ell=\mu _{0} \mathrm{I} \quad \qquad \ldots . \ldots (3) \end{equation*} $$

(4) Faraday’s Law of Electromagnetic Induction

The induced emf is numerically equal to the time rate of change of magnetic flux.

$$ \begin{align*} \varepsilon _{\text {ind }}=\oint \mathbf{E} \cdot \mathrm{d} \ell & =-\frac{\mathrm{d} \phi _{\mathrm{m}}}{\mathrm{dt}} \\ & =-\frac{\mathrm{d}}{\mathrm{dt}} \oint \mathbf{B} \cdot \mathrm{d} \mathbf{S} \quad \qquad \ldots . \ldots (4) \end{align*} $$

The Displacement Current

Maxwell pointed a fundamental inadequecy of the above four equations. Maxwell argued that if a varying magnetic flux produces an induced electric field and therefore an induced emf, a varying electric flux should also produce an induced magnetic field due to an induced current. In other words Maxwell demanded “Parity” between electric and magnetic fields.

Maxwell also shared that the Ampere’s circuital law as expressed in Eqn (3); in general; is not valid. As expressed in Eqn (3); the law is valid for direct currents (d.c). But is is not applicable to alternating currents.

We know that a direct current cannot flow through a capacitor. However an a.c current can flow through a capacitor, though an insulator is present between plates of the capacitor. Obviously conduction electrons ,that move in wires conneting a.c source to capacitor, cannot flow from one plate of the capacitor to the other. How circuit is completed when a.c current flows?

To overcome these difficulties Maxwell proposed that a new current, now known as displacement current $I _{d}$ is produced due to a varying electric flux. Maxwell showed that:

$$ \begin{equation*} \mathrm{I} _{\mathrm{d}}=\varepsilon _{0} \frac{\mathrm{d} \phi _{\mathrm{E}}}{\mathrm{dt}} \quad \qquad \ldots . \ldots (5) \end{equation*} $$

The modified Ampere’s circuital law is

$$ \begin{equation*} \oint \mathbf{B} \cdot \mathrm{d} \ell=\mu _{0}\left(\mathrm{I} _{\mathrm{c}}+\mathrm{I} _{\mathrm{d}}\right) \quad \qquad \ldots . \ldots (6) \end{equation*} $$

where $I _{c}$ is conduction current and $I _{d}$ is displacement current.

Though displacement current is not due to movement of free electrons; it has same effects as conduction current. We say displacement current can produce a magnetic field in the same manner as a conduction current.

Existence of Displacement Currnet

We discuss briefly, by using idea of displacement current, how do we get a continuity of current through a capacitor connected across an a.c. Let a capacitor (parallel-plate) of capacity $\mathrm{C}$ be connected across an a.c voltage given by

$$ \mathrm{V}(\mathrm{t})=\mathrm{V} _{0} \sin \omega \mathrm{t} $$

The capacity $\mathrm{C}=\frac{\mathrm{A} \varepsilon _{0}}{\mathrm{~d}}$; where $\mathrm{A}$ is aera of cross-section and $\mathrm{d}$ is distance between plates of capacitor. The instantaneous charge $\mathrm{q}(\mathrm{t})$ on capacitor is:

$$ \begin{align*} \mathrm{q}(\mathrm{t})= & C \mathrm{~V} _{(\mathrm{t})} \\ & =\left[\frac{\mathrm{A} \varepsilon _{0}}{\mathrm{~d}}\right] \mathrm{V} _{0} \sin \omega \mathrm{t} \quad \qquad \ldots . \ldots (7) \end{align*} $$

The conduction current $\mathrm{I} _{c}$; in the wires connecting a.c source to capacitor; by defintion, is

$$ \begin{equation*} I _{c}=\frac{d q(t)}{d t}=\frac{A \varepsilon _{0}}{d} V _{0} \omega \cos \omega t \quad \qquad \ldots . \ldots (8) \end{equation*} $$

According to Maxwell, in between plates of capacitor there is a varying electric flux which produces displacement current $I _{d}$. The instantaneous electric flux $\phi _{\mathrm{E}}$; in between plates of capacitor, is

$$ \begin{aligned} \mathrm{\phi}_{\mathrm{E}}=\mathrm{AE}(\mathrm{t})=\mathrm{A} \frac{\mathrm{V}(\mathrm{t})}{\mathrm{d}} \\ =\left(\frac{\mathrm{A}}{\mathrm{d}}\right) \mathrm{V} _{0} \sin \omega \mathrm{t} \end{aligned} $$

The displacement current, $\mathrm{I} _{\mathrm{d}}$, is

$$ \begin{gather*} \mathrm{I} _{\mathrm{d}}=\varepsilon _{0} \frac{\mathrm{d} \phi _{\mathrm{E}}}{\mathrm{dt}}=\left(\frac{\mathrm{A} \varepsilon _{0}}{\mathrm{~d}}\right) \frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{V} _{0} \sin \omega \mathrm{t}\right) \\ =\left(\frac{\mathrm{A} \varepsilon _{0}}{\mathrm{~d}}\right) \omega \mathrm{V} _{0} \cos \omega \mathrm{t} \quad \qquad \ldots . \ldots (9) \end{gather*} $$

From eqns (8) and (9) we have

$$ I _{d}=I _{c} $$

Let $I=I _{d}+I _{c}$ be the total current in circuit. In conducting wires $I _{d} \rightarrow 0$, because $\phi _{E} \rightarrow 0$ and $I=I _{c}$. In between plates of capacitor $\mathrm{I} _{\mathrm{c}}=0$ and $\mathrm{I}=\mathrm{I} _{\mathrm{d}}$. There is a continuity of a.c current in a capacitor.

Electromagnetic Waves

Maxwell, using modified Ampere’s law and the other three equations showed that the electric field (E) and the magnetic field (B) satisfy the equation.

$$ \nabla^{2} \mathbf{E}=\left(\frac{1}{\mu _{0} \varepsilon _{0}}\right) \frac{\partial^{2} \mathbf{E}}{\partial \mathrm{t}^{2}} $$

$$\text{and} \quad \nabla^{2} \mathbf{B}=\left(\frac{1}{\mu _{0} \varepsilon _{0}}\right) \frac{\partial^{2} \mathbf{B}}{\partial \mathrm{t}^{2}}$$

These two equatinos are standard equations of wave motion. The speed of waves $V$ is

$$ \mathrm{v}=\frac{1}{\sqrt{\mu _{0} \varepsilon _{0}}} $$

For free space

$$ \varepsilon _{0}=\frac{1}{4 \pi \times 9 \times 10^{9}} \text { SI Unit } $$

and

$$ \mu _{0}=4 \pi \times 10^{-7} \text { SI Unit } $$

$$ \begin{aligned} \therefore \quad \mathrm{v}= & \frac{1}{\left[4 \pi \times 10^{-7} \times \frac{1}{4 \pi \times 9 \times 10^{9}}\right]^{1 / 2}}=\left[9 \times 10^{16}\right]^{1 / 2} \\ \\ & =3 \times 10^{8} \mathrm{~m} / \mathrm{s} \end{aligned} $$

$\mathrm{v}$ is same as speed of light $=\mathrm{c}$, in free space. Therefore Maxwell postulated that light is electric and magnetic fields propogating in sapce! Since electric and magnetic fields can exist in vaccum, light can propagate in vaccum!

Origin of Electromagnetic Wave

Maxwell explained theoretically, the production of EM waves with the help of the two equations, namely:

$\oint \mathbf{B} \cdot \mathrm{d} \ell=\mu _{0} \varepsilon _{0} \frac{\mathrm{d} \phi _{\mathrm{E}}}{\mathrm{dt}}$

and $\oint \mathbf{E} \cdot \mathrm{d} \ell=-\frac{\mathrm{d} \phi _{\mathrm{B}}}{\mathrm{dt}}$

In order to produce a magnetic field, we require a time varying electric field in space and a time-varying electric field can only be produced by an accelerating or oscillating electric change. Hence an oscillating electric change acts as the source of EM waves. As long as the charge continues to oscillate, EM waves are produced from those oscillations.

Nature of Electromagnetic Waves

An electromagnetic wave produced by oscillating charge (and transmitted with the help of a dipole antenna) consists of time-varying E and B fields, the variation in one sustaining the production of the other. As we know, the varition in E-field is the source of displacement current $\mathrm{I} _{\mathrm{d}}$, which actually produces the $\mathrm{B}$-field. Using Bio-savart law, it can be shown that the direction of B is perpendicular to the direction of the displacement current. As the displacement current is always along the direction of E-field, it is clear that these two fields are mutually perpendicular. E and B-fields of EM waves are mutually perpendicular and they are also perpendicular to the direction of propagation of the wave. Thus EM-waves show transverse nature.

Electromagnetic waves propagate, in space with the speed equal to that of light (c). If the electric field $\mathbf{E}$ is along the $\mathbf{y}$-axis and the magnetic field $\mathbf{B}$ is along the $\mathrm{z}$-axis, then the direction of wave propagation is along the vector $\mathbf{E} \times \mathbf{B}$, that is, along the positive $x$-axis as shown in the figure below. Both $\mathbf{E}$ and $\mathbf{B}$ vary sinusoidally with time and space and have same frequency. $\mathbf{E} _{\mathrm{y}}$ and $\mathbf{B} _{\mathrm{z}}$ vary always in same phase.

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Since $\mathrm{E} _{\mathrm{y}}$ and $\mathrm{B} _{\mathrm{z}}$ are progressing along the $x$-axis, with speed equal to $\mathrm{v}$, in space, we can mathematically represent these sinusoidal field variation as under:

$ \mathrm{E} _{\mathrm{y}}=\mathrm{E} _{0} \sin (\mathrm{k} x-\omega \mathrm{t}) $

or $\quad \mathrm{E} _{\mathrm{y}}=\mathrm{E} _{0} \sin \left\{2 \pi\left(\frac{x}{\lambda}-{vt}\right)\right\}$

or $\mathrm{E} _{\mathrm{y}}=\mathrm{E} _{0} \sin \left\{2 \pi\left(\frac{x}{\lambda}-\frac{\mathrm{t}}{\mathrm{T}}\right)\right\}$

and $\mathrm{B} _{\mathrm{z}}=\mathrm{B} _{0} \sin (\mathrm{k} x-\omega \mathrm{t})$

or $\mathrm{B} _{z}=\mathrm{B} _{0} \sin \left\{2 \pi\left(\frac{x}{\lambda}-{vt}\right)\right\}$

or $\mathrm{B} _{z}=\mathrm{B} _{0} \sin \left\{2 \pi\left(\frac{x}{\lambda}-\frac{\mathrm{t}}{\mathrm{T}}\right)\right\}$

Here, $\mathrm{E} _{0}$ and $\mathrm{B} _{0}$ are the respective amplitude or peak values of electric and magnetic fields. It may be recalled here that:

$\omega$ is angular frequency of the EM wave

$v$ is linear frequency

$ \omega=2 \pi v $

$\mathrm{v}$ is wave velocity (along $x$-axis, here)

$ \mathrm{v}=\mathrm{v} \lambda $

$\lambda$ is wave length

$\mathrm{k}$ is angular wave number

$ \mathrm{k}=\frac{2 \pi}{\lambda} $

and $\quad \mathrm{v}=\frac{\omega}{\mathrm{k}}$

Using Maxwell’s equations it can be proved that:

$\mathrm{v}=\frac{\omega}{\mathrm{k}}=\frac{1}{\sqrt{\mu _{0} \varepsilon _{0}}}$, for EM waves, in free space.

or $\mathrm{v}=\frac{1}{\sqrt{\mu _{0} \varepsilon _{0}}}=\mathrm{c}$

It can also be proved, from Maxwell’s equations, that the electric field amplitude $\left(\mathrm{E} _{0}\right)$ and the magnetic field amplitude $\left(\mathrm{B} _{0}\right)$ in an electromagnetic wave are related as:

$$\frac{\mathrm{E} _{0}}{\mathrm{~B} _{0}}=\mathrm{c}$$

As we know, both E and B fields in EM wave vary is same phase, equation (15.21) can also be written as:

$$ \frac{E}{B}=C $$

[Here $\mathrm{E}$ and $\mathrm{B}$ are the instantaneous values of electric and magnetic fields, at a particular time instant $(\mathrm{t})$ ].

It is well known to all that light, an electromagnetic wave, does propagate through glass and water. We have also seen, in earlier uits, that electric and magnetic field can be established in material medium and that the total electric field inside a medium is described in terms of permittivity $(\varepsilon)$ where $\varepsilon=\varepsilon _{\mathrm{r}} \varepsilon _{0}$ and the magnetic field insdie the medium is described in terms of permeability $(\mu)$, where $\mu=\mu _{\mathrm{r}} \mu _{0}$.

Thus the speed of electromegnetic waves, and therefore light, can be written in terms of $\mu$ and $\varepsilon$ as:

$$ \mathrm{v}=\frac{1}{\sqrt{\mu \varepsilon}} $$

Here $v$ is the speed of EM waves in a medium for which $\mu$ and $\varepsilon$ are respectively permeability and permittivity of the material of the medium. It can also be written as:

$$ \begin{aligned} & v=\frac{1}{\sqrt{\mu _{\mathrm{r}} \mu _{0} \varepsilon _{\mathrm{r}} \varepsilon _{0}}} \\ & =\frac{1}{\sqrt{\mu _{\mathrm{r}} \varepsilon _{\mathrm{r}}} \times \sqrt{\mu _{0} \varepsilon _{0}}} \\ & v=\frac{c}{\sqrt{\mu _{\mathrm{r}} \varepsilon _{\mathrm{r}}}} \quad \qquad \left(\because \frac{1}{\sqrt{\mu _{0} \varepsilon _{0}}}=c\right) \\ \text { or } \quad v & =\frac{c}{\sqrt{\mu _{\mathrm{r}} \mathrm{K}}} \end{aligned} $$

Here $\mathrm{K}=\varepsilon _{\mathrm{r}}$ is the dielectric constant of the material.

Heinrich RudolfHertz not only proved that accelerating / oscillating charges produced electromagnetic waves, but also demonstrated these waves, which had wavelength ten million times that of the light waves, could be diffracted, refracted and polarised. Further Hertz produced stationary electromagnetic waves and determined their wavelength, by measuring the distance between two successive nodes. As the frequency of the wave (equal to the frequency of the oscillator) is known, he then calculated the speed of these waves using the equation $\mathrm{v}=\mathrm{v} \lambda$ and found that the waves travelled with the speed of light.

The electric and the magnetic fields of an electromagnetic wave are plane polarised. This nature of EM waves can be easily understood in the response of a portable AM radio to a broadcasting station. If the radio has a telescopic antenna it responds to the electric field of the signal. When this antenna is kept horizontal the signal strength will be greatly diminished.

Electromagnetic waves do carry energy and momentum, like other waves. To establish this, consider a plane perpendicular to the direction of propagation of the electromagnetic wave. If there are electric charges (like the free electrons in a conductor) in this plane, they will be set and sustained in motion by the electric and magnetic fields of the EM wave. The charges thus acquire energy and momentum from these waves. We have seen in earlier units, that, in a region of free space, with electric field $\mathrm{E}$, there is an energy density of $\overline{\mathrm{U}} _{\mathrm{E}}=\frac{1}{2} \varepsilon _{0} \mathrm{E}^{2}$. Similarly, if there is a magnetic field, the associated magnetic energy density is

$\overline{\mathrm{U}} _{\mathrm{B}}=\frac{1}{2} \frac{\mathrm{B}^{2}}{\mu _{0}}$.

Since electromagnetic waves contain both electric and magnetic fields, there is non-zero energy density associated with the electromagnetic waves and therefore, they carry energy.

The great technological importance of EM waves stems from the fact that they carry energy from one place to another. The radio, TV and mobile signals carry energy. Light carry energy from the sun to the earth, thus making things visible to human eye, helping leaves to photo-synthesis and enabling solar panels to produce electricity.

Since electromagnetic waves carry momentum, they can also exert pressure, called radiation pressure. If the total energy transferred to a surface, in time $t$, is $U$, the toal momentum imparted to the surface, for complete absorption is

$$ \mathrm{P}=\frac{\mathrm{U}}{\mathrm{c}} $$

This radiation pressure, exerted by sunlight, is found to be of the order of $7 \times 10^{-6} \mathrm{~N}-\mathrm{m}^{-2}$. Thus, on a surface of area $10 \mathrm{~cm}^{2}$, sunlight exerters a force (radiation force) of around $7 \times 10^{-9} \mathrm{~N}$.

Suppose EM waves are incident normally on the surface of a perfectly black block of surface area A and thickness $d$. The total energy transferred to this block, over a cycle of these EM waves is equal to:

$$ \begin{equation*} \left[\frac{1}{2} \in _{0} \mathrm{E} _{\mathrm{rms}}^{2}+\frac{1}{2} \frac{\mathrm{B} _{\mathrm{rms}}^{2}}{\mu _{0}}\right] ~~ {(Ad)} \end{equation*} $$

Characterstics and Properties of EM Waves

1. Electromagnetic waves are produced by accelerating or oscillating electric charges.

2. They are self-sustaining oscillations of electric and magnetic fields, which are coupled to each other.

3. The time-varying electric and magnetic fields are mutually perpendicular and are perpendicular to the direction of propagation of the wave.

Hence electromagnetic waves are transverse is nature.

4. The coupled sinusoidal electric and magnetic field variations are occuring in same phase, so that they attain their maximum or peak values at the same time.

5. Electromagnetic waves do not require any material medium for propagation.

6. They propagate in free space with a speed equal to $\mathrm{c}$, given by

$$ \mathrm{c}=\frac{1}{\sqrt{\mu _{0} \varepsilon _{0}}} \quad \qquad \left(\mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}\right) $$

7. When electromagnetic wave propagate through a material medium, the speed becomes $v$, given by

$$ \mathrm{v}=\frac{1}{\sqrt{\mu \varepsilon}}=\frac{\mathrm{c}}{\sqrt{\mu _{\mathrm{r}} \varepsilon _{\mathrm{r}}}}=\frac{\mathrm{c}}{\sqrt{\mu _{\mathrm{r}} \mathrm{K}}} $$

(v for any material medium is lesser than $\mathrm{c}$ )

8. The speed of electromagnetic waves is also obtained as ratio of the magnitude of electric field to that of the magnetic field at any instant

$ \qquad $ i.e. $\mathrm{c}=\frac{\mathrm{E} _{0}}{\mathrm{~B} _{0}}=\frac{\mathrm{E}}{\mathrm{B}}$

9. EM waves do not carry any charge and so they are not deflected by external static electric or magnetic fields.

10. All the electromagnetic waves, irrespective of their frequency (low or high), travel with the speed equal to $\mathrm{c}$, in space.

11. The speed of EM waves follows the general expression for wave-speed, $v=v \lambda$.

12. Experiments by Hertz demonstrated that elecromagnetic waves of any frequency (or wavelength) undergo refraction, diffraction, polarisation and can form standing waves.

13. Electromagnetic waves carry energy which is shared equally by the electric field $(\mathbf{E})$ and magnetic field $(\mathbf{B})$ vectors.

In a region of space where $\mathbf{E}$ and $\mathbf{B}$ fields vary sinusoidally, the average energy density $\left(\overline{\mathrm{U}} _{\mathrm{E}}\right)$ associated with the electric field (E) and the average energy density $\left(\overline{\mathrm{U}} _{B}\right)$ associated with the magnetic field (B) are given, respectively, as:

$$ \overline{\mathrm{U}} _{\mathrm{E}}=\frac{1}{2} \varepsilon _{0} \mathrm{E}^{2} $$

$$ \text{and} \quad \overline{\mathrm{U}} _{\mathrm{B}}=\frac{1}{2} \frac{\mathrm{B}^{2}}{\mu _{0}}$$

Here $\mathrm{E}$ and $\mathrm{B}$ represented, obviously, the rms values of the sinusoidally varying electric and magnetic fields,

where $\mathrm{E}=\mathrm{E} _{\mathrm{rms}}=\frac{\mathrm{E} _{0}}{\sqrt{2}}$ and $\mathrm{B}=\mathrm{B} _{\mathrm{rms}}=\frac{\mathrm{E} _{0}}{\sqrt{2}}$.

Therefore, $\overline{\mathrm{U}} _{\mathrm{E}}=\frac{1}{4} \varepsilon _{0} \mathrm{E} _{0}^{2}$

and $\quad \overline{\mathrm{U}} _{\mathrm{B}}=\frac{1}{4} \frac{\mathrm{B} _{0}^{2}}{\mu _{0}}$

The total average energy density possed by the electromagnetic waves is therefore,

$$ \overline{\mathrm{U}}=\overline{\mathrm{U}} _{\mathrm{E}}+\overline{\mathrm{U}} _{\mathrm{B}}=\frac{1}{4} \varepsilon _{0} \mathrm{E} _{0}^{2}+\frac{1}{4} \frac{\mathrm{B} _{0}^{2}}{\mu _{0}} $$

$$\text{But} ~ ~ \overline{\mathrm{U}} _{\mathrm{E}}+\overline{\mathrm{U}} _{\mathrm{B}} ~ , ~ \text{as proved below.}$$

$$ \begin{aligned} \overline{\mathrm{U}} _{\mathrm{B}} & =\frac{1}{4} \frac{\mathrm{B} _{0}^{2}}{\mu _{0}}=\frac{1}{4} \times \frac{1}{\mu _{0}}\left(\frac{\mathrm{E} _{0}^{2}}{\mathrm{C}^{2}}\right) \\ & =\frac{1}{4} \frac{\mathrm{E} _{0}^{2}}{\mu _{0}} \times \mu _{0} \varepsilon _{0} \quad\left(\because \frac{\mathrm{E} _{0}}{\mathrm{~B} _{0}}=\mathrm{c}\right) \\ & =\frac{1}{4} \varepsilon _{0} \mathrm{E} _{0}^{2} \\ & =\overline{\mathrm{U}} _{\mathrm{E}} \end{aligned} $$

Therefore, total energy density $(\overline{\mathrm{U}})$ can be written as:

$ \qquad \overline{\mathrm{U}}=\overline{\mathrm{U}} _{\mathrm{E}}+\overline{\mathrm{U}} _{\mathrm{B}} $

or $\quad \overline{\mathrm{U}}=2 \overline{\mathrm{U}} _{\mathrm{E}}$

or $ \quad \overline{\mathrm{U}}=\frac{1}{2} \varepsilon _{0} \mathrm{E} _{0}^{2}$

14. Electromagnetic waves carry energy and linear momentum( $\mathrm{P})$. If $\mathrm{U}$ is the total energy transferred to a surface (total absorption), is time $t$, then

$$ \mathrm{P}=\frac{\mathrm{U}}{\mathrm{c}} $$

(In case the waves are totally reflected from the surface, backwards, than the momentum transferred to the surface will be zero)

15. Since electromagnetic waves possess momentum, they can exert radiation pressure.

$$ \text { Radiation pressure }=\frac{\text { Linear momentum transferred by the EM waves }}{\text { Area of surface } \times \text { time }} $$

16. The intensity of electromagnetic radiation (I), during the propagation of the EM wave, at a point, is defined as the amount of energy passing through unit normal area around that point in unit time.

That is:

$ \quad \mathrm{I}=\frac{\text { Energy } / \text { Time }}{\text { Area }} $

or $\quad \mathrm{I}=\frac{\text { Power }}{\text { Area }}$

If $\bar{U}$ is the total average energy density carried by the electromagnetic wave, then total energy passing through unit normal area in one second will be:

$$ \begin{aligned} & \mathrm{I}=\overline{\mathrm{U}} \mathrm{c}, \text { where } \overline{\mathrm{U}}=\frac{1}{2} \varepsilon _{0} \mathrm{E} _{0}^{2} \\ \therefore \quad \mathrm{I} & =\frac{1}{2} \varepsilon _{0} \mathrm{cE} _{0}^{2} \end{aligned} $$

Since $\frac{1}{2} \varepsilon _{0} \mathrm{c}$ is a constant, we can write:

$$ \mathrm{I} \propto \mathrm{E} _{0}^{2} $$

17. If $\mathbf{E}$ and $\mathbf{B}$ are the electric and magnetic field vectors, then the velocity vector ( $\mathbf{v})$ of wave propagation is along $\mathbf{E} X \mathbf{B}$. i.e. $\mathbf{v}$ is in the direction of $\mathbf{E} X \mathbf{B}$.

It may be noted that a vector called poynting vector (S) gives the magnitude and direction of energy transported by the EM waves per second per unit normal area, and $\mathbf{S}$ is given as:

$ \mathbf{S}=\frac{\mathbf{E X B}}{\mu _{0}} $

Now, by definition, $\mathrm{I}=|\mathbf{S}|$

$\therefore \quad \mathrm{I}=|\mathbf{S}|=\frac{|\mathbf{E X B}|}{\mu _{0}}$

or $\quad \mathrm{I}=\mathrm{S}=\frac{\mathrm{EB}}{\mu _{0}}$

or $\quad \mathrm{I}=\mathrm{S}=\frac{1}{2} \frac{\mathrm{E} _{0} \times \mathrm{B} _{0}}{\mu _{0}}$

or $\quad \mathrm{I}=\mathrm{S}=\frac{1}{2} \frac{\mathrm{E} _{0}}{\mu _{0}} \cdot \frac{\mathrm{E} _{0}}{\mathrm{C}}$

or $\quad \mathrm{I}=\mathrm{S}=\frac{1}{2} \varepsilon _{0} \mathrm{CE} _{0}^{2} \hspace{2cm} \left(\because \mu _{0} \mathrm{C}=\frac{1}{\varepsilon _{0} \mathrm{C}}\right)$

18. The total energy transferred to a black surface, over a cycle, by normally incident EM waves is equal to :

$\left[\frac{1}{2} \in _{0} \mathrm{E} _{\mathrm{rms}}^{2} + \frac{1}{2} \frac{\mathrm{B} _{\mathrm{rms}}^{2}}{\mu _{0}}\right] (\mathrm{Ad})$ .

$ ( \mathrm{A}=$ surface area and $\mathrm{d}=$ thickness of the body)

Example-1:

A plane electromagnetic wave of frequency $25 \mathrm{MHz}$ travels in free space along the $x$-diretion. At a particular point in space and time, $\mathbf{E}=6.3 \hat{\mathrm{j}} \mathrm{Vm}^{-1}$. What is $\mathbf{B}$ at this point?

Show Answer

Solution :

Given : $\mathbf{v}=\hat{c} \hat{i}=3 \times 10^{8} \hat{i} \mathrm{~ms}^{-1}$

$ \mathbf{E}=6.3 \hat{\mathrm{j}} \mathrm{Vm}^{-1} $

As we know that $(\mathbf{E X} \mathbf{B})$ is along $\mathbf{v}$ and only $(\hat{j} \times \hat{k})$ gives $\hat{\dot{i}}$. Hence $\mathbf{B}=B \hat{k}$.

Now, $B=\frac{E}{v}=\frac{E}{c}$

$ \begin{aligned} & =\frac{6.3}{3 \times 10^{8}} \mathrm{~T} \\ & =2.1 \times 10^{-8} \mathrm{~T} \end{aligned} $

$\therefore \quad \mathbf{B}=\left(2.1 \times 10^{-8} \hat{\mathrm{k}}\right)$ tesla

Example-2:

The magnetic field in a plane electromagnetic wave is given as:

$$ \mathrm{B} _{\mathrm{y}}=2 \times 10^{-7} \sin \left(1.5 \times 10^{11} \mathrm{t}+500 x\right) \mathrm{T} $$

(a) What is the wave length and frequency of the wave?

(b) What is the expression for the electric field?

Show Answer

Solution :

Given $B _{y}=2 \times 10^{-7} \sin \left(1.5 \times 10^{11} t+500 x\right)$ T from this expression we obtain:

(i) $\quad \mathbf{v}=\mathrm{c}(-\hat{\mathrm{i}})=3 \times 10^{8}(-\hat{\mathrm{i}}) \mathrm{ms}^{-1}$

(ii) $\quad \omega=1.5 \times 10^{11} \mathrm{rad} . \mathrm{s}^{-1}$

(iii) $ \quad \mathrm{k}=500 \mathrm{rad}-\mathrm{m}^{-1}$

(iv) $\quad \mathrm{B} _{0}=2 \times 10^{-7}$ tesla

When we compare the above given equation with: $\mathrm{B} _{\mathrm{y}}=\mathrm{B} _{0}(\omega \mathrm{t}+\mathrm{kx})$

(a) $\quad \mathrm{k}=\frac{2 \pi}{\lambda}$

$ \begin{aligned} \therefore \quad \lambda & =\frac{2 \pi}{\mathrm{k}}=\frac{2 \times 3.14}{500}=2 \times 2 \times 3.14 \times 10^{-3} \mathrm{~m} \\ & =12.56 \times 10^{-3} \mathrm{~m} \\ & =1.256 \mathrm{~cm} \end{aligned} $

Also, $\omega=2 \pi v$

$ \begin{aligned} \therefore \quad v & =\frac{\omega}{2 \pi}=\frac{1.5 \times 10^{11}}{2 \times 3.14} \\ & =23.9 \times 10^{9} \mathrm{~Hz} \\ & =23.9 \mathrm{GHz} \end{aligned} $

(b) $ \quad \mathrm{E} _{0}=\mathrm{B} _{0} \mathrm{c}=2 \times 10^{-7} \times 3 \times 10^{8}$

$ \quad\quad=60 \mathrm{Vm}^{-1} $

Since the wave velocity (v) is along $(-\hat{\mathrm{i}})$ and $\mathbf{B}$ is along $\hat{\mathbf{j}}, \mathbf{E}$ has to be along the positive z-axis (or along $\hat{\mathrm{k}}$ )

$\therefore \quad \mathrm{E} _{\mathrm{z}}=\mathrm{E} _{0} \sin (\omega \mathrm{t}+\mathrm{k} x)$

or $\mathrm{E} _{\mathrm{z}}=60 \sin \left(1.5 \times 10^{11} \mathrm{t}+500 x\right) \mathrm{Vm}^{-1}$

Example-3:

Light with an energy flux of $18 \mathrm{Wcm}^{-2}$ falls on a non-reflecting surface, at normal incidence. If the surface has an area of $20 \mathrm{~cm}^{2}$, find the average force exerted on the surface by the incident light.

Show Answer

Solution :

Given : Energy flux $=18 \mathrm{Wcm}^{-2}$

$$ =18 \frac{\mathrm{J}}{\mathrm{s}-\mathrm{cm}^{2}} $$

Therefore, energy falling on $20 \mathrm{~cm}^{2}$ in one second is

$$\mathrm{U}=\text{Flux} \times \text{Area} \times \text{Time}$$

$$ \begin{aligned} & =18 \frac{\mathrm{J}}{\mathrm{s} \times \mathrm{cm}^{2}} \times 20 \mathrm{~cm}^{2} \times 1 \mathrm{~s} \\ & =360 \mathrm{~J} \end{aligned} $$

Hence, total momentum delivered (for complete absorption) on to $20 \mathrm{~cm}^{2}$ area, in 1 second is

$$ \mathrm{P}=\frac{\mathrm{U}}{\mathrm{c}}=\frac{360}{3 \times 10^{8}}=1.2 \times 10^{-6} \mathrm{~N}-\mathrm{s} $$

Therefore the average force exerted on the surface is:

$$ \begin{aligned} & \mathrm{F}=\frac{\mathrm{p}}{\mathrm{t}}=\frac{1.2 \times 10^{-6} \mathrm{~N}-\mathrm{s}}{1 \mathrm{~s}} \\ & =1.2 \times 10^{-6} \mathrm{~N} \end{aligned} $$

Example-4:

Calculate the amplitude of electric and magnetic fields produced by the radiation coming from a 100 watt bulb, at a distance of $3 \mathrm{~m}$ from the bulb. (Assume that the luminous efficiency of the bulb is $2.5 \%$ and that it is a point object).

Show Answer

Solution :

The bulb, as a point source,radiates light in all directions unformly. At a distance of $3 \mathrm{~m}$ from the bulb, the energy radiated by the bulb, and the energy recieved by a spherical surface, of area $\mathrm{A}=4 \pi \mathrm{r}^{2}=4 \pi(3)^{2}=$ $113 \mathrm{~m}^{2}$, will be the same.

Therefore, the intensity of light at $3 \mathrm{~m}$ is:

$$ \begin{gathered} \mathrm{I}=\frac{\text { Power }}{\text { Area }}=\frac{100 \mathrm{~W} \times 2.5 \%}{113 \mathrm{~m}^{2}} \\ \\ =0.022 \mathrm{Wm}^{-2} \end{gathered} $$

Half of this intensity is provided by the electric field and half is provided by the magnetic field of the radiation.

Now, we have, $\mathrm{I}=\frac{1}{2} \varepsilon _{0} \mathrm{c} \mathrm{E} _{0}^{2}$

$\therefore \mathrm{E} _{0}=\sqrt{\frac{2 \mathrm{I}}{\mathrm{c} \varepsilon _{0}}}=\sqrt{\frac{2 \times 0.022}{8.854 \times 10^{-12} \times 3 \times 10^{8}}}$

or $\mathrm{E} _{0}=4.07 \mathrm{Vm}^{-1}$

Hence, $\mathrm{B} _{0}=\frac{\mathrm{E} _{0}}{\mathrm{c}}=\frac{4.07}{3 \times 10^{8}}=1.4 \times 10^{-8} \mathrm{~T}$

(Note that, even though the magnetic field amplitude is extremely small, as compared to electric field magnitude, the energy associated with both of then are equal. Further $\mathrm{E} _{0}=4.07 \mathrm{Vm}^{-1}$ for a $100 \mathrm{~W}$ electric bulb, used by us for reading, at a distance of $3 \mathrm{~m}$ from it, is much higher than that of the TV or FM waves used by us, which is of the order of $10^{-6} \mathrm{Vm}^{-1}$ ).

Example-5:

Suppose that the electric field amplitude of an electromagnetic wave is $\mathrm{E} _{0}=120 \mathrm{NC}^{-1}$ and its frequency is $\mathrm{v}=50.0 \mathrm{MHz}$

(a) Determine the values of $\mathrm{B} _{0}, \omega, \mathrm{k}$ and $\lambda$.

(b) Obtain the expression for $\mathbf{E}$ and $\mathbf{B}$.

Show Answer

Solution :

Given : $\mathrm{E} _{0}=120 \mathrm{NC}^{-1}\left(\right.$ or $\left.\mathrm{Vm}^{-1}\right)$

$v=50.0 \mathrm{MHz}$

$ =50 \times 10^{6} \mathrm{~Hz} $

(a) $~$ (i) $\mathrm{B} _{0}=\frac{\mathrm{E} _{0}}{\mathrm{c}}=\frac{120}{3 \times 10^{8}}=4 \times 10^{-7} \mathrm{~T}$

$\quad$ (ii) $\quad \omega=2 \pi \mathrm{v}=2 \times 3.14 \times 50 \times 10^{6}=3.14 \times 10^{8} \mathrm{rad}-\mathrm{s}^{-1}$

$\quad$ (iii) $\mathrm{k}=\frac{\omega}{\mathrm{v}}=\frac{\omega}{\mathrm{c}}=\frac{3.14 \times 10^{8}}{3 \times 10^{8}}=1.05 \mathrm{rad}-\mathrm{m}^{-1}$

$\quad$ (iv) $\lambda=\frac{\mathrm{c}}{\mathrm{v}}=\frac{3 \times 10^{8}}{5 \times 10^{6}}=6.0 \mathrm{~m}$

(b) Suppose the wave propagate in the positive $x$-direction. Then, the expression for $\mathbf{E}$ is

$$ \begin{aligned} \mathbf{E}= & \mathrm{E} _{0} \sin (\omega \mathrm{t}-\mathrm{k} x) \hat{\mathrm{j}} \\ & =120 \sin \left(3.14 \times 10^{8} \mathrm{t}-1.05 x\right) \hat{\mathrm{j}} \mathrm{Vm}^{-1} \end{aligned} $$

Expression for $\mathbf{B}$ is:

$$ \begin{aligned} \mathbf{B}= & \mathrm{B} _{0} \sin (\omega \mathrm{t}-\mathrm{k} x) \hat{\mathrm{k}} \\ & =4 \times 10^{-7} \sin \left(3.14 \times 10^{8} \mathrm{t}-1.05 x\right) \hat{\mathrm{k}} \text { tesla } \end{aligned} $$

Example-6:

Suppose that the electric field part of an electromagnetic wave in vacuum is :

$ \mathbf{E}=3.1 \cos \left(1.8 \mathrm{y}+5.4 \times 10^{8} \mathrm{t}\right) \hat{\mathrm{i}} \mathrm{Vm}^{-1} $

Obtain the:

(i) direction of propagation

(ii) wavelength $(\lambda)$

(iii) frequency $(v)$

(iv) amplitude of the mangetic field part

(v) expression for the magnetic field part of the wave

Show Answer

Solution :

Given: $\mathbf{E}=3.1 \cos \left(1.8 \mathrm{y}+5.4 \times 10^{8} \mathrm{t}\right) \hat{\mathrm{i}} \mathrm{Vm}^{-1}$

The standard equation for $\mathbf{E}$ is

$$ \mathbf{E}=\mathrm{E} _{0} \cos (\mathrm{ky}+\omega \mathrm{t}) \hat{\mathrm{i}} $$

Comparing the two we get:

(i) The direction of wave propagation is $(-\hat{\mathrm{j}})$ or along the negative $\mathrm{y}$-direction.

(ii) $\mathrm{k}=1.8 \mathrm{rad}-\mathrm{m}^{-1}$

$$ \therefore \quad \lambda=\frac{2 \pi}{\mathrm{k}}=\frac{2 \times 3.14}{1.8}=3.49 \mathrm{~m} $$

(iii) $\quad \omega=5.4 \times 10^{8} \mathrm{rad}-\mathrm{s}^{-1}$

$$ \therefore \quad v=\frac{\omega}{2 \pi}=\frac{5.4 \times 10^{8}}{2 \times 3.04}=85.9 \times 10^{6} \mathrm{~Hz} $$

(iv) $\mathrm{E} _{0}=3.1 \mathrm{~N}-\mathrm{C}^{-1}$

$$ \therefore \quad B _{0}=\frac{E _{0}}{c}=\frac{3.1}{3 \times 10^{8}}=1.03 \times 10^{-8} \mathrm{~T} $$

(v) $\mathbf{B}=\mathrm{B} _{0} \cos (\mathrm{kj}+\omega \mathrm{t}) \hat{\mathrm{k}}$ $ \hspace{2cm}(\because \mathbf{E} \times \mathbf{B}$ is along $\mathbf{v}$ or $\hat{\mathrm{i}} \times \hat{\mathrm{k}}=-\hat{\mathrm{j}})$

$=1.03 \times 10^{-8} \cos \left(1.08 \mathrm{y}+5.4 \times 10^{8} \mathrm{t}\right) \hat{\mathrm{k}} \mathrm{T}$

Electromagnetic Spectrum

After Hertz’s experimental demonstration regarding electromagnetic waves, establihing Maxwell’s theory, several types of electromagnetic waves were discovered, in addition to the familiar and well known visible light wave. Now we are all aware that EM waves include visible light, ultra violet radiation, infrared waves, $x$-rays, gamma rays, micro waves and radio waves.

The orderly distribution and classification of electromagnetic waves, according to their frequency or wavelength is known as the electromagnetic spectrum.

The electromagnetic spectrum has a wide range of wavelength, varying from about $10^{-14} \mathrm{~m}$ for gamma rays to $6 \times 10^{2} \mathrm{~m}$ for radio waves. It may be pointed out that only a small region of the complete electromagnetic spectrum is visible to human eye, since the waves in this ‘visible region’ can only produce some sensation on the retina of human eye. These radiations in the visible range (starting with $\lambda \approx 400 \mathrm{~nm}$ for violet and ending with $\lambda \approx 700 \mathrm{~nm}$ for red) produce sensation of different colours on the regina.

The following figure, and the table below it, give the details of the various regions of the electromagnetic spectrum with their range of frequency and wavelength, their production as well as the means of their detection. It may be noticed that there is no sharp division between the different regions.

The Electromagnetic Spectrum

$~$

Different types of EM waves and their brief discription

S. No.	Type	Wavelength range	Frequency range, (Hz)	Production	Detection
1.	Radio waves	$>0.1 \mathrm{~m}$ $\left(0.1 \mathrm{~m}\right.$ to $\left.6 \times 10^{2} \mathrm{~m}\right)$	$3 \times 10^{9}$ to $5 \times 10^{5}$	Oscillating circuit or rapid acceleration and decelerations of electrons in aerials	Receiver’s aerials
2.	Microwaves	$0.1 \mathrm{~m}$ to $1 \mathrm{~mm}$ $\left(10^{-3} \mathrm{~m}\right.$ to $\left.0.1 \mathrm{~m}\right)$	$3 \times 10^{11}$ to $3 \times 10^{9}$	Khystron valve or magnetron valve	Point contact diodes
3.	Infrared waves	$1 \mathrm{~mm}$ to $700 \mathrm{~nm}$ $\left(7 \times 10^{7} \mathrm{~m}\right.$ to $\left.10^{-3} \mathrm{~m}\right)$	$4 \times 10^{14}$ to $3 \times 10^{11}$	Vibration of atoms and molecules	Thermopiles, Bolometer, Infrared photographic films
4.	Visible light waves	$800 \mathrm{~nm}$ to $400 \mathrm{~nm}$ $\left(4 \times 10^{-7} \mathrm{~m}\right.$ to $\left.8 \times 10^{7} \mathrm{~m}\right)$	$8 \times 10^{14}$ to $4 \times 10^{14}$	Excitation of valency electrons in atoms, electrons in atoms emit light when they move from one energy level to a lower energy level	The human eye, Photocells, Photog- raphic film/plate
5.	Ultravoilet waves	$400 \mathrm{~nm}$ to $1 \mathrm{~nm}$ $\left(1 \times 10^{-9} \mathrm{~m}\right.$ to $\left.4 \times 10^{-7} \mathrm{~m}\right)$	$1 \times 10^{17}$ to $8 \times 10^{14}$	Excitation of atoms, spark and arc lamps, inner shell electrons in atoms moving from one energy level to a lower level	Photocells, photographic film
6.	$x$-rays	$\operatorname{lnm}$ to $10^{-3} \mathrm{~nm}$ $\left(10^{-12} \mathrm{~m}\right.$ to $\left.10^{-9} \mathrm{~m}\right)$	$3 \times 10^{20}$ to $1 \times 10^{17}$	$x$-rays tubes or excitation of inner shell electrons	Photographic film, Gaiger tubes, Ionisation chamber
7.	Gamma rays	$<10^{3} \mathrm{~nm}$ $\left(<10^{-12} \mathrm{~m}\right)$	$5 \times 10^{22}$ to $3 \times 10^{20}$	Radioactive decay of the nucleus, Nuclear origin	Photographic film, Ionisation chamber

The Electromagnetic Spectrum

$~$

Main Regions of EM Spectrum and uses and Applications of these Region

The entire electromagnetic spectrum is broadly classified in the following regions, in the order of the decreasing wave length (or increasing frequency). We now describe each region, briefly, including some elementary facts about the use of various electromagnetic waves.

(i) Radio Waves

They are produced by the oscillating charges at the ends of a dipole antenna connected to an oscillating voltage source. These radio waves have a frequency range from about $5 \times 10^{5} \mathrm{~Hz}$ to $10^{9} \mathrm{~Hz}$. These waves are used as carrier waves to transport message signals (audio and video) in radio, TV and mobile communication systems.

In early radio-communication, amplitude modulated (AM) waves were used. AM radio were used for this purpose. There are two bands of AM waves (a) Medium wave band (MW bnad), with frequency ranging from $540 \mathrm{KHz}$ to $1710 \mathrm{KHz}$ and are propagated as ‘ground waves’. The MW band of AM radios can receive these waves. (b) Short wave band (SW band), with a frequency range from $1710 \mathrm{KHz}$ to $54 \mathrm{MHz}$. These wave are utilized in sky-wave propagation and are received by the $\mathrm{SW}$ bands of AM radios.

TV waves employ EM waves in the frequency range from $54 \mathrm{MHz}$ to $890 \mathrm{MHz}$. The frequency modulated (FM) radio band extends from $88 \mathrm{MHz}$ to $108 \mathrm{MHz}$.

The ultra High frequency (UHF) band extends from $300 \mathrm{MHz}$ to $3000 \mathrm{MHz}$ (or $3 \mathrm{GHz}$ ) and it is used for cellular or mobile phone communication.

(ii) Microwaves

They are short-wave length radio waves, having frequency range from $1 \mathrm{GHz}$ to $300 \mathrm{GHz}$. These waves are produced by special vacuum tubes like Klystrons, Mangetrons, Gunndiodes etc.

By virture of their short wave length, microwaves can be transmitted through large distances, without much spreading, like a narrow beam. Thus they are very effectively used in RADAR systems, aircraft navigation and satellite communication.

Microwaves are used in speed guns to measure the speed of fast moving objects and detecting the service speeds of tennis and squash balls, cricket balls, automobile vehicles etc.

Microwaves are also used in Microwave oven for cooking purposes. (Microwave oven works on the principle of resonance between the elcgric field vibrations of the microwaves and electric field variations around the highly polarised water molecules (or dipoles) due to the vibrational and rotational motions of the water molecules. They both have the same or almost same frequencies, producing resonance. It may be noted that at resonance maximum absorption of energy takes place. Hence water moelculas absorbes energy from microwaves due to resonance. In the microwave oven, the energy absored by water molecules heats up these molecules and they then conduct (transfer) the heat to the other food particles, heating the entire food).

(iii) Infrared Waves

These waves have frequency range from $3 \times 10^{11} \mathrm{~Hz}$ to $4 \times 10^{14} \mathrm{~Hz}$.

They are produced or emitted by all hot objects and molecules. These waves are easily absorbed by objects in the surroundings, there by cansing heating effect in them. Because of this, infrared (IR) radiations are some times referred to as ‘heat waves’.

Infrared waves (or radiations) are used for a variaty of applications like:

• Physical therapy for treating muscular strain (infrared lamps are used for this).
• Solar water heater, solar cooker, solar cells, remote sensing satellites.
• Drying purposes (fruits and vegetables).
• Night vision goggles
• Imprared LEDs, used in remote controles to operate electronic systems in TV sets, video recorders, hi-fi system, ACs, set-top boxes etc.
• Green houses to keep plants warm.

(Infrared waves trapped by green house gases like $\mathrm{CO} _{2}$ and water vapour, helps keep the earth warm. But excessive green house gases can cause global warming).

(iv) Visible Light

This is the most familiar form of electromagnetic waves. Even though this region forms a significant part of solar spectrum, it is in fact only a very small part of the whole electromagnetic spectrum. Human eye is sensitive only to this part of the EM spectrum.

The visible light waves have a frequency range from $4 \times 10^{14} \mathrm{~Hz}$ to $8 \times 16^{14} \mathrm{~Hz}$ (or wave length range from $400 \mathrm{~nm}$ to $800 \mathrm{~nm}$ ).

The visible light waves are produced by de-excitation of valance electrons from higher energy to lower energy levels.

(v) Ultra Violet Waves (UV Rays)

This region has a range of frequencies from $8 \times 10^{14} \mathrm{~Hz}$ to $5 \times 10^{16} \mathrm{~Hz}$.

Sun is the most aboudant source of U-V rays. They are also produced by special U-V lamps, very hot objects, mercury arcs and welding iron arcs.

Over-exposure to $U-V$ rays from sun can cause very dangerous harmful effects on humans on earth (like skin cancer etc). But the ozone layer in the earth’s atmosphere (at around $40-50 \mathrm{~km}$ ) absorbs most of these radiations from sun, thereby protecting us from harmful effects. However, the ozone depletion caused by chloroflouro carbons (CFCs) emitted by refrigerators, ACs and airosoles is

adversely affecting the absorption of $\mathrm{U}-\mathrm{V}$ rays by atmosphere and the matter is of international concern for the word community.

Ultravoilet radiations have wide range of uses like:

• in preserving food stuff, by killing germs and bacteria using U-V rays.
• in burglar alarms.
• In the detection of forged documents.
• in finger print analysis.
• in checking and testing mineral samples (through flourescence caused by $U-V$ rays).
• U-V lamps used in water purifiers.
• In LASIK

(It may be noted that UV rays are easily absorbed by ordinary glass and therefore special glass goggles or face masks with glass windows help welders protect their eyes from the UV rays produced by welding arcs).

(vi) $\mathbf{x}$-rays

This part of the EM waves has the range of frequencies from about $10^{16} \mathrm{~Hz}$ to $3 \times 10^{21} \mathrm{~Hz}$. The $\mathrm{x}-$ rays are produced by bombarding a metal target by high energy electrons. They possess high penatrating power.

$\mathrm{x}$-rays also have wide range of uses and application.

• $\mathrm{x}$-rays are used as diagnosting tool for medical application.
• They are used as treatment tool for certain forms of cancer.
• These rays are used for detection of bone fractures, unauthourised presence of hidden objects like precious metals and stones, bullets etc. inside human body as well as explosive and drugs hidden in the body.
• These rays are used for scientific research in crystal structure etc.
• They are used in engineering, to detect and test any faults, cracks and flows in structures built with metals and concrete.
• They are also used to test the uniformity of insulating materials and to detect leakage of oils in underground pipes.

(It may be noted that over-exposure to $\mathrm{x}$-rays can damage or even destroy living tissues and organisms).

(vii) Gamma Rays ( $\gamma$-rays)

These rays have a frequency range from about $3 \times 10^{18} \mathrm{~Hz}$ to $5 \times 10^{22} \mathrm{~Hz}$. They are highly energetic and are produced or emitted by radio-active nuclei. They have extremely high penetrating power and can cause huge biological damage to tissues and organisms.

They are used in medical field to destroy cancer cells (radio-therapy).

Soft $\gamma$-rays are used to kill micro-organisms to help preserve food-stuff for longer period. They are used in the study of the structure of nucleus of atoms and also to detect and test faults in constructed structures.

Example-7:

Use Wien’s formula : $\lambda _{\mathrm{m}} \mathrm{T}=0.29 \mathrm{~cm}-\mathrm{K}$ to find the temperature of a hot body which can possibly emit microwaves of wavelength $1 \mu \mathrm{m}$.

Show Answer

Solution :

Given: $\lambda _{\mathrm{m}} \mathrm{T}=0.29 \mathrm{~cm}-\mathrm{K}$

$ \begin{aligned} & \lambda _{\mathrm{m}}=1 \mu \mathrm{m}=1 \times 10^{-6} \mathrm{~m}=1 \times 10^{-4} \mathrm{~cm} \\ \therefore \quad \mathrm{T}= & \frac{0.29 \mathrm{~cm}-\mathrm{K}}{10^{-4} \mathrm{~cm}} \\ = & 2900 \mathrm{~K} \end{aligned} $

Example-8:

Given below are some famous numbers associated with electromagnetic radiation in different contexts in physics. State the part of electromagnetic spectrum to which each belongs to:

(i) $21 \mathrm{~cm}$ (wavelength emitted by atomic hydrogen in interstellor space).

(ii) $1057 \mathrm{~Hz}$ (frequency of radiation arising from two close energy levels in hydrogens known as lamb shift).

(iii) $2.7 \mathrm{~K}$ (temperature associated with isotropic radiation filling all space : thought to be a relic of the big bang origin of the universe).

(iv) $5890 \AA$ and $5896 \AA$ (double lines of sodium).

(v) $14.4 \mathrm{KeV}$ (energy of a particular transition in $\mathrm{Fe}^{56}$ nucleous associated with a famous high resolution spectroscopic method-Mossbauer spectroscopy).

Show Answer

Solution :

(i) Given wave length corresponds to short wavelength radio-waves.

(ii) High frequency end of radio waves (short radio waves).

(iii) $\mathrm{T}=2.7 \mathrm{~K}$ gives $\lambda=\frac{0.29 \mathrm{~cm}}{\mathrm{~T}}=0.11 \mathrm{~cm}$, which corresponds to microwave region.

(iv) $\mathrm{D} _{1}$ and $\mathrm{D} _{2}$ lines in the visible region (yellow).

(v) Given $\mathrm{E}=14.4 \mathrm{KeV}=14.4 \times 10^{3} \times 1.6 \times 10^{-19} \mathrm{~J}$

$$ \begin{aligned} & =14.4 \times 1.6 \times 10^{-16} \mathrm{~J} \\ \therefore \quad v & =\frac{E}{h}=\frac{14.4 \times 1.6 \times 10^{-16}}{6.6 \times 10^{-34}} \mathrm{~Hz}=3.5 \times 10^{12} \mathrm{MHz} \end{aligned} $$

Which lies high frequency $x$-rays or soft $\gamma$-ray region.

Example-9:

Electromagnetic waves with wavelenth

(i) $\lambda _{1}$ is used in satellite communication.

(i) $\lambda _{2}$ is used to kill germs in wter purifiers.

(iii) $\lambda _{3}$ is used to detect leakage of oil in undergound pipe lines.

(iv) $\lambda _{4}$ is used to improve visibility in runways during fog and mist condition.

(a) Identify and name the part of EM spectrum to which these radiation belong.

(b) Arrange them in the ascending order of their wavelength magnitude.

(c) Write one more application for each of them.

Show Answer

Solution :

(a) $ ~ \lambda _{1} \rightarrow$ microwave

$ \quad \lambda _{2} \rightarrow$ ultraviolet radiation

$ \quad \lambda _{3} \rightarrow \mathrm{x}$-rays

$ \quad \lambda _{4} \rightarrow$ Infrared radiation

(b) $ ~ \lambda _{3}<\lambda _{2}<\lambda _{4}<\lambda _{1}$

(c) $ ~ \lambda _{1} \rightarrow$ microwaves - used in radar (or microwave oven).

$ \quad \lambda _{2} \rightarrow$ U-V radiation - used in LASIK eye surgery.

$ \quad \lambda _{3} \rightarrow \mathrm{x}$-rays - used in radio-graphy.

$\quad \lambda _{4} \rightarrow$ infrared - used in remote controls.

EXERCISE

1. The magnetic field of a beam emerging from a filter facing a flood light is given as:

$$ B=12 \times 10^{-8} \sin \left(1.20 \times 10^{7} z-3.60 \times 10^{15} t\right) \text { tesla } $$

What is the average intensity of the bean?

Show Answer

Answer: $1.71 \mathrm{~W} . \mathrm{m}^{-2}$

Solution:

Given $B=12 \times 10^{-8} \sin \left(1.20 \times 10^{7} z-3.6 \times 10^{15} t\right)$ tesla. The standard equation for $B$ of EM-wave is:

$\mathrm{B}=\mathrm{B} _{0} \sin (\mathrm{kz}-\omega \mathrm{t})$.

Therefore, $\mathrm{B} _{0}=12 \times 10^{-8} \mathrm{~T}$

Now, $I _{a v}=\frac{1}{2} c \frac{B _{0}^{2}}{\mu _{0}}=\frac{\frac{1}{2} \times 3 \times 10^{8} \times\left(12 \times 10^{-8}\right)^{2}}{4 \pi \times 10^{-7}} \mathrm{~W} / \mathrm{m}^{2}$

$ =1.71 \mathrm{~W}-\mathrm{m}^{-2} $

2. Poyting vector $\mathbf{S}$ is defined as:

$$ \mathbf{S}=\frac{1}{\mu _{0}}(\mathbf{E} \times \mathbf{B}) $$

Show graphically the variation of $S$ with $t$.

Show Answer

Solution:

For an EM-wave propagating along $x$-direction, the electric field $\mathbf{E}$ is along the $y$-direction and the magnetic field $\mathrm{B}$ is along the $\mathrm{z}$-direction, such that $\mathbf{E} \times \mathbf{B}$ is along the direction of wave velocity (i.e. along the $x$-direction).

Now, $\mathbf{E} _{\mathrm{y}}=\mathrm{E} _{0} \sin (\omega \mathrm{t}-\mathrm{k} x)$

and $\quad \mathbf{B} _{z}=\mathrm{B} _{0} \sin (\omega \mathrm{t}-\mathrm{k} x)$

$\therefore \quad \mathbf{S}=\frac{\mathbf{E} \times \mathbf{B}}{\mu _{0}}=\frac{\mathrm{E} _{0} \mathrm{~B} _{0}}{\mu _{0}} \sin ^{2}(\omega \mathrm{t}-\mathrm{k} x) \hat{\mathrm{i}}$

Hence, the graph of $|\mathbf{S}|$ verses $t$ varies as square of give function, as shown below.

$~$

3. Show that the radiation pressure exerted by an EM-wave of intensity I on a surface, kept in vaccum, for complete absorption is $\frac{\mathrm{I}}{\mathrm{c}}$.

Show Answer

Solution:

Force $\mathrm{F}=\frac{\Delta \mathrm{P}}{\Delta \mathrm{t}}=$ rate of change of momentum imparted.

$\therefore \quad$ Pressure $\mathrm{P}=\frac{\mathrm{F}}{\mathrm{A}}$

or $\mathrm{P}=\frac{\left(\frac{\Delta \mathrm{P}}{\Delta \mathrm{t}}\right)}{\mathrm{A}}$

or $\mathrm{P}=\frac{\left(\frac{\Delta \mathrm{P}}{\Delta \mathrm{t}}\right)}{\mathrm{A}} \times \frac{\mathrm{c}}{\mathrm{c}}$

or $\quad \mathrm{P}=\frac{1}{\mathrm{Ac} \Delta \mathrm{t}}(\Delta \mathrm{P} \times \mathrm{c})$

or $\quad \mathrm{P}=\frac{1}{\operatorname{Ac} \Delta \mathrm{t}}(\Delta \mathrm{U}) \qquad \quad (\Delta \mathrm{U}=\Delta \mathrm{P} \times \mathrm{c}$ is energy imported to the surface)

But $\quad \mathrm{I}=\frac{\text { Energy imported }}{\text { Area } \times \text { time }}$

$ =\frac{\Delta \mathrm{U}}{\mathrm{A} \Delta \mathrm{t}} $

Hence, $P=\frac{I}{c}$

4. For frequency $v=4 \times 10^{8} \mathrm{~Hz}$, sea water has a permittivity of $\varepsilon=80 \varepsilon _{0}$ and permeability $\mu=\mu _{0}$. The resitivity of sea water is $\rho=0.25 \Omega-\mathrm{m}$. Imagine a parallel plate capacitor immersed is sea water and driven by an A.C voltage of $\mathrm{V}=\mathrm{V} _{0} \sin (2 \pi {vt})$. What fraction of conduction current density is its conduction current density?

Show Answer

Answer: $\frac{4}{9}$

Solution:

Let $\mathrm{d}$ be the distance between the plates of the capacitor. The applied voltage is

$$ V=V _{0} \sin (2 \pi v t) $$

Therefore, the field $\mathrm{E}$ is

$$ \mathrm{E}=\frac{\mathrm{V}}{\mathrm{d}}=\frac{\mathrm{V} _{0}}{\mathrm{~d}} \sin (2 \pi {vt}) $$

Hence, the conduction current density is

$$ J _{c}=\frac{E}{\rho}=\frac{V _{0}}{\rho d} \sin (2 \pi v t) $$

or $\mathrm{J} _{\mathrm{c}} = \mathrm{J} _{0 _{\mathrm{C}}} \sin (2 \pi {vt})$

$\therefore$ The maximum conduction current density $\left(\mathrm{J} _{\mathrm{o} _{\mathrm{C}}}\right)$ is

$ \mathrm{J} _{0 _{\mathrm{c}}}=\frac{\mathrm{V} _{0}}{\rho \mathrm{d}} $

Now, the displacement current density is

$ \mathrm{J} _{\mathrm{d}}=\frac{\mathrm{i} _{\mathrm{d}}}{\mathrm{A}}=\frac{1}{\mathrm{~A}} \varepsilon \frac{\mathrm{dQ} _{\mathrm{E}}}{\mathrm{dt}}=\in \frac{\mathrm{dE}}{\mathrm{dt}} \quad\left(\because \mathrm{Q} _{\mathrm{E}}=\mathrm{EA}\right) $

$\text { or } \mathrm{J} _{\mathrm{d}}=\varepsilon \frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{V} _{0}}{\mathrm{~d}} \sin (2 \pi {vt})\right) $

$\text { or } \mathrm{J} _{\mathrm{d}}=\varepsilon \frac{\mathrm{V} _{0}}{\mathrm{~d}} \cdot 2 \pi \mathrm{v} \cos (2 \pi {vt}) $

$\text { or } \mathrm{J} _ {\mathrm{d}}=\mathrm{J} _ {0_ {\mathrm{d}}} \cos (2 \pi {vt})$

where $\mathrm{J} _ {0_{\mathrm{d}}}=\varepsilon \frac{\mathrm{V} _ {0}}{\mathrm{~d}} \cdot 2 \pi \mathrm{v}$ is the maximum displacement current density.

Hence, $\frac{\mathrm{J} _ {0_{\mathrm{d}}}}{\mathrm{J} _{0 \mathrm{c}}}=\frac{\varepsilon \frac{\mathrm{V} _{0}}{\mathrm{~d}} \cdot 2 \pi \mathrm{v}}{\left(\frac{\mathrm{V} _{0}}{\rho \mathrm{d}}\right)}$

$ \begin{aligned} & \quad=2 \pi v \rho \varepsilon \\ & \text { or } \quad \frac{\mathrm{J} _ {0 _{\mathrm{d}}}}{\mathrm{J} _{0 \mathrm{c}}}=2 \pi \varepsilon _{\mathrm{r}} \in _{0} \cdot v \rho \\ & =\frac{1}{9 \times 10^{9} \times 2} \times 80 \times 4 \times 10^{8} \times 0.25 \\ & =\frac{4}{9} \end{aligned} $

Answer: $\frac{\mathrm{J} _ {0_ {\mathrm{d}}}} {\mathrm{J} _ {0_ {\mathrm{c}}}} =\frac{4}{9}$

5. If $\varepsilon _{0}$ and $\mu _{0}$ are the electric permittivity and magnetic permeability of free space and $\varepsilon$ and $\mu$ are the corresponding quantities for a given medium, then express the refractive index ’ $n$ ’ of the medium in terms these quantities.

Show Answer

Answer: $\sqrt{\frac{\mu \varepsilon}{\mu _{0} \varepsilon _{0}}}$

Solution:

Defractive index $n=\frac{\text { Speed of light in air }}{\text { Speed of light in medium }}$

$ =\frac{\mathrm{c}}{\mathrm{v}} $

Now, $c=\frac{1}{\sqrt{\mu _{0} \varepsilon _{0}}}$

$ \begin{array}{r} \mathrm{v}=\frac{1}{\sqrt{\mu \varepsilon}} \\ \therefore \quad \mathrm{n}=\sqrt{\frac{\mu \varepsilon}{\mu _{0} \varepsilon _{0}}} \end{array} $

$($ It may be noted that :

$ \begin{gathered} \mu=\mu _{\mathrm{r}} \mu _{0} \\ \varepsilon=\varepsilon _{\mathrm{r}} \varepsilon _{0} \\ =\mathrm{K} \varepsilon _{0} \\ \therefore \quad \mathrm{n}=\sqrt{\frac{\mu _{\mathrm{r}} \mu _{0} \mathrm{~K} \varepsilon _{0}}{\mu _{0} \varepsilon _{0}}} \\ \text { or } \left.\mathrm{n}=\sqrt{\mu _{0} \mathrm{~K}}\right) \end{gathered} $

Question Bank

Key Learning Points

1. Maxwell’s equation for displacement current $\left(i _{d}\right)$ in space (i.e. $\left.i _{d}=\varepsilon _{0} \frac{d _{E}}{d t}\right)$ gave the first equation of Maxwell’s electromagnetic wave thoery i.e. $\oint \mathbf{B} \cdot \boldsymbol{d} \ell=\mu _{0} \varepsilon _{0} \frac{\mathrm{d} \phi _{\mathrm{E}}}{\mathrm{dt}}$, which is the modified Amepere’s circuital law applicable for an imaginary loop in space where there is a time-varying electrica field existing.

2. The above equation along with Faraday’s equation for emf induced in space $(\varepsilon=\oint \mathbf{E} \cdot \mathbf{d} \ell)$, namely $\oint \mathbf{E} . \mathbf{d} \ell=\frac{-\mathrm{d} \phi _{\mathrm{B}}}{\mathrm{dt}}\left(\operatorname{read} \varepsilon=\frac{-\mathrm{d} \phi _{\mathrm{B}}}{\mathrm{dt}}\right)$, predicted the possible production of coupled time-varying electric and magnetic fields, the variation in one field acting as the source for the other field.

$$ \text{his} \oint \mathbf{B} \cdot \mathbf{d} \ell=\mu _{0} \varepsilon _{0} \frac{\mathrm{d} \phi _{\mathrm{E}}}{\mathrm{dt}} \text{and} $$

$$ \oint \mathbf{E} \cdot \mathbf{d} \ell=\frac{-\mathrm{d} \phi _{\mathrm{B}}}{\mathrm{dt}} $$

together, as Maxwell’s first two equation in electromagnetic wave theory predicted the production EM waves, which travel with the speed of light.

3. Along with the above two equations, the equations for Gauss’s theorem in electrostatics and magnetism (namely $\oint \mathbf{E} . \mathbf{d} \mathbf{S}=\frac{1}{\varepsilon _{0}}(\mathrm{q})$ and $\oint \mathbf{B} . \mathbf{d S}=0$ ) were used by Maxwell to develop the complete thoery of EM waves.

4. Hertz experiments to produce and detect EM waves demonstrated the validity of Maxwell’s theory.

5. The main properties of Em waves are:

(i) They are made up of coupled time varying electric and magnetic fields (simusoidally varying).

(ii) The variation is E-field produces the B-field and vice-versa.

(iii) $\mathrm{E}$ and $\mathrm{B}$ fields one mutually perpendicular and both are perpendicular to the direction of propagation of the EM wave, such that: $\mathbf{E} \times \mathbf{B}$ is along $\mathbf{v}$ (the wave velocity). (This established the transverse nature of the EM-waves.

(iv) $\mathbf{E}$ and $\mathbf{B}$ vary in same phase.

(v) The equation for the sinusoidal $\mathbf{E}$ and $\mathbf{B}$ fields propagating in space along the $x$-direction can be expressed as: $\mathbf{E}=\mathrm{E} _{0} \sin (\omega \mathrm{t}-\mathrm{k} x) \hat{\mathrm{j}}$ and

$\mathbf{B}=\mathrm{B} _{0} \sin (\omega \mathrm{t}-\mathrm{k} x) \hat{\mathrm{k}}$

where, $\omega=2 \pi v$ is the angular frequency of the wave and $\mathrm{k}=\frac{2 \pi}{\lambda}$ is the angular wave number.

Also, $\frac{\omega}{\mathrm{k}}=v$ is the wave velocity, which is along the $x$-direction, here.

(vi) $\mathrm{v}=\mathrm{c}$ for free-space, where $\mathrm{c}=\frac{1}{\sqrt{\mu _{0} \varepsilon _{0}}}$.

(vii) For any material medium,

$$ \mathrm{v}=\frac{1}{\sqrt{\mu \varepsilon}} \frac{\mathrm{c}}{\sqrt{\mu _{\mathrm{r}} \varepsilon _{\mathrm{r}}}} $$

(viii) The wave velocity is also equal to the ratio of E-field to B-field at any instant of time.

i.e. $v=\frac{E}{B}=\frac{E _{0}}{B _{0}}$

(ix) EM-waves carry energy. Where these waves are absorbed, the enery gets transferred to the material which absorbs the waves.

(x) The average energy density carried by the $\mathbf{E}$ and $\mathbf{B}$ fields are respectively,

$ \overline{\mathbf{U}} _{\mathrm{E}}=\frac{1}{2} \varepsilon _{0} \mathrm{E} _{\mathrm{rms}}^{2} \text { and } \overline{\mathbf{U}} _{\mathrm{B}}=\frac{1}{2} \frac{\mathrm{B} _{\mathrm{rms}}^{2}}{\mu _{0}} \text { or } $

or $\quad \overline{\mathbf{U}} _{\mathrm{E}}=\frac{1}{4} \varepsilon _{0} \mathrm{E} _{0}^{2}$ and $\overline{\mathbf{U}} _{\mathrm{B}}=\frac{1}{4} \frac{\mathrm{B} _{0}^{2}}{\mu _{0}}$

(xi) $\mathbf{U} _{\mathrm{E}}=\mathbf{U} _{\mathrm{B}}$ (i.e. both $\mathrm{E}$ and $\mathrm{B}$ fields carry same amount of energy).

(xii) The total energy density of an EM-wave is

$ \overline{\mathbf{U}}=\overline{\mathbf{U}} \mathrm{E}+\overline{\mathbf{U}} \mathrm{B} $

or $\overline{\mathbf{U}}=2 \overline{\mathbf{U}} _{\mathrm{E}}$

or $\mathbf{U}=\frac{1}{2} \varepsilon _{0} \mathrm{E} _{0}^{2}$

(xiii) The intensity (I) of EM-waves is defined as the total energy flowing through unit normal area in one second.

$$ \begin{aligned} & \therefore \quad \mathrm{I}=\overline{\mathbf{U}} \times \mathrm{c} \\ & \text { or } \quad \mathrm{I}=\frac{1}{2} \varepsilon _{0} \mathrm{cE} _{0}^{2} & \hspace{1.5cm} (\mathrm{or} \mathrm{I} \propto \mathrm{E}^2 _ 0) \end{aligned} $$

(xiv) Since EM-waves carry energy and momentum these waves can exert pressure (called radiation pressure) on the incident surface for total absorption,

$$ \begin{aligned} & \text { Momentum, } \mathrm{P}=\frac{\mathrm{U}}{\mathrm{c}}=\frac{\text { Total energy absorbed }}{\text { Speed of light in free space }} \\ & \text { and radiation pressure }=\frac{\text { Change in momentum }}{\text { Surface area }} \\ & \hspace{3.6cm} =\frac{\mathrm{P}}{\mathrm{A}} \end{aligned} $$

(xv) Pointing vector $\mathbf{S}$ is defined as:

$ \mathrm{S}=\frac{1}{\mu _{0}}(\mathbf{E} \times \mathbf{B}) $

Also, $\mathrm{I}=|\mathbf{S}|=\left|\frac{\mathbf{E} \times \mathbf{B}}{\mu _{0}}\right|$; the direction of this vector gives the direction of propagation of the EM-wave.

(xiv) To produce a continuous train ofEM waves, we need an oscillating / accelerating charge. This is done with the help of dipole antenna connected to oscillating voltage source. The frequency of EM waves is exactly equal to the frequency of the oscillating charge or oscillating voltage source.

6. The electromagnetic spectrum consists of different regions; namely. Radio waves, micro waves, infrared radiations, visible light, ultraviolet rays, $x$-rays and gamma rays, in the ascending order of their frequency.

Easy

Source of Electromagnetic Waves

1. Electromagnetic waves are produced by

(1) charges at rest, having the same linear charge density along two mutually perpendicular directions.

(2) equal charges, moving with equal uniform velocities, along two mutually perpendicular directinos.

(3) planer rectangular coils having a steady current flowing through them.

(4) charges that move with a continuously changing velocity

Show Answer

Correct answer: (4)

Solution:

Maxwell’s theory, of EM waves, tells us that charges, when accelerated, radiate EM waves. It is only charges, moving with a changing velocity, that can be regarded as accelerated charges. Therefore, it is only such charges that can be a source of EM waves.

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Equations for EM Waves

2. An electromagnetic wave, propagating along the $x$-axis, can be represented by the equations:

(1) $\quad \mathrm{E} _{\mathrm{y}}=\mathrm{E} _{0} \sin (\mathrm{ky}-\omega \mathrm{t}) ; \mathrm{B} _{\mathrm{z}}=\mathrm{B} _{0} \sin (\mathrm{kz}-\omega \mathrm{t})$

(2) $\quad\mathrm{E} _{\mathrm{z}}=\mathrm{E} _{0} \sin (\mathrm{ky}-\omega \mathrm{t}) ; \mathrm{B} _{\mathrm{y}}=\mathrm{B} _{0} \sin (\mathrm{ky}-\omega \mathrm{t})$

(3) $\quad \mathrm{E} _{\mathrm{y}}=\mathrm{E} _{0} \sin (\mathrm{k} x-\omega \mathrm{t}) ; \mathrm{B} _{\mathrm{z}}=\mathrm{B} _{0} \sin (\mathrm{k} x-\omega \mathrm{t})$

(4) $\quad\mathrm{E} _{\mathrm{z}}=\mathrm{E} _{0} \sin (\mathrm{ky}-\omega \mathrm{t}) ; \mathrm{B} _{\mathrm{y}}=\mathrm{B} _{0} \sin (\mathrm{kz}-\omega \mathrm{t})$

Show Answer

Correct answer: (3)

Solution:

For an EM wave, propagating along the $x$-direction, the electric and magnetic fields have to be along the $y(/ z)$ and $z(/ y)$ directions. Further there magnitudes show a simusoidal variation, for different points along their direction of propagation, i.e, along the $x$-axis. This simusoidal dependence, on $x$, corresponds only to the equations given in option (3).

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Features of EM Waves

3. From the following, the only incorrect statement, about EM waves, is the statement:

(1) “EM waves can correspond only to self-sustaining oscillations of electric and magnetic fields in vaccum”.

(2) ‘We do not need the hypothesis of the (hypothetical) medium ’the ether’, to understand the production, and propagation, of EM waves”.

(3) “EM waves can propagate through material media”.

(4) “The standard, of length, the metre, is defined in terms of the (strictly) constant value of the velocity of EM waves in vaccum”.

Show Answer

Correct answer: (1)

Solution:

Statement (1) is incorrect, because of the presence of the word ‘only’ in it. EM waves can correspond to self sustaining oscillations of electric and magnetic fields, in vaccum, as well as in material media.

Statement (2), (3) and (4) are all (now) well known correct statements about EM waves.

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Relating E and B Vector

4. The electric field, in a plane $E M$ wave, has a value $\left(9 \mathrm{Vm}^{-1}\right) \hat{\mathbf{j}}$, at a point $(\mathbf{a}, \mathbf{b}, c)$ in space at a time, $t _{0}$. The velocity of this wave is given by $c=\left(3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right) \hat{\mathrm{i}}$. The magnetic field, in this wave, at the point $(a, b, c)$, and at time $t _{0}$, would be given by

(1) $\mathbf{B}=\left(3 \times 10^{-8} \mathrm{~T}\right)(-\hat{\mathrm{k}})$

(2) $\mathbf{B}=\left(27 \times 10^{8} \mathrm{~T}\right)(\hat{\mathrm{k}})$

(3) $\mathbf{B}=\left(3 \times 10^{-8} \mathrm{~T}\right)(\hat{\mathrm{k}})$

(4) $\mathbf{B}=\left(27 \times 10^{8} \mathrm{~T}\right)(-\hat{\mathrm{k}})$

Show Answer

Correct answer: (3)

Solution:

The magnitude of the magnetic field, in an EM wave, is related to the corresponding magnitude of electric field, through the relation

$$ B=\frac{E}{c} $$

Hence the magnitude of the magnetic field (at the same point and at the same time), equals $\left(\frac{9}{3 \times 10^{8}}\right) \mathrm{T}$, i.e. $3 \times 10^{-8} \mathrm{~T}$.

Further, the direction of propagation (i.e., the direction $(\hat{\mathrm{n}})$ in which energy is transferred, by the EM wave) is related to the directions of $\mathbf{E}$ and $\mathbf{B}$, as $\hat{n}=\frac{(\mathbf{E} \times \mathbf{B})}{|\mathbf{E} \times \mathbf{B}|}$.

Here $\hat{\mathrm{n}}=\hat{\mathrm{i}}$ (given). We, therefore have

$$ \hat{\mathrm{j}} \times \frac{(\mathbf{B})}{|\mathbf{B}|}=\hat{\mathrm{i}} $$

This gives $\hat{k}$ as the direction of $\mathbf{B}$.

Hence $\mathbf{B}=\left(3 \times 10^{-8}\right) \hat{\mathrm{k}}$

Difficult

Relating Directions of E and B

5. The magnetic field, (at a particular point $x _{0}$ at a time $t _{0}$ ) in a plane EM wave, is given by

$$ \mathbf{B}=\mathbf{B} _{\mathrm{y}}=\left[\mathrm{B} _{0} \sin \left(\mathrm{k} x _{0}+\omega \mathrm{t} _{0}\right)\right] \hat{\mathrm{j}} $$

The electric field, for this EM wave again at the point $x _{0}$ and at the time $t _{0}$, would be represented by

(1) $\quad \mathbf{E}=\mathbf{E} _{\mathrm{z}}=\left[\mathrm{cB} _{0} \sin \left(\mathrm{k} x _{0}+\omega \mathrm{t} _{0}\right)\right](\hat {\mathrm{k}})$

(2) $\quad \mathbf{E}=\mathbf{E} _{z}=\left[\mathrm{cB} _{0} \sin \left(\mathrm{kx} _{0}+\omega \mathrm{t} _{0}\right)\right](-\hat {\mathrm{k}})$

(3) $\quad \mathbf{E}=\mathbf{E} _{z}=\left[\mathrm{CB} _{0} \sin \left(\mathrm{kx} _{0}-\omega \mathrm{t} _{0}\right)\right](\hat {\mathrm{k}})$

(4) $\mathbf{E}=\mathbf{E} _{z}=\left[\mathrm{cB} _{0} \sin \left(\mathrm{kx} _{0}-\omega \mathrm{t} _{0}\right)\right](-\hat {\mathrm{k}})$

Show Answer

Correct answer: (1)

Solution:

A close look, at the equation for $\mathbf{B}$, (where there is a +sign between the $x$ and $t$ terms) reveals that the EM wave is propagating along the $(-x)$ axis, i.e., has the direction $(-\hat{\mathrm{i}})$.

The direction of $\mathbf{E} \times \mathbf{B}$, i.e. $\left[\frac{(\mathbf{E})}{|\mathbf{E}|} \times \hat{\mathbf{j}}\right]$, must, therefore, be $(-\hat{\mathrm{i}})$. Hence the direction of $\mathbf{E}$ has to be along $(+\hat{\mathrm{k}})$. We also cannot have a representation of $\mathbf{E}$ (where there is a (-ve) sing between the $x$ and $\mathrm{t}$ terms) which implies that the EM wave is propagating along the $(+x)$ axis. The amplitude of $|\mathbf{E}|$ is $\mathrm{cB} _{0}$. Hence the overall correct representation of $\mathbf{E}$ is the one given in option(1).

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Wavelength Estimation

6. The electric field (in $\frac{\mathrm{V}}{\mathrm{m}}$ ), in a plane EM wave propagating along the $x$-axis, is given by

$$ \mathbf{E}=\mathbf{E} _{z}=\left[9 \sin \left(10^{3} x-3 \times 10^{11} \mathrm{t}\right)\right] \hat{\mathrm{k}} $$

The wavelength, $\lambda _{0}$, of this wave, and the amplitude $B _{0}$, of its magnetic field, are given by

(1) $\lambda _{0}=1.26 \times 10^{-2} \mathrm{~m} ; \mathrm{B} _{0}=3 \times 10^{-8} \mathrm{~T}$

(2) $\lambda _{0}=0.63 \times 10^{-2} \mathrm{~m} ; \mathrm{B} _{0}=3 \times 10^{-8} \mathrm{~T}$

(3) $\lambda _{0}=1.26 \times 10^{-2} \mathrm{~m} ; \mathrm{B} _{0}=27 \times 10^{8} \mathrm{~T}$

(4) $\lambda _{0}=0.63 \times 10^{-2} \mathrm{~m} ; \mathrm{B} _{0}=27 \times 10^{8} \mathrm{~T}$

Show Answer

Correct answer: (2)

Solution:

Comparing the given equation with its standard form, namely, $\mathrm{E} _{z}=\mathrm{E} _{0} \sin (\mathrm{k} x-\omega \mathrm{t})$, we find that

$$ \begin{aligned} & \mathrm{k}=\frac{2 \pi}{\lambda}=10^{3} \mathrm{~m} \\ \therefore \quad \lambda & =\frac{2 \pi}{10^{3}} \mathrm{~m}=6.284 \times 10^{-3} \mathrm{~m} \\ & \cong 0.63 \mathrm{~cm} \end{aligned} $$

Also, the amplitude, $\mathrm{B} _{0}$, of the magnetic field, equals $\frac{\mathrm{E} _{0}}{\mathrm{c}}$. Hence

$$ \mathrm{B} _{0}=\frac{9}{3 \times 10^{8}} \mathrm{~T}=3 \times 10^{-8} \mathrm{~T} $$

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Frequency Estimation

7. The magnetic field, (in tesla) in a plane EM wave, propagating along the $x$-axis, is given by

$$ \mathbf{B}=\mathbf{B} _{y}=3 \times 10^{-8} \sin \left(10^{3} x-3 \times 10^{11} \mathrm{t}\right) $$

The frequency of the EM wave, and the part of the EM spectrum, to which this wave could belong, are given, respectively, by

(1) $23.9 \mathrm{GHz}$; (short radio waves / microwaves)

(2) $23.9 \mathrm{GHz}$; (infrared waves)

(3) $47.7 \mathrm{GHz}$; (infrared waves)

(4) $47.7 \mathrm{GHz}$; (short radio waves / microwaves)

Show Answer

Correct answer: (4)

Solution: Comparing the given equation with its standard form, namely, $\mathrm{B} _{\mathrm{y}}=\mathrm{B} _{0} \sin (\mathrm{k} x-\omega \mathrm{t})$, we find that

$$ \omega(=2 \pi v)=3 \times 10^{11} \mathrm{~s}^{-1} $$

$$ \begin{aligned} \therefore \quad v & =\frac{3 \times 10^{11}}{2 \pi} \mathrm{Hz}=47.7 \times 10^{9} \mathrm{~Hz} \\ & =47.7 \mathrm{GHz} \end{aligned} $$

This frequency corresponds to the (short radio wave / microwave) part of the EM spectrum. (The infrared waves have frequencies of the order of $10^{12} \mathrm{~Hz}$ to $10^{14} \mathrm{~Hz}$ ).

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Estimating the Velocity of EM Waves

8. The electric field (in SI Units), in a plane EM wave, propagating along the $x$-axis, is given by

$$ \mathbf{E}=\mathbf{E z}=27 \sin \left(1.02 \times 10 x-2.95 \times 10^{15} \mathrm{t}\right) $$

The speed of propagation of this wave (as per this data), and the part of the EM spectrum, to which this wave corresponds, are given respectively, by

(1) $3.01 \times 10^{8} \mathrm{~m} / \mathrm{s}$; (visible)

(2) $2.89 \times 10^{8} \mathrm{~m} / \mathrm{s}$; (visible)

(3) $3.01 \times 10^{8} \mathrm{~m} / \mathrm{s}$; (infrared)

(4) $2.89 \times 10^{8} \mathrm{~m} / \mathrm{s}$; (infrared)

Show Answer

Correct answer: (2)

Solution:

Comparing the given equation, with its standard form, namely $\mathrm{E} _{\mathrm{z}}=\mathrm{E} _{0} \sin (\mathrm{k} x-\omega \mathrm{t})$, we find that

$$ \mathrm{k}\left(=\frac{2 \pi}{\lambda}\right)=1.02 \times 10^{7} \mathrm{~m}^{-1} \text { and } \omega(=2 \pi \lambda)=2.95 \times 10^{15} \mathrm{~s}^{-1} $$

$\therefore$ We get $\lambda=\frac{2 \pi}{1.02} \times 10^{-7} \mathrm{~m}=6.160 \times 10^{-7} \mathrm{~m}$

$ =6160 \AA $

This corresponds to the green colour in the visible part of EM spectrum.

Also speed of propagation $=v \lambda=\frac{\omega}{\mathrm{k}}$

$$ \begin{aligned} & =\frac{2.95 \times 10^{15}}{1.02 \times 10^{7}} \mathrm{~m} / \mathrm{s} \\ & \simeq 2.89 \times 10^{8} \mathrm{~m} / \mathrm{s} \end{aligned} $$

Average

Writing the Equation for Electric Field of an EM-Wave

9. An EM wave has a wavelength $6.284 \times 11^{-7} \mathrm{~m}$ and a frequency $4.77 \times 10^{14} \mathrm{~Hz}$. It is propagating along the $(+x)$ - axis and its magnet field oscillations are along the $(-\mathrm{y})$ - axis. The electric field, for this wave, can be represented [in its standard form $\left.\mathrm{E}=\mathrm{E} _{0} \sin (\mathrm{k} x \mp \omega \mathrm{t})\right]$ as

(1) $\mathrm{E} _{\mathrm{y}}=\mathrm{E} _{0} \sin \left(10^{7} x+2.995 \times 10^{15} \mathrm{t}\right)$

(2) $\mathrm{E} _{\mathrm{z}}=\mathrm{E} _{0} \sin \left(2.997 \times 10^{15} x+10^{7} \mathrm{t}\right)$

(3) $\mathrm{E} _{\mathrm{y}}=\mathrm{E} _{0} \sin \left(2.997 \times 10^{15} x-10^{7} \mathrm{t}\right)$

(4) $\mathrm{E} _{\mathrm{z}}=\mathrm{E} _{0} \sin \left(10^{7} x+2.997-10^{15} \mathrm{t}\right)$

Show Answer

Correct answer: (4)

Solution:

We have

$$ \mathrm{k}=\frac{2 \pi}{\lambda} \text { and } \omega=2 \pi \mathrm{v} $$

$\therefore$ In the given case

$$ \mathrm{k}=\frac{2 \pi}{6.284 \times 10^{-7}} \mathrm{~m}^{-1}=10^{7} \mathrm{~m}^{-1} $$

and $\omega=2 \pi \mathrm{v}=2 \pi\left(4.77 \times 10^{14}\right) \mathrm{s}^{-1}=\left(2.997 \times 10^{15}\right) \mathrm{s}^{-1}$

We, therefore, use these values of $k$ and $\omega$ to write the equation in its standard form. We need to use the ( - ve) sign between the $x$ and $t$, terms as the wave is propagating along the $+x$-axis. Further, the elctric field has to be along the $z$-axis, the third of the three perpendicular directions, satisfying the condition that the direction of propagation is along EXB.

Difficult

Writing the Equation for the Magnetic Field of an EM-Wave

10. An EM-wave has a wavelength of $3142 \AA$ and a frequency of $9.54 \times 10^{14} \mathrm{~Hz}$. Its electrical field oscillations are along the $x$-axis and it is transmitting energy along the $(+z)$-axis. The equation, representing its magnetic field oscillations in the standard form (like $\mathrm{B} _{z}=\mathrm{B} _{0} \sin (\mathrm{k} x \mp \omega \mathrm{t})$, would be:

(1) $\mathrm{B} _{\mathrm{y}}=\mathrm{B} _{0} \sin \left[2 \times 10^{7} x-5.99 \times 10^{15} \mathrm{t}\right]$

(2) $\mathrm{B} _{\mathrm{y}}=\mathrm{B} _{0} \sin \left[2 \times 10^{7} \mathrm{z}-5.99 \times 10^{15} \mathrm{t}\right]$

(3) $\mathrm{B} _{\mathrm{y}}=\mathrm{B} _{0} \sin \left[5.99 \times 10^{15} \mathrm{z}-2 \times 10^{7} \mathrm{t}\right] $

(4) $\mathrm{B} _{\mathrm{y}}=\mathrm{B} _{0} \sin \left[5.99 \times 10^{15} x-2 \times 10^{7} \mathrm{t}\right]$

Show Answer

Correct answer: (2)

Solution:

The direction of transmission of energy (given to be along the $\mathrm{z}$-axis) is also the direction of the cross product $\mathbf{E} \times \mathbf{B}$. Since $\mathbf{E}$ is (given) oscillating along the $x$-axis and $\mathbf{E} \times \mathbf{B}$ is along the $z$-axis, the oscillations of $\mathbf{B}$ has to be along the $y$-axis. From the given data

$$ \mathrm{k}=\frac{2 \pi}{\lambda}=\frac{2 \pi}{3142 \times 10^{-10}}=2 \times 10^{7} \mathrm{~m}^{-1} $$

and $\omega=2 \pi \upsilon =2 \pi \times 9.54 \times 10^{14} \mathrm{~s}^{-1} \simeq 5.99 \times 10^{15} \mathrm{~s}^{-1}$

The direction of propagation, being the $(+z)$-axis, we would write

$$ B=B _{0} \sin \left(2 \times 10^{7} z-5.99 \times 10^{15} t\right) $$

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Electromagnetic Spectrum

11. The oscillating electric field ( $\mathrm{E} _{1}$ ) of an $\mathbf{E M}$ wave (1), and the oscillating magnetic field ( $\mathbf{B} _{2}$ ), of an EM wave (2), are represented, respectively, by

$ \mathrm{E} _{1}=\mathrm{E} _{0} \sin \left(10^{13} x-3 \times 10^{21} \mathrm{t}\right) $

and $\mathrm{B} _{2}=\mathrm{B} _{0} \sin \left(6 \times 10^{5} \mathrm{z}-1.8 \times 10^{14} \mathrm{t}\right)$

The parts of the EM spectrum, to which waves (1) and (2), are likely to belong, are

(1) UV rays and microwaves, respectively

(2) $\gamma$-rays and infrared rays, respectively

(3) $x$-rays and radio waves, respectively

(4) UV rays and microwaves, respectively

Show Answer

Correct answer: (2)

Solution:

For wave (1), $\mathrm{k}=\left(10^{13}\right) \mathrm{m}^{-1}$

$\therefore \quad \lambda=\frac{2 \pi}{\mathrm{k}}=2 \pi \times 10^{-13} \mathrm{~m}$

This wavelength value corresponds to $\gamma$-rays.

For wave (2), $\mathrm{k}=\left(6 \times 10^{5}\right) \mathrm{m}^{-1}$

$\therefore \quad \lambda=\frac{2 \pi}{6 \times 10^{5}} \approx 1.02 \times 10^{-5} \mathrm{~m}$

$=10200 \AA$

This wavelength value corresponds to infrared rays.

Hence waves (1) and (2) are likely to be $\gamma$-rays and infrared rasy, respectively.

Average

EM Spectrum

12. The equation $\mathrm{E}=\mathrm{E} _{0} \sin (\mathrm{k} x-\omega \mathrm{t})$ could represent the oscillations of the electric field of a microwave if $\mathrm{k}$ and $\omega$ have the values

(1) $10^{10} \mathrm{~m}^{-1}$ and $3 \times 10^{18} \mathrm{~s}^{-1}$, respectively

(2) $10^{7} \mathrm{~m}^{-1}$ and $3 \times 10^{15} \mathrm{~s}^{-1}$, respectively

(3) $6 \times 10^{2} \mathrm{~m}^{-1}$ and $1.8 \times 10^{11} \mathrm{~s}^{-1}$, respectively

(4) $6 \times 10^{-2} \mathrm{~m}^{-1}$ and $1.8 \times 10^{7} \mathrm{~s}^{-1}$, respectively

Show Answer

Correct answer: (3)

Solution:

For option (1), $\lambda \simeq 2 \pi \times 10^{-10} \mathrm{~m}$ ( $x$-rays)

For option (2), $\lambda \simeq 2 \pi \times 10^{-7} \mathrm{~m}$ (visible light)

For option (3), $\lambda \simeq 1.05 \times 10^{-2} \mathrm{~m} \simeq 1.02 \mathrm{~cm}$ (microwaves)

For option (4), $\lambda \simeq 1.05 \times 10^{2} \mathrm{~m}$ (radiowaves)

(In each case, the value of $\omega$ is ( $\simeq \mathrm{ck})$, as it should be)

Hence only option (3) could correspond to the representatin of microwaves.

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Related Values of $\omega$ and $k$, for an EM Wave

13. A student writes the following expression for the oscillations of the magnetic field of an em wave: $B=B _{0} \sin \left(10^{13} x-3 \times 10^{20} t\right)$

Assuming that his choice of the value, of the coefficient of $x$, is correct, the value, of his quoted coefficient of $t$, is

(1) less than the correct value by (nearly) $2.7 \times 10^{21}$

(2) more than the correct value by $2.7 \times 10^{21}$

(3) less than the correct value by (nearly) $3 \times 10^{19}$

(4) more than the correct value by $3 \times 10^{19}$

Show Answer

Correct answer: (1)

Solution:

As per the stand form of representation, namely $B=B _{0} \sin \left(10^{13} x-3 \times 10^{20} t\right)$

the values of the coefficient of $x$ and $\mathrm{t}$, correspond to ’ $\mathrm{k}$ ’ and ’ $\omega$ ‘, respectively.

Since $\frac{\omega}{\mathrm{k}}=\mathrm{c}$ (the velocity of em waves), we have $\omega=\mathrm{ck}=\left(3 \times 10^{8}\right) \mathrm{k}$

$\therefore$ If $\mathrm{k}$ is $10^{13} \mathrm{~m}^{-1}$, we should have $\omega=3 \times 10^{21} \mathrm{~s}^{-1}$

The given value is $\omega=3 \times 10^{20} \mathrm{~s}^{-1}=0.3 \times 10^{21} \mathrm{~s}^{-1}$

$\therefore$ The given value is less than the correct value by $(3-0.3) \times 10^{21} \mathrm{~s}^{-1}\left(\right.$ or $\left.2.7 \times 10^{21} \mathrm{~s}^{-1}\right)$

Average

Related Values of $\omega$ and $k$, for an EM Wave

14. A student writes the following expression, for the oscillation of the electric field of an em wave: $\mathrm{E} _{x}=\mathrm{E} _{0} \sin \left(6 \times 10^{11} \mathrm{z}-1.8 \times 10^{19} \mathrm{t}\right)$

Assuming that she has correctly written the value of the coefficient of $t$, the value of the coefficient of $x$, given by her, is

(1) less than its correct value by (nearly) $5.4 \times 10^{11} \mathrm{~m}^{-1}$

(2) more than its correct value by (nearly) $5.4 \times 10^{11} \mathrm{~m}^{-1}$

(3) less than its correct value by (nearly) $6.6 \times 10^{11} \mathrm{~m}^{-1}$

(4) more than its correct value by (nearly) $6.6 \times 10^{11} \mathrm{~m}^{-1}$

Show Answer

Correct answer: (2)

Solution:

As per the standard representation, namely $\mathrm{E} _{x}=\mathrm{E} _{0} \sin (\mathrm{kz}-\omega \mathrm{t})$, the value, of the coefficient of $\mathrm{z}$, corresponds to $\mathrm{k}$ and that, of the coefficient of t, corresponds to $\omega$. We also have $\mathrm{k}=\frac{\omega}{\mathrm{c}}$

$\therefore$ For the given {correct} value of $\omega\left(=1.8 \times 10^{19}\right)$, we should have $\mathrm{k}=\frac{1.8 \times 10^{19}}{3 \times 10^{8}} \mathrm{~m}^{-1}=6 \times 10^{10} \mathrm{~m}^{-1}$

The given value is $\mathrm{k}=6 \times 10^{11} \mathrm{~m}^{-1}=60 \times 10^{10} \mathrm{~m}^{-1}$

This is more than the correct value by $(60-6) \times 10^{10} \mathrm{~m}^{-1}=\left(54 \times 10^{10}\right) \mathrm{m}^{-1}=5.4 \times 10^{11} \mathrm{~m}^{-1}$

Average

Identifying the Part of the EM Spectrum

15. The oscillation, of the magnetic field, of an EM wave, propagating along the $x$-axis, are represented by $B=B _{0} \sin (k x-\omega t)$

This EM wave could be visible light if the values of $k$ and $\omega$ are (nearly)

(1) $10^{10} \mathrm{~m}^{-1}$ and $3 \times 10^{18} \mathrm{~s}^{-1}$, respectively

(2) $10^{8} \mathrm{~m}^{-1}$ and $3 \times 10^{16} \mathrm{~s}^{-1}$, respectively

(3) $10^{7} \mathrm{~m}^{-1}$ and $3 \times 10^{15} \mathrm{~s}^{-1}$, respectively

(4) $10^{3} \mathrm{~m}^{-1}$ and $3 \times 10^{11} \mathrm{~s}^{-1}$, respectively

Show Answer

Correct answer: (3)

Solution:

We have $\mathrm{k}=\frac{2 \pi}{\lambda}$

$ \therefore \quad \lambda=\frac{2 \pi}{\mathrm{k}} $

For option(1)

$ \lambda \simeq 2 \pi \times 10^{-10} \mathrm{~m}^{-1} \simeq 6.14 \stackrel{\circ}{\AA} \qquad(x-\text { rays }) $

For option(2)

$ \lambda \simeq 2 \pi \times 10^{-8} \mathrm{~m} \simeq 614 \stackrel{\circ}{\AA} \qquad \text { (uv }- \text { rays) } $

For option(3)

$ \lambda \simeq 2 \pi \times 10^{-7} \mathrm{~m} \simeq 6140 \AA \qquad \text{(visible light)} $

For option (4)

$ \lambda \simeq 2 \pi \times 10^{-3} \mathrm{~m} \simeq 6.14 \mathrm{~mm} \qquad \text { (microwaves) } $

(In each case, the value of $\omega(=\mathrm{ck})$, has been correctly quoted).

Hence only option (3) could correspond to visible light.

Average

Characterstics of EM Waves

16. In the figure drawn here, the points A, B, C, D, E, F and G, on the $x$-axis, correspond to radiowaves; infrared waves, visible light, $x$-rays; microwaves, uv rays and $\gamma$-rays, in that order. The points P, Q, R, S, T, U, V, shown in the figure correspond to certain values, of a varible $y$, of these waves, on the $y$-axis. The variable $y$, could be the

$~$

(1) wavelength $(\lambda)$, or the angular frequency $(\omega)$, of these waves

(2) frequency ( $v$ ), or the wave number ( $k$ ), of these waves

(3) angular frequency $(\omega)$, or the velocity (v) of these waves in vaccum

(4) velocity (v), in vaccum, or the wave number ( $k$ ), of these waves in vaccum

Show Answer

Correct answer: (2)

Solution:

For the given order of points A, B, C, D, E, F, G; the frequency increases for the first four waves; then decreases (to a value between the first two radius); then increases but to a value between the third and fourth values and finally increases to a value higher than all the other values. The wave number $\mathrm{k}\left(=\frac{2 \pi}{\lambda}\right)$ showss a similar behaviour (This is in accord with the changes in the values of variable $\mathrm{Y}$, as show in the figure). The wavelength $\lambda$ behaves is an exactly opposite manner while $v$ has the same value for all the waves.

Hence option(2) is the only correct option.

Average

Characterstics of EM Waves

17. An em wave is propagating along the positive $x$-axis. The representation, of the oscillations of its electric field, and the direction of oscillations of its $E$ and $B$ vectors, could then be

(1) $\mathrm{E}=\mathrm{E} _{0} \sin (\mathrm{k} x-\omega \mathrm{t}) ;\left(\mathbf{E}=\mathbf{E} _{\mathrm{y}}\right.$ and $\left.\mathbf{B}=\mathbf{B} _{z}\right)$

(2) $\mathrm{E}=\mathrm{E} _{0} \sin (\mathrm{k} x-\omega \mathrm{t}) ;\left(\mathbf{E}=\mathbf{E} _{z}\right.$ and $\left.\mathbf{B}=\mathbf{B} _{\mathrm{y}}\right)$

(3) $\mathrm{E}=\mathrm{E} _{0} \sin (\mathrm{k} x+\omega \mathrm{t}) ;\left(\mathbf{E}=\mathbf{E} _{z}\right.$ and $\left.\mathbf{B}=\mathbf{B} _{\mathrm{y}}\right)$

(4) $\mathrm{E}=\mathrm{E} _{0} \sin (\mathrm{k} x+\omega \mathrm{t}) ;\left(\mathbf{E}=\mathbf{E} _{\mathrm{y}}\right.$ and $\left.\mathbf{B}=\mathbf{B} _{z}\right)$

Show Answer

Correct answer: (1)

Solution:

The wave is propagating along the (+ ve) $x$-axis. Hence we need to have a $(-\mathrm{ve})$ sign between the ’ $x$ ’ and ’ $t$ ’ terms in the representative equation of the oscillations of its electric field.

Further, the cross product $(\mathbf{E} \times \mathbf{B})$ must also have the direction of propagation of the em. Waves, i.e. the $(+\mathrm{ve}) x$-axis. This is possible if we take $\left(\mathbf{E}=\mathbf{E} _{y}\right)$ and $(\mathbf{B}=\mathbf{B z})$; the other choice $\left(\mathbf{E}=\mathbf{E} _{z}\right.$ and $\left.\mathbf{B}=\mathbf{B} _{y}\right)$ gives the direction of $(\mathbf{E} \times \mathbf{B})$ as the $(-\mathrm{ve}) x$-axis.

Hence option (1), consistent with both these requirements, is the correct option.

Easy

Electromagnetic Spectrum

18. The angular frequencies, for three em waves $\mathbf{E} _{1}, \mathbf{E} _{2}, \mathbf{E} _{3}$ have values $\omega _{1}, \omega _{2}$ and $\omega _{3}$, respectively. If $\omega _{1}$ is greater than both $\omega _{2}$ and $\omega _{3}$ and $\omega _{2}$ is less than $\omega _{3}$, the three em waves could be, respectively,

(1) $x$-rays, visible light, microwaves

(2) $\gamma$ - rays, uv - rays, microwaves

(3) visible light, infrared rays, radiowaves

(4) uv - rays, radiowaves, visible light

Show Answer

Correct answer: (4)

Solution:

We are given that $\omega _{1}>\left(\omega _{2}\right.$ or $\left.\omega _{3}\right)$ and $\omega _{2}<\omega _{3}$

$\therefore$ The angular frequencies are in the order $\omega _{1}>\omega _{3}>\omega _{2}$

The three em waves must, therefore, have their frequencies as per this requirement. In all the options given, $\omega _{1}$ is greater than both $\omega _{2}$ and $\omega _{3}$. However, in the first three options, $\omega _{2}<\omega _{3}$. It is only in the fourth option that we have $\omega _{2}<\omega _{3}$. Hence option (4) is the correct choice.

Average

Electromagnetic Spectrum

19. The wave number ( $k$ ), for three em waves $\left(x _{1}, x _{2}\right.$ and $x _{3}$ ) has value $k _{1}, k _{2}$ and $k _{3}$, respectively. It is given that $k _{1}$ is less than both $k _{2}$ and $k _{3}$ while $k _{2}$ is greater than $k _{3}$. The three em waves, $\left(x _{1}, x _{2}, x _{3}\right)$ could then be

(1) infrared rays, uv rays, visible light

(2) visible light, uv rays, $\gamma$ - rays

(3) radiowaves, visible light, uv rays

(4) microwaves, visible light, $\gamma$ - rays

Show Answer

Correct answer: (1)

Solution:

We have $\mathrm{k}=\frac{2 \pi}{\lambda}$

$\therefore$ We must have $\lambda _{1}$ more than both $\lambda _{2}$ and $\lambda _{3}$.

Also we must have $\lambda _{2}$ less than $\lambda _{3}$. Hence the three em waves must have their wavelenghts in the order.

$$ \lambda _{1}>\lambda _{3}>\lambda _{2} $$

This ordering of wavelenghts is valid only in option (1). In all the other three options, $\lambda _{2}$ is more than $\lambda _{3}$.

Hence option (1) is the correct choice.

Difficult

Electromagnetic Spectrum

20. The velocity (v), in a given material medium, has values $v _{1}, v _{2}$ and $v _{3}$, respectively for three em waves $\left(y _{1}, y _{2}\right.$ and $\left.y _{3}\right)$. It is given that $v _{3}$ is more than both $v _{1}$ and $v _{2}$ but $v _{2}$ is less than $v _{1}$. The three em waves $\left(\mathbf{y} _{1}, \mathbf{y} _{2}, \mathbf{y} _{3}\right)$ could than be, respectively

(1) $\gamma$ - rays, uv rays, visible light

(2) uv rays, visible light, microwaves

(3) infrared rays, uv rays, radiowaves

(4) $x$-rays, visible light, infrared rays

Show Answer

Correct answer: (3)

Solution:

We know that the refractive index of a medium decreases with increase in wavelength $\left(\because \mu \simeq \mathrm{A}+\frac{\mathrm{B}}{\lambda^{2}}\right)$.

Further, since $\mu=\frac{\text { speed of light in vaccum }}{\text { speed of light in medium }}$, a decrease, in the value of $\mu$, implies a higher value of the speed of light in the medium. We, therefore, can conclude, that:

‘Higher wavelength implies a higher speed of light in the medium.’

We are given that $\mathrm{v} _{3}>\left(\mathrm{v} _{1}\right.$ or $\left.\mathrm{v} _{2}\right)$ and $\left(\mathrm{v} _{2}<\mathrm{v} _{1}\right)$. Thus $\mathrm{v} _{3}>\mathrm{v} _{1}>\mathrm{v} _{2}$. This implies $\lambda _{3}>\lambda _{1}>\lambda _{2}$. This sequence of wavelengths values is valid only in option (3).

Hence option (3) is the correct choice.

Difficult

Electromagnetic Spectrum

21. An em wave is propagating along the $(+) x$-axis. Let there be a point charge initially at rest located in some plane parallel to the $y-z$ plane. At a time instant, just (infinitesimally) greater than the instant at which the em wave reaches this plane, this charge could experience forces, due to the $E$ and $B$ fields, of the em wave, along the

(1) $y$-axis and $x$-axis, respectively

(2) $z$-axis and $y$-axis, respectively

(3) $x$-axis and $y$-axis, respectively

(4) y-axis and z-axis, respectively

Show Answer

Correct answer: (1)

Solution:

When the wave propagates along the $(+) x$-axis, its $\mathbf{E}$ and $\mathbf{B}$ fields oscillate in the yz plane; the $\mathbf{E}$ field along the $\mathrm{y}$-axis and $\mathbf{B}$ field along the $\mathrm{z}$-axis. (This is because $\mathbf{E} \times \mathbf{B}$ has the same direction as the direction of propagation, i.e. $+x$-axis, in this case).

The charge, initially at rest, experiences a force due to the $\mathbf{E}$ field, along the y-axis. If thus acuiqres a velocity along the $y$-axis. The magnetic field $\mathbf{B}$ (along the $z$-axis) would then exert a force along the $x$-axis (the direction of $\mathbf{v} \times \mathbf{B}$ ). This could happen at a time instant just infinitesimally greater than the time instant at which the em wave reaches the plane containing the charge).

Easy

Transfer of Energy by EM Waves

22. EM waves can transport energy;

(1) however, it is only the $\mathbf{E}$ field that contributes to this transfer of energy

(2) however, it is only the $\mathbf{B}$ field that contributes to this transfer of energy

(3) both the $\mathbf{E}$ and $\mathbf{B}$ fields contribute to the transfer of energy; their contributions are in the ratio $2: 1$

(4) both the $\mathbf{E}$ and $\mathbf{B}$ fields contribute to the transfer of energy; their contributions are in the ratio $1: 1$

Show Answer

Correct answer: (4)

Solution:

EM waves, (having energy associated with their electric and magnetic fields) can transport energy from one place to another.

Both the electric and mangetic fields contribute, in this transfer of energy. Further, the energy of an EM wave, is shared equally between its electric and magnetic fields. Hence the ratio of their contributions, in the transfer of energy, equals $(1: 1)$.

Easy

Transfer of Energy and Momentum by EM Waves

23. (1) EM waves can transport energy (through their $\mathbf{E}$ field) but not momentum

(2) EM waves can transport energy (through their $\mathbf{B}$ field) but not momentum

(3) EM waves can transport energy as well as momentum (through their $\mathbf{E}$ field only)

(4) EM waves can transport energy as well as momentum (through their $\mathbf{E}$ and $\mathbf{B}$ fields )

Show Answer

Correct answer: (4)

Solution:

EM waves (having energy, associated with their electric and magnetic fields) can transport energy from one place to another.

It has been shown that they can trasnfer momentum also.

$\left(\right.$ Total momentum delivered $\left.=\frac{(\text { Total energy transferred })}{c}\right)$

The electric, and the magnetic fields, of an em wave, contribute equally in this transfer of energy and momentum.

Easy

Energy Density of the E and B Fields

24. The (instantaneous) energy density (energy per unit volume) of the electric and magnetic fields, of an em wave, are equal, respectively, to

(1) $\frac{1}{2} \frac{\mathrm{E}^{2}}{\varepsilon _{0}}$ and $\frac{1}{2} \mu _{0} \mathrm{~B}^{2}$; with $\frac{1}{2} \frac{\mathrm{E}^{2}}{\varepsilon _{0}} \neq \frac{1}{2} \frac{\mathrm{B}^{2}}{\mu _{0}}$

(2) $\frac{1}{2} \frac{\mathrm{E}^{2}}{\varepsilon _{0}}$ and $\frac{1}{2} \mu _{0} \mathrm{~B}^{2}$; with $\frac{1}{2} \frac{\mathrm{E}^{2}}{\varepsilon _{0}}=\frac{1}{2} \frac{\mathrm{B}^{2}}{\mu _{0}}$

(3) $\frac{1}{2} \varepsilon _{0} \mathrm{E}^{2}$ and $\frac{1}{2} \frac{\mathrm{B}^{2}}{\mu _{0}}$; with $\frac{1}{2} \varepsilon _{0} \mathrm{E}^{2}=\frac{1}{2} \frac{\mathrm{B}^{2}}{\mu _{0}}$

(4) $\frac{1}{2} \varepsilon _{0} \mathrm{E}^{2}$ and $\frac{1}{2} \frac{\mathrm{B}^{2}}{\mu _{0}}$; with $\frac{1}{2} \varepsilon _{0} \mathrm{E}^{2} \neq \frac{1}{2} \frac{\mathrm{B}^{2}}{\mu _{0}}$

Show Answer

Correct answer: (3)

Solution:

The (instantaneous) energy density (energy per unit volume), of the electric and magnetic fields, of an em wave, is given by;

(Volume) energy density of electric field $=\frac{1}{2} \varepsilon _{0} \mathrm{E}^{2}$

and (volume) energy density of magnetic field $=\frac{1}{2} \frac{\mathrm{B}^{2}}{\mu _{0}}$

Here $\mathrm{E}$ and $\mathrm{B}$ refer to the instantaneous values of the electric and magnetic fields.

The energy, of an em wave, in shared equally between its electric and magnetic fields. We, therefore, also have

$$ \frac{1}{2} \varepsilon _{0} \mathrm{E}^{2}=\frac{1}{2} \frac{\mathrm{B}^{2}}{\mu _{0}},(\text { at all instants }) $$

Easy

Transfer of Energy, Momentum and ‘Pressure’by EM Waves

25. It has been experimentally demonstrated that

(1) EM waves can transport energy but not momentum

(2) EM waves can transport energy but not momentum; however, they do exert a ‘radiation pressure’

(3) EM waves can transport both energy and momentum; however, they do not exert any ‘radiation pressure’

(4) EM waves can transport both energy and momentum; they also exert a ‘radiation pressure’

Show Answer

Correct answer: (4)

Solution:

EM waves (having energy associated with their electric and magnetic fields) can transfer energy from one place to another.

It has been shown that they can transfer momentum also (Total momentum delivered in time $\Delta \mathrm{t}$ )

$=\Delta \mathrm{p}=\frac{\text { Total energy transferred in time }(\Delta \mathrm{t})}{\mathrm{c}}$

The average rate of transfer of momentum $\left(=\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}}\right)$, equals the force exerted on the surface (absorbing the radiant energy).

This force, per unit area of the surface, would then equal the ‘radiation pressure’.

Average

Energy Transfer by EM Waves

26. The total energy transfered to a (perfectly black) surface of area $A$ (located normal to the dirction of propagation of the EM wave) and thickness d, by an em wave, over one complete cycle, can be expressd as

(1) $\left[\frac{1}{2} \varepsilon _{0}(\overline{\mathrm{E}})^{2}+\frac{1}{2} \mu _{0}(\overline{\mathrm{B}})^{2}\right]( \mathrm {Ad})$

(2) $\left[\frac{1}{2} \varepsilon _{0}(\overline{\mathrm{E}})^{2}+\frac{1}{2} \frac{(\overline{\mathrm{B}})^{2}}{\mu _{0}}\right]( \mathrm {Ad} )$

(3) $\left[\frac{1}{2} \varepsilon _{0} \mathrm{E} _{\mathrm{rms}}^{2}+\frac{1}{2} \frac{\mathrm{B} _{\mathrm{rms}}^{2}}{\mu _{0}}\right]( \mathrm {Ad})$

(4) $\left[\frac{1}{2} \varepsilon _{0} \mathrm{E} _{\mathrm{rms}}^{2}+\frac{1}{2} \mu _{0} \mathrm{~B} _{\mathrm{rms}}^{2}\right]( \mathrm {Ad})$

Show Answer

Correct answer: (3)

Solution:

We know that

(Volume) energy density of electric field $=\frac{1}{2} \varepsilon _{0} \mathrm{E}^{2}$

(Volume) energy density of magnetic field $=\frac{1}{2}\left(\frac{\mathrm{B}^{2}}{\mu _{0}}\right)$

The energy, of the em wave, is shared equally between its electric and magnetic fields.

Over one complete cycle, we need to use $\mathrm{E} _{\mathrm{rms}}$ and $\mathrm{B} _{\mathrm{rms}}$ in the above expressions.

[We cannot use $\overline{\mathrm{E}}$ and $\overline{\mathrm{B}}$, as these equal zero each over one complete cycle].

The volume absorbing this energy $=(\mathrm{Ad})$

Hence total energy transferred over one complete cycle

$$ =\left[\frac{1}{2} \varepsilon _{0} \mathrm{E} _{\mathrm{rms}}^{2}+\frac{1}{2} \frac{\mathrm{B} _{\mathrm{rms}}^{2}}{\mu _{0}}\right][\mathrm{Ad}] $$

Difficult

Force due to EM Wave

27. An electromagnetic wave (of frequency $v$ ), propagating along the $x$-axis, is incident normally on a perfect black surface of area $A$ and thickness $d$. The average force, exerted on the surface, over one complete cycle, of this EM wave, is

(1) $\left[\frac{1}{2} \varepsilon _{0} \mathrm{E} _{\mathrm{rms}}^{2}+\frac{1}{2} \frac{\mathrm{B} _{\mathrm{ms}}^{2}}{\mu _{0}}\right]\left(\frac{\mathrm{Avd}}{\mathrm{c}}\right)$

(2) $\left[\frac{1}{2} \varepsilon _{0} \mathrm{E} _{\mathrm{rms}}^{2}+\frac{1}{2} \frac{\mathrm{B} _{\mathrm{ms}}^{2}}{\mu _{0}}\right]( \mathrm {cAvd})$

(3) $\left[\frac{1}{2} \varepsilon _{0} \mathrm{E} _{\mathrm{ms}}^{2}+\frac{1}{2} \frac{\mathrm{B} _{\mathrm{rms}}^{2}}{\mu _{0}}\right]\left(\frac{\mathrm{c}}{\mathrm{Avd}}\right)$

(4) $\left[\frac{1}{2} \varepsilon _{0} \mathrm{E} _{\mathrm{rms}}^{2}+\frac{1}{2} \frac{\mathrm{B} _{\mathrm{rms}}^{2}}{\mu _{0}}\right]\left(\frac{\mathrm{cAd}}{v}\right)$

Show Answer

Correct answer: (1)

Solution:

We know that

(Volume) energy density of electric field $=\frac{1}{2} \varepsilon _{0} \mathrm{E}^{2}$

(Volume) energy density of magnetic field $=\frac{1}{2}\left(\frac{\mathrm{B}^{2}}{\mu _{0}}\right)$

The energy, of the em wave, is shared equally between its electric and magnetic fields.

Over one complete cycle, we need to use $\mathrm{E} _{\mathrm{rms}}\left(=\frac{\mathrm{E} _{0}}{\sqrt{2}}\right)$ and $\mathrm{B} _{\mathrm{rms}}\left(=\frac{\mathrm{B} _{0}}{\sqrt{2}}\right)$ in the above expressions.

$\therefore$ Total energy transferred over one complete cycle

$$ =\mathrm{U}=\left[\frac{1}{2} \varepsilon _{0} \mathrm{E} _{\mathrm{rms}}^{2}+\frac{1}{2} \frac{\mathrm{B} _{\mathrm{rms}}^{2}}{\mu _{0}}\right][\mathrm{Ad}] $$

$\therefore$ Momentum delivered $=\mathrm{p}=\frac{\mathrm{U}}{\mathrm{c}}=\left[\frac{1}{2} \varepsilon _{0} \mathrm{E} _{\mathrm{rms}}^{2}+\frac{1}{2} \frac{\mathrm{B} _{\mathrm{rms}}^{2}}{\mu _{0}}\right]\left[\frac{\mathrm{Ad}}{\mathrm{c}}\right]$ over one cycle.

Time taken $=$ Time period $=\frac{1}{v}$

$\therefore$ Average force $=\frac{\mathrm{p}}{\left(\frac{1}{\mathrm{v}}\right)}=\mathrm{pu}$

$$ =\left[\frac{1}{2} \varepsilon _{0} \mathrm{E} _{\mathrm{rms}}^{2}+\frac{1}{2} \frac{\mathrm{B} _{\mathrm{rms}}^{2}}{\mu _{0}}\right]\left[\frac{\mathrm{vAd}}{\mathrm{c}}\right] $$

Difficult

Energy of an EM Wave

28. The oscillatory electric field, of an em wave, propagating along the $+z-$ axis, is represented by $\mathbf{E} _{x}=\mathbf{E} _{0} \sin (\mathbf{k z}-\omega \mathbf{t})$.

This em wave is incident on a perfectly black circular disc, of diameter $d$ and height $h$, located parallel to the $x-y$ plane. The total energy absorbed by this surface, over one complete cycle of this em wave, would equal

(1) $\left(\frac{\pi \mathrm{d}^{2} \mathrm{~h} \mathrm{E} _{0}^{2}}{8 \mu _{0} \mathrm{c}^{2}}\right)$

(2) $\frac{\pi \mathrm{d}^{2} \mathrm{~h} \mathrm{E} _{0}^{2}}{16 \mathrm{c}^{2} \mu _{0}}\left[1+\frac{\mu _{0} \varepsilon _{0}}{\mathrm{c}^{2}}\right]$

(3) $\frac{\pi \mathrm{d}^{2} \mathrm{~h} \mathrm{E} _{0}^{2}}{4 \mathrm{c}^{2}}\left[\varepsilon _{0}+\frac{1}{\mathrm{c}^{2} \mu _{0}}\right]$

(4) $\frac{\pi \mathrm{d}^{2} \mathrm{~h} \mathrm{E} _{0}^{2}}{8 \varepsilon _{0} \mathrm{c}^{2}}$

Show Answer

Correct answer: (1)

Solution:

We know that

(Volume) energy density of electric field $=\frac{1}{2} \varepsilon _{0} \mathrm{E}^{2}$

(Volume) energy density of magnetic field $=\frac{1}{2}\left(\frac{\mathrm{B}^{2}}{\mu _{0}}\right)$

The energy of the EM wave is shared equally between the electric and the magnetic fields.

Over one complete cycle, we need to use $\mathrm{E} _{\mathrm{rms}}$ and $\mathrm{B} _{\mathrm{rms}}$ in the above expressions.

Also $\mathrm{E} _{\mathrm{rms}}=\frac{\mathrm{E} _{0}}{\sqrt{2}}$ and $\mathrm{B} _{\mathrm{rms}}=\frac{\mathrm{B} _{0}}{2}$

We also have $\mathrm{B} _{0}=\left(\frac{\mathrm{E} _{0}}{\mathrm{c}}\right)$

The volume of the circuluar disc $=\left(\frac{\pi \mathrm{d}^{2}}{4}\right) \mathrm{h}$

$\therefore$ Total energy delivered over one complete cycle

$$ =\left[\frac{1}{2} \varepsilon _{0} \mathrm{E} _{\mathrm{rms}}^{2}+\frac{1}{2} \frac{\mathrm{B} _{\mathrm{rms}}^{2}}{\mu _{0}}\right]\left(\frac{\pi \mathrm{d}^{2}}{4}\right) \mathrm{h} $$

$$ \begin{aligned} & =\frac{1}{16}\left(\pi \mathrm{d}^{2} \mathrm{~h}\right)\left[\varepsilon _{0} \mathrm{E} _{0}^{2}+\frac{\mathrm{E} _{0}^{2}}{\mathrm{c}^{2} \mu _{0}}\right]=\frac{\pi \mathrm{d}^{2} \mathrm{~h}}{16} \varepsilon _{0} \mathrm{E} _{0}^{2}\left[1+\frac{1}{\mathrm{c}^{2} \varepsilon _{0} \mu _{0}}\right] \\ \\ & =\frac{\pi \mathrm{d}^{2} \mathrm{~h} \varepsilon _{0} \mathrm{E} _{0}^{2}}{8}=\frac{\pi \mathrm{d}^{2} \mathrm{hE} _{0}^{2}}{8 \mu _{0} \mathrm{c}^{2}} \quad\left(\because \mu _{0} \varepsilon _{0}=\frac{1}{\mathrm{c}^{2}}\right) \end{aligned} $$

Difficult

Energy Transfer by an EM Wave

29. The oscillatory magnetic field, of an em wave, propagating along the (-) $x$-axis, is represented by $B _{y}=B _{0} \sin (k x+\omega t)$

This em wave is incident on a perfectly black circular disc (or radius $r$ and thickness $d$ ) kept parallel to the $y-z$ plane. The total energy, transferred to this disc, over one complete cycle, by this em wave, would equal

(1) $\frac{\mathrm{B} _{0}^{2}}{4}\left[\varepsilon _{0}+\frac{1}{\mu _{0} \mathrm{c}^{2}}\right]\left(\pi \mathrm{r}^{2} \mathrm{~d}\right)$

(2) $\frac{\mathrm{B} _{0}^{2}}{8}\left[\frac{\varepsilon _{0}}{\mathrm{c}^{2}}+\mu _{0}\right]\left(\pi \mathrm{r}^{2} \mathrm{~d}\right)$

(3) $\frac{\mathrm{B} _{0}^{2}}{8}\left[\varepsilon _{0}+\frac{1}{\mu _{0} \mathrm{c}^{2}}\right]\left(\pi \mathrm{r}^{2} \mathrm{~d}\right)$

(4) $\frac{\mathrm{B} _{0}^{2}}{4}\left[\varepsilon _{0} \mathrm{c}^{2}+\frac{1}{\mu _{0}}\right]\left(\pi \mathrm{r}^{2} \mathrm{~d}\right)$

Show Answer

Correct answer: (4)

Solution:

We know that

(Volume) energy density of the electric field $=\frac{1}{2} \varepsilon _{0} \mathrm{E}^{2}$

(Volume) energy density of the magnetic field $=\frac{1}{2} \frac{B^{2}}{\mu _{0}}$

The energy, of the em wave, is shared equally between the electric + magnetic fields. Over one complete cycle, we need to use $\mathrm{E} _{\mathrm{rms}}\left(=\frac{\mathrm{E} _{0}}{\sqrt{2}}\right)$ and $\mathrm{B} _{\text {rms }}\left(=\frac{\mathrm{B} _{0}}{\sqrt{2}}\right)$ in the above expressions.

Also $\mathrm{E} _{0}=\mathrm{cB} _{0}$ and $\mathrm{c}=\frac{1}{\sqrt{\mu _{0} \varepsilon _{0}}}$

$\therefore$ Total energy, transferred to the disc, over one complete cycle

$$ \begin{array}{r} =\left[\frac{1}{2} \varepsilon _{0} \mathrm{E} _{\mathrm{rms}}^{2}+\frac{1}{2} \frac{\mathrm{B} _{\mathrm{rms}}^{2}}{\mu _{0}}\right]\left(\pi \mathrm{r}^{2} \mathrm{~d}\right) \\ \\ =\left[\frac{1}{2} \varepsilon _{0} \frac{\mathrm{c}^{2} \mathrm{~B} _{0}^{2}}{2}+\frac{1}{2} \mu _{0} \frac{\mathrm{B} _{0}^{2}}{2}\right] \pi \mathrm{r}^{2} \mathrm{~d} \\ \\ \quad=\frac{\mathrm{B} _{0}^{2}}{4}\left[\varepsilon _{0} \mathrm{c}^{2}+\frac{1}{\mu _{0}}\right]\left(\pi \mathrm{r}^{2} \mathrm{~d}\right) \end{array} $$

Difficult

Pressure due to EM Wavess

30. The oscillatory electric field, of an em wave propagating along the $(+x)$ direction, is represented by $\mathbf{E} _{\mathrm{y}}=\mathrm{E} _{0} \sin (k x-\omega t)$

This em wave is incident on a perfectly black surface (of area $A$ and thickness $d$ ), located in the $y-z$ plane. The average, rediation pressure, due to this em wave, on this surface, over one complete cycle, would equal

(1) $\frac{1}{2} \mathrm{E} _{0}^{2}\left[\varepsilon _{0}+\frac{1}{\mu _{0} \mathrm{c}^{2}}\right]\left(\frac{\mathrm{A} \omega \mathrm{d}}{2 \pi}\right)$

(2) $\left[\frac{1}{4 \mathrm{c}} \mathrm{E} _{0}^{2}\left(\varepsilon _{0}+\frac{\mathrm{c}^{2}}{\mu _{0}}\right)\right]\left(\frac{\omega \mathrm{d}}{2 \pi}\right)$

(3) $\left(\frac{\mathrm{E} _{0}^{2} \omega \mathrm{d}}{4 \pi \varepsilon _{0} \mathrm{c}^{3}}\right)$

(4) $\left[\frac{\mathrm{E} _{0}^{2} \omega \mathrm{d}}{4 \pi \mu _{0} \mathrm{c}^{3}}\right]$

Show Answer

Correct answer: (4)

Solution:

We know that

(Volume) energy density of electric field $=\frac{1}{2} \varepsilon _{0} \mathrm{E}^{2}$

(Volume) energy density of magnetic field $=\frac{1}{2}\left(\frac{\mathrm{B}^{2}}{\mu _{0}}\right)$

The energy of the em wave is shared equally between its electric and magnetic fields.

Over, one complete cycle, we need to use $\mathrm{E} _{\mathrm{rms}}=\frac{\mathrm{E} _{0}}{\sqrt{2}}$ and $\mathrm{B} _{\mathrm{rms}}=\frac{\mathrm{B} _{0}}{2}$ is the above expressions.

Also $\mathrm{B} _{0}=\frac{\mathrm{E} _{0}}{\mathrm{c}}$

$\therefore$ Total energy, transferred to the surface, over one complete cycle

$$ \begin{equation*} =\mathrm{U}=\left[\frac{1}{2} \varepsilon _{0} \mathrm{E} _{\mathrm{rms}}^{2}+\frac{1}{2} \frac{\mathrm{B} _{\mathrm{rms}}^{2}}{\mu _{0}}\right][\mathrm{Ad}]=\frac{1}{4} \mathrm{E} _{0}^{2}\left[\varepsilon _{0}+\frac{1}{\mu _{0} \mathrm{c}^{3}}\right] {(Ad)} \end{equation*} $$

$\therefore$ Momentum delivered $=\frac{\mathrm{U}}{\mathrm{c}} ;$ Time period $($ for one cycle $)=\frac{2 \pi}{\omega}$

$\therefore$ Average force $\mathrm{F}=\frac{\mathrm{U}}{\mathrm{c}} \div\left(\frac{2 \pi}{\omega}\right)=\frac{\omega \mathrm{U}}{2 \pi \mathrm{c}}$

Average radiation pressure $=\frac{\mathrm{F}}{\mathrm{A}}=\frac{\omega \mathrm{U}}{2 \pi \mathrm{cA}}$

$$ =\frac{\mathrm{E} _{0}^{2}}{4 \mathrm{c}}\left[\varepsilon _{0}+\frac{1}{\mu _{0} \mathrm{c}^{2}}\right]\left(\frac{\omega \mathrm{d}}{2 \pi}\right)=\left(\frac{\mathrm{E} _{0}^{2} \omega \mathrm{d}}{4 \pi \mu _{0} \mathrm{c}^{3}}\right) \hspace{2cm} \left(\because \mathrm{c}^{2}=\frac{1}{\mu _{0} \varepsilon _{0}}\right) $$

Average f Intensity of an EM Wave

31. The intensity (I) of an em wave (having $E _{0}$ and $B _{0}$ as the ‘peak values’ for its electric and

magnetic fields), can be expressed in the form:(1) $\frac{\varepsilon _{0} \mathrm{E} _{0}^{2}}{2 \mathrm{c}}$ or $\frac{\mu _{0} \mathrm{cB} _{0}^{2}}{2}$

(2) $\frac{\mathrm{c} \varepsilon _{0} \mathrm{E} _{0}^{2}}{2}$ or $\frac{\mu _{0} \mathrm{cB} _{0}^{2}}{2}$

(3) $\frac{\mathrm{c}}{2} \varepsilon _{0} \mathrm{E} _{0}^{2}$ or $\frac{\mathrm{cB} _{0}^{2}}{2 \mu _{0}}$

(4) $\frac{\varepsilon _{0} \mathrm{E} _{0}^{2}}{2 \mathrm{c}}$ or $\frac{\mathrm{cB} _{0}^{2}}{2 \mu _{0}}$

Show Answer

Correct answer: (3)

Solution:

The intensity (at a given point), of an em wave, equals the energy received there per unit area per unit time.

Consider an em wave, propagating along the $x$-axis; consider a cylinder, of unit area of cross section, having a length $\mathrm{c}(\mathrm{c}=$ velocity of em waves). The axis of this cylinder is assumed to be parallel to the direction of propagation of the em wave.

The volume of this cyilnder is $(1 \times \mathrm{c})=\mathrm{c}$

The em wave (covering a distance $\mathrm{c}$ in one second) would transfer a total energy.

$ \mathrm{U}=\left[\frac{1}{2} \varepsilon _{0} \mathrm{E} _{\mathrm{rms}}^{2}+\frac{1}{2} \frac{\mathrm{B} _{\mathrm{rms}}^{2}}{\mu _{0}}\right][\mathrm{c}] $

Through the cross section (= unity) of this cylinder in one second.

Hence energy transferred per unit area per unit time.

$ \begin{aligned} =\text { intensity of wave }=\frac{U}{1 \times 1}=U \\ \therefore \quad I & =\left[\frac{1}{2} \varepsilon _{0} \mathrm{E} _{\mathrm{rms}}^{2}+\frac{1}{2} \frac{\mathrm{B} _{\mathrm{rms}}^{2}}{\mu _{0}}\right] \mathrm{c} \end{aligned} $

$ \begin{aligned} & =\frac{\mathrm{c}}{2}\left[\varepsilon _{0} \frac{\mathrm{E} _{0}^{2}}{2}+\frac{\mathrm{B} _{0}^{2}}{2 \mu _{0}}\right] \\ \\ & =\frac{\mathrm{c}}{4 \mu _{0}}\left[\mu _{0} \varepsilon _{0} \mathrm{E} _{0}^{2}+\frac{\mathrm{E} _{0}^{2}}{\mathrm{c}^{2}}\right] \\ \\ & =\frac{\mathrm{c}}{4 \mu _{0} \mathrm{c}^{2}} 2 \mathrm{E} _{0}^{2} \hspace{2cm} \left(\because \mu _{0} \varepsilon _{0}=\frac{1}{\mathrm{c}^{2}}\right) \\ \\ & =\frac{\mathrm{E} _{0}^{2}}{2 \mu _{0} \mathrm{c}}=\frac{1}{2}\left(\varepsilon _{0} \mathrm{E} _{0}^{2}\right) \mathrm{c}=\frac{1}{2} \varepsilon _{0} \mathrm{c}^{3} \mathrm{~B} _{0}^{2} \\ \\ & =\frac{1}{2} \frac{\mathrm{cB} _{0}^{2}}{\mu _{0}} \end{aligned} $

Average

Peak Value of Electric Field in an EM Wave

32. The peak value of the electric field, in the light radiated by a given (point) source of power $P$, at a point distance $r$, from the source equals.

(1) $\left(\frac{\varepsilon _{0} \mathrm{P}}{4 \pi \mathrm{c}}\right)^{1 / 2} \frac{1}{\mathrm{r}}$

(2) $\left(\frac{\mathrm{P}}{2 \pi \varepsilon _{0} \mathrm{c}}\right)^{1 / 2}\left(\frac{1}{\mathrm{r}}\right)$

(3) $\left(\frac{2 \pi \mathrm{P}}{\varepsilon _{0} \mathrm{c}}\right)^{1 / 2}\left(\frac{1}{\mathrm{r}}\right)$

(4) $\left(2 \pi \mathrm{c} \varepsilon _{0} \mathrm{P}\right)^{1 / 2}\left(\frac{1}{\mathrm{r}}\right)$

Show Answer

Correct answer: (2)

Solution:

The intensity, I, of the light radiated by a (point) source of power $\mathrm{P}$, at a point distance $\mathrm{r}$ from the source, is

$$ \mathrm{I}=\frac{\mathrm{P}}{4 \pi \mathrm{r}^{2}} $$

The intensity also equals $\mathrm{c}$ times the total energy density of the relevant em wave. This implies that

$$ \begin{aligned} \mathrm{I} & =\mathrm{c}\left[\frac{1}{2} \varepsilon _{0} \mathrm{E} _{\mathrm{rms}}^{2}+\frac{1}{2} \frac{\mathrm{B} _{\mathrm{rms}}^{2}}{\mu _{0}}\right] \\ \\ \frac{\mathrm{P}}{4 \pi \mathrm{r}^{2}} & =\mathrm{c}\left[2 \times\left(\frac{1}{2} \varepsilon _{0} \mathrm{E} _{\mathrm{rms}}^{2}\right)\right] \\ \\ & =\mathrm{c}\left[\varepsilon _{0} \mathrm{E} _{\mathrm{rms}}^{2}\right] \\ \\ & =\mathrm{c}\left[\varepsilon _{0} \frac{\mathrm{E} _{0}^{2}}{2}\right] \end{aligned} $$

This gives $\mathrm{E} _{0}^{2}=\frac{\mathrm{P}}{2 \pi \varepsilon _{0} \mathrm{cr}^{2}}$

$\therefore \quad \mathrm{E} _{0}=\left(\frac{\mathrm{P}}{2 \pi \varepsilon _{0} \mathrm{c}}\right)^{1 / 2}\left(\frac{1}{\mathrm{r}}\right)$

Average

Peak Value of the Magnetic Field in an EM Wave

33. Agiven (point) source, of power $P$, radiates light, which has a peak value, $B _{0}$, for its magnetic field, at a point distant $r$ from the source. The graph, showing the dependence of $B _{0}$ on $r$, would, then, be the graph labelled as graph

$~$

$~$

(1) $\mathrm{K}$

(2) $\mathrm{L}$

(3) $\mathrm{M}$

(4) $\mathrm{N}$

Show Answer

Correct answer: (2)

Solution:

The intensity, I, of the light from a (point) source of power $\mathrm{P}$, at a point distant $\mathrm{r}$ from it, is

$$ \mathrm{I}=\frac{\mathrm{P}}{4 \pi \mathrm{r}^{2}} $$

The intensity I also equals $\mathrm{c}$ times the total energy density of the relevant em wave. This implies that

$$ \begin{aligned} & \mathrm{I}=\mathrm{c}[\left.\frac{1}{2} \varepsilon _{0} \mathrm{E} _{\mathrm{rms}}^{2}+\frac{1}{2} \frac{\mathrm{B} _{\mathrm{rms}}^{2}}{\mu _{0}}\right] \\ \\ &= \mathrm{c}\left[2 \times \frac{1}{2} \frac{\mathrm{B} _{\mathrm{rms}}^{2}}{\mu _{0}}\right] \hspace{2cm}\left(\because \frac{1}{2} \frac{\mathrm{B} _{\mathrm{rms}}^{2}}{\mu _{0}}=\frac{1}{2} \varepsilon _{0} \mathrm{E} _{\mathrm{rms}}^{2}\right) \\ \\ &=\frac{\mathrm{c}}{\mu _{0}} \mathrm{~B} _{\mathrm{rms}}^{2}=\frac{\mathrm{c}}{2 \mu _{0}} \mathrm{~B} _{0}^{2} \hspace{2cm}\left(\because \mathrm{B} _{\mathrm{rms}}=\frac{\mathrm{B} _{0}}{\sqrt{2}}\right) \\ \\ & \therefore \quad \mathrm{B} _{0}^{2}=\frac{2 \mu _{0}}{\mathrm{c}} \mathrm{I}=\frac{2 \mu _{0}}{\mathrm{c}} \frac{\mathrm{P}}{4 \pi \mathrm{r}^{2}}=\left(\frac{\mu _{0} \mathrm{P}}{2 \pi \mathrm{c}}\right) \frac{1}{\mathrm{r}^{2}} \\ \\ & \therefore \quad \mathrm{B} _{0}=\left(\frac{\mu _{0} \mathrm{P}}{2 \pi \mathrm{c}}\right)^{1 / 2} \frac{1}{\mathrm{r}}=\text { const. } \frac{1}{2} \end{aligned} $$

Thus $\mathrm{B} _{0} \propto \frac{1}{\mathrm{r}}$

This dependance of $\mathrm{B} _{0}$ on $\mathrm{r}$, is shown in the graph labelled as graph $\mathrm{L}$.

Average

Comparison of Peak Value, of Electric Fields of Two EM Waves

34. The peak value of the electric field, in the light radiated by a point source of power $P _{1}$, at a point, distant $r _{1}$ from the source, is $E _{01}$. The corresponding value, for a point source of power $P _{2}$, at a point, distant $r _{2}$ from the source, is $E _{02}$. The ratio $\left(\frac{E _{01}}{E _{02}}\right)$ equals

(1) $\left(\frac{\mathrm{P} _{1} \mathrm{r} _{1}^{2}}{\mathrm{P} _{2} \mathrm{r} _{2}^{2}}\right)^{1 / 2}$

(2) $\left(\frac{\mathrm{P} _{2} \mathrm{r} _{2}^{2}}{\mathrm{P} _{1} \mathrm{r} _{1}^{2}}\right)^{1 / 2}$

(3) $\left(\frac{\mathrm{P} _{1} \mathrm{r} _{2}^{2}}{\mathrm{P} _{2} \mathrm{r} _{1}^{2}}\right)^{1 / 2}$

(4) $\left(\frac{\mathrm{P} _{1} \mathrm{r} _{1}}{\mathrm{P} _{2} \mathrm{r} _{2}}\right)^{\frac{1}{2}}$

Show Answer

Correct answer: (3)

Solution:

The peak value, of the electric field ( $\mathrm{E} _{0}$ ) of an em wave, (at a point distant $r$ from the source) is related to the power $(\mathrm{P})$ of the source as

$$ \begin{aligned} & \frac{\mathrm{P}}{4 \pi \mathrm{r}^{2}}=\mathrm{c}\left[2 \times \frac{1}{2} \varepsilon _{0}\left(\frac{\mathrm{E} _{0}}{\sqrt{2}}\right)^{2}\right] \quad(=\text { Intensity) } \\ & \therefore \quad \mathrm{E} _{0}^{2}=\frac{\mathrm{P}}{2 \pi \mathrm{r}^{2} \mathrm{c} \varepsilon _{0}}=\left(\frac{1}{2 \pi \mathrm{c} \varepsilon _{0}}\right) \frac{\mathrm{P}}{\mathrm{r}^{2}} \end{aligned} $$

$\therefore$ The required ratio is

$\left[\frac{E _{01}}{E _{02}}\right]^{2}=\frac{P _{1}}{P _{2}} \cdot\left(\frac{r _{2}}{r _{1}}\right)^{2}$

$\therefore \frac{\mathrm{E} _{01}}{\mathrm{E} _{02}}=\left(\frac{\mathrm{P} _{1} \mathrm{r} _{2}^{2}}{\mathrm{P} _{2} \mathrm{r} _{1}^{2}}\right)^{1 / 2}$

Average

Comparison of Peak Value, of Electric Fields of Two EM Waves

35. The ratio, of the rms values of the magnetic fields in the em waves, radiated by two point sources, at points distance $r$ and $R$, respectively, from the two sources, is ’ $s$ ‘. If the power of one of the sources is $P$, that of the other, would be

(1) $\left(\frac{\mathrm{PR}^{2}}{\mathrm{r}^{2} \mathrm{~s}}\right)$

(2) $\left(\frac{\operatorname{Pr}^{2}}{\mathrm{R}^{2} \mathrm{~s}}\right)$

(3) $\left(\frac{\mathrm{sR}^{2}}{\mathrm{Pr}^{2}}\right)$

(4) $\left(\frac{\mathrm{r}^{2} \mathrm{~s}}{\mathrm{PR}^{2}}\right)$

Show Answer

Correct answer: (1)

Solution:

The rms value of the magnetic field ( $\mathrm{B} _{\mathrm{rms}}$ ), of an em wave (at a point distant $\mathrm{r}$ from the source), is related to the power, $\mathrm{P}$, of the source, as

$$ \begin{aligned} & \frac{\mathrm{P}}{4 \pi \mathrm{r}^{2}}=\mathrm{c}\left[2 \times \frac{1}{2} \frac{\mathrm{B} _{\mathrm{rms}}^{2}}{\mu _{0}}\right] \quad(=\text { intensity }) \\ \\ \therefore ~ ~ & \mathrm{B} _{\mathrm{rms}}^{2}=\frac{\mathrm{P}}{4 \pi \mathrm{r}^{2}} \frac{\mu _{0}}{\mathrm{c}} \end{aligned} $$

For the given data, we then have

$$ \mathrm{s}=\left(\frac{\mathrm{P}}{4 \pi \mathrm{r}^{2}} \frac{\mu _{0}}{\mathrm{c}}\right) \times\left(\frac{\mathrm{c}}{\mu _{0}} \frac{4 \pi \mathrm{r}^{2}}{\mathrm{P}^{\prime}}\right) $$

where $\mathrm{P}^{\prime}$ is the power of the other source.

$\therefore \mathrm{s}=\left(\frac{\mathrm{P}}{\mathrm{P}} \frac{\mathrm{R}^{2}}{\mathrm{r}^{2}}\right)$

$\therefore \quad \mathrm{P}^{\prime}=\left(\frac{\mathrm{P}}{\mathrm{r}^{2}} \frac{\mathrm{R}^{2}}{\mathrm{~s}}\right)$

Easy

Momentum Transfer due to EM Waves

36. An em wave, originating from a source, is radiating out $10 \mathrm{~J}$ of energy per second. At a distance of $50 \mathrm{~cm}$ from the source, a perfectly reflecting surface, of area $8 \pi \mathrm{cm}^{2}$, has these em waves incident on it. The average force, exerted on the surface, during a 30 seconds span would be

(1) zero

(2) $\left(\frac{1}{3750 \mathrm{c}}\right) \mathrm{N}$

(3) $\left(\frac{1}{125 \mathrm{c}}\right) \mathrm{N}$

(4) $\left(\frac{1}{2500 \mathrm{c}}\right) \mathrm{N}$

Show Answer

Correct answer: (1)

Solution:

The surface, being a perfectly reflecting surface, does not absorbe the incident em at all. Hence the energy absorbed, and, therefore, the momentum delivered to the surface, are zero each. The average force (= rate of change of momentum) is, therefore, zero.

Average

Relating between rms Value of Electric Field in an EM Wave + its Intensity

37. The rms value, of the electric field, in the em waves, produced a point source, at a distance of $\left(\frac{4}{\sqrt{\pi}}\right)$ in from the point source, is $2 \mathrm{Vm}^{-1}$. Taking $\varepsilon _{0}=8.85 \times 10^{-12} \mathrm{SI}$ units and $\mathrm{C}=3 \times$ $10^{8} \mathrm{~ms}^{-1}$, we would find the power of the source to be (nearly)

(1) $2.3 \mathrm{nW}$

(2) $1.15 \mathrm{nW}$

(3) $2.3 \mathrm{pW}$

(4) $1.15 \mathrm{pW}$

Show Answer

Correct answer: (1)

Solution:

Let $\mathrm{P}$ be the power of the source. The intensity, at a distance $\left(\frac{4}{\sqrt{\pi}}\right) \mathrm{m}$ is then

$$ \mathrm{I}=\frac{\mathrm{P}}{4 \pi\left(\frac{16}{\pi}\right)} \frac{\mathrm{W}}{\mathrm{m}^{2}}=\frac{\mathrm{P}}{64} \frac{\mathrm{W}}{\mathrm{m}^{2}} $$

$$ \text{This equals} \left[2 \times \frac{1}{2} \varepsilon _{0} \mathrm{E} _{\mathrm{rms}}^{2}\right]$$

$$ \begin{aligned} \therefore \quad \mathrm{P} & =\left[64 \times 8.85 \times 10^{-12} \times 4\right] \mathrm{W} \\ & =2265.6 \times 10^{-12} \mathrm{~W} \\ & \approx 2.3 \mathrm{nW} \end{aligned} $$

Average

Relation between the Peak Value of the Magnetic Field of an EM Wave and its Intensity

38. The peak value of the magnetic field, at a certain point, for an em wave is $9 \mathrm{nT}$. (It is give that $\mu _{0}=4 \pi \times 10^{-7}$ SI units and $c=3 \times 10^{8} \mathrm{~ms}^{-1}$ ). The peak value, of its electric field, and its intensity, at the same point, would be (nearly)

(1) $2.7 \frac{\mathrm{V}}{\mathrm{m}}$ and $4.8 \times 10^{-3} \frac{\mathrm{W}}{\mathrm{m}^{2}}$

(2) $5.4 \frac{\mathrm{V}}{\mathrm{m}}$ and $9.7 \times 10^{-3} \frac{\mathrm{W}}{\mathrm{m}^{2}}$

(3) $5.4 \frac{\mathrm{V}}{\mathrm{m}}$ and $4.8 \times 10^{-3} \frac{\mathrm{W}}{\mathrm{m}^{2}}$

(4) $2.7 \frac{\mathrm{V}}{\mathrm{m}}$ and $9.7 \times 10^{-3} \frac{\mathrm{W}}{\mathrm{m}^{2}}$

Show Answer

Correct answer: (4)

Solution:

$\mathrm{E} _{0}=$ Peak value of electric field $=\mathrm{cB} _{0}$

$$ =3 \times 10^{8} \times 9 \times 10^{-9} \frac{\mathrm{V}}{\mathrm{m}}=2.7 \frac{\mathrm{V}}{\mathrm{m}} $$

$$ \text{Also intensity} =\mathrm{c}\left[2 \times \frac{1}{2 \mu _{0}} \mathrm{~B} _{\mathrm{rms}}^{2}\right]$$

$$ \begin{aligned} & =\mathrm{c}\left[\frac{1}{2 \mu _{0}} \mathrm{~B} _{0}^{2}\right] \quad\left[\because \mathrm{B} _{\mathrm{rms}}=\frac{\mathrm{B} _{0}}{\sqrt{2}}\right] \\ & =\frac{3 \times 10^{8}}{8 \pi \times 10^{-7}} \times 81 \times 10^{-18} \frac{\mathrm{W}}{\mathrm{m}^{2}} \\ & =\frac{243}{8 \pi} \times 10^{-3} \frac{\mathrm{W}}{\mathrm{m}^{2}} \\ & \simeq 9.7 \times 10^{-3} \frac{\mathrm{W}}{\mathrm{m}^{2}} \end{aligned} $$

Average

Intensity of EM Wave + Force Exerted on a Surface

39. EM Waves having a uniform and constant, intensity I, are incident normally on a (perfectly black) surface of area $A$, for a time $t$. The average force (F), exerted on the surface by these em waves, depends on the time $(t)$ in the manner shown in graph (W)

$~$

(1) $\mathrm{W}$

(2) $\mathrm{X}$

(3) $\mathrm{Y}$

(4) Z

Show Answer

Correct answer: (1)

Solution:

When EM Waves, of intensity I, are incident normally, on a surface of area A for a time t, the total energy absorbed by the (black) surface is

$$ \mathrm{U}=\mathrm{I} \times \mathrm{A} \times \mathrm{t} $$

Therefore, the total momentum, delivered to the surface, is

$$ \mathrm{F}=\frac{\mathrm{p}}{\mathrm{t}}=\frac{\mathrm{I} \times \mathrm{A} \times \mathrm{t}}{\mathrm{c} \times \mathrm{t}}=\frac{\mathrm{IA}}{\mathrm{c}} $$

The average force thus does not depend on the time, $t$, for which the em waves are incident on. Hence the graph, between $\mathrm{F}$ and $\mathrm{t}$, has the form shown in graph $\mathrm{W}$.

Aveage

Radiation Pressure due to an EM Wave

40. EM waves, of a uniform intensity $I\left(\frac{W}{m^{2}}\right)$, are incident normally on a perfect black surface, of area $10 \mathrm{~m}^{2}$, for a time span of $100 \mathrm{~s}$. The graph between the ‘average radiation pressure’ $p _{R}$ (exerted on the surface by these $e m$ waves) and the intensity (I), of the incident radiation, would then be the graph, labelled as graph (K)

$~$

(1) $\mathrm{K}$; and its slope would be zero

(2) L; and its slope would be (nearly) $3.33 \times 10^{-9} \mathrm{SI}$ units

(3) $\mathrm{M}$; and its slope would be (nearly) $-5.55 \times 10^{-8} \mathrm{SI}$ units

(4) $\mathrm{N}$; and its slope would be (nearly) $+\left(6.66 \times 10^{-6}\right) \mathrm{SI}$ units

Show Answer

Correct answer: (2)

Solution:

When em waves, of intensity I, are incident normally, on a surface of area A, for a time t, the total energy, absorbed by the black surface, is

$$ \mathrm{U}=\mathrm{I} \times \mathrm{A} \times \mathrm{t}=(\mathrm{I} \times 10 \times 100) \mathrm{J}=(1000 \mathrm{I}) \mathrm{J} $$

The take momentum, delivered to the surface, is $\frac{\mathrm{U}}{\mathrm{c}}$

$$ =\frac{(1000 \mathrm{I})}{\mathrm{c}} \mathrm{kg} \mathrm{ms}^{-1} $$

$\therefore$ Average force, exerted on the surface, is $\left(\frac{1000 \mathrm{I}}{\mathrm{c} \times 100}\right) \mathrm{N}$

$=\left(\frac{1 \mathrm{I}}{\mathrm{c}}\right) \mathrm{N}$

$\therefore$ Average ‘radition pressure’ $=\left[\left(\frac{10 \mathrm{I}}{\mathrm{c}}\right) \times\left(\frac{1}{10}\right)\right] \frac{\mathrm{N}}{\mathrm{m}^{2}}$

$\therefore \quad \mathrm{p} _{\mathrm{R}}=\frac{\mathrm{I}}{\mathrm{c}}$

The graph between $\mathrm{p} _{\mathrm{R}}$ and $\mathrm{I}$ is, therefore, a straight line with a slope $=\frac{1}{\mathrm{c}}$

$\therefore$ Slope $=\frac{1}{3 \times 10^{8}}$ SI units

$$ \simeq 3.33 \times 10^{-9} \text { SI units } $$

Difficult

Radiation Pressure

41. Electromagnetic waves of intensity I, are incident normally on a material surface. If the reflected, transmitted and absorbed parts of these waves (by the material) are in the same ratio, then the radiation pressure $p$ experienced by the surface equals

(1) $\frac{\mathrm{I}}{2 \mathrm{c}}$

(2) $\frac{\mathrm{I}}{\mathrm{c}}$

(3) $\frac{\mathrm{I}}{3 \mathrm{c}}$

(4) $\frac{2 \mathrm{I}}{3 \mathrm{c}}$

Show Answer

Correct answer: (2)

Solution:

Let $U$ be the total energy of the EM waves received by the surface of one A in time $t$.

Therefore, intensity, I = Energy recieved per unit normal area in one second.

$$ \text { or } \mathrm{I}=\frac{\mathrm{U}}{\mathrm{At}} $$

Now, momentum imparted for computer absorption (i.e. if the surface is a perfectly block-body surface), is : $\mathrm{P}=\frac{\mathrm{U}}{\mathrm{c}}$

Pressure (p) exerted on the surface will then be:

$$ \mathrm{p}=\frac{\mathrm{F}}{\mathrm{A}}=\frac{\text { Change in momentum } / \text { time }}{\mathrm{A}} $$

or $\mathrm{p}=\frac{\mathrm{p} / \mathrm{t}}{\mathrm{A}}=\frac{\mathrm{U}}{\mathrm{cAt}}=\frac{\mathrm{I}}{\mathrm{c}}$

or $\mathrm{p}=\frac{\mathrm{I}}{\mathrm{c}}$ (for complete absorption)

Now, for complete / total reflection, where the EM waves get bounced back with speed c, the change in momentum is, zero

For complete transmission, $\mathrm{p}=0$

Now, here, $\frac{1}{3}$ gets reflected, $\frac{1}{3}$ gets transmitted and $\frac{1}{3}$ gets absorbed.

$\therefore$ The pressure exerted by each part, respectively will be 0,0 and $\frac{1 / 3}{c}$.

Hence, the total pressure on the surface, here, will be;

Pressure $=0+0+\frac{\mathrm{I}}{3 \mathrm{c}}$

$ =\frac{\mathrm{I}}{3 \mathrm{c}} $

Correct answer is : $\mathrm{p}=\frac{\mathrm{I}}{3 \mathrm{c}}$

Difficult

Poynting Vector

42. An infinitely long thin wire has a uniform charge density of $+\lambda \mathbf{C m}^{-1}$. The wire is placed along the $z$-axis and moved along its length with a uniform velocity $v=v \hat{k}$. The magnitude and direction of poynting vector, at $p(0, a, 0)$ in this situation, equal

(1) $\mathrm{S}=\frac{\lambda \mathrm{v}^{2}}{4 \pi^{2} \varepsilon _{0} \mathrm{a}^{2}}$, along $(+\hat{\mathrm{k}})$

(2) $\mathrm{S}=\frac{\lambda^{2} \mathrm{v}^{2}}{4 \pi^{2} \varepsilon _{0} \mathrm{a}^{2}}$, along $(-\hat{\mathrm{k}})$

(3) $\mathrm{S}=\frac{\lambda^{2} \mathrm{v}}{4 \pi^{2} \varepsilon _{0} \mathrm{a}^{2}}$, along $(+\hat{\mathrm{k}})$

(4) $\mathrm{S}=\frac{\lambda^{2} \mathrm{v}}{4 \pi^{2} \varepsilon _{0} \mathrm{a}^{2}}$, along $(-\hat{\mathrm{k}})$

Show Answer

Correct answer: (3)

Solution:

The figure given aside represents the situation. Due to the uniform $\mathbf{v}$ along $\hat{k}$, there will be a current $\mathrm{I}=\lambda \mathrm{V}$, along $\hat{\mathrm{k}}$ Now, consider a point $\mathrm{P}$, on the $\mathrm{y}$-axis, at a distance of ’ $\mathrm{a}$ ’ from the wire.

The electric field at $\mathrm{P}$, due to the charge on the wire is $: \mathbf{E}=\frac{\lambda}{2 \pi \varepsilon _{0} \mathrm{a}} \hat{\mathrm{j}}$

The magnetic field at $P$, due to the current $I=\lambda v$ along $\hat{k}$, is $\mathbf{B}=\frac{\mu _{0} \mathrm{I}}{2 \pi \mathrm{a}}(-\hat{\mathrm{i}})=\frac{\mu _{0} \lambda \mathrm{v}}{2 \pi \mathrm{a}}(-\hat{\mathrm{i}})$

$\therefore \mathbf{S}=\frac{1}{\mu _{0}}(\mathbf{E} \times \mathbf{B})$

$~$

$\text{or} \quad \mathbf{S}=\frac{1}{\mu _{0}}\left[\frac{\lambda}{2 \pi \varepsilon _{0} \mathrm{a}} \times \frac{\mu _{0} \lambda \mathrm{v}}{2 \pi \mathrm{a}}\right](\hat{\mathrm{j}} \times-\hat{\mathrm{i}})$

$ \mathbf{S}=\frac{\lambda^{2} \mathrm{v}}{4 \pi^{2} \varepsilon _{0} \mathrm{a}^{2}}(\hat{\mathrm{k}}) $

or $\mathrm{S}=\frac{\lambda^{2} \mathrm{v}}{4 \pi^{2} \varepsilon _{0} \mathrm{a}^{2}}$,along $(+\hat{\mathrm{k}})$

Correct answer: (3)

Average

Intensity of EM Wave / Poynting Vector

43. The SI unit of the poynting vector (S) defined by the formula: $S=\frac{1}{\mu _{0}}(E \times B)$ is

(1) volt-tesla

(2) volt - metre - ampere $^{-1}$

(3) joule - metre $^{-2}$

(4) watt - metre $^{-2}$

Show Answer

Correct answer: (4)

Solution:

The magnitude of $\mathbf{S}$ is same as intensity (I) of EM wave.

$\therefore|\mathbf{S}|=\mathrm{I}=$ energy transmitted through unit normal area in one second.

$$=\frac{\mathrm{U} / \mathrm{Time}} {\text { Area }} \qquad \quad (\mathrm{U}=\text { Energy })$$

$\therefore$ Unit of $|\mathbf{S}|=$ unit of $\mathrm{I}=\frac{\mathrm{y} / \mathrm{s}}{\mathrm{m}^{2}}=\mathrm{Wm}^{-2}$

$ =\text { watt }- \text { metre }^{-2} $

Answer (4) is correct.

Difficult

Speed EM Waves

44. An electromagnetic wave of frequency $3 \mathrm{MHz}$ passes from vaccum in to dielectric medium of constant $K=4$ and permeability $\mu _{0}$. The wavelength of this wave in the dielectric is:

(1) $200 \mathrm{~m}$

(2) $100 \mathrm{~m}$ (Inorrect)

(3) $25 \mathrm{~m}$

(4) $25 \mathrm{~m}$

Show Answer

Correct answer: (3)

Solution:

Given, $\mathrm{v}=3 \mathrm{M} \mathrm{Hz}=3 \times 10^{6} \mathrm{~Hz}$

$ \mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}(\text { is vaccum }) $

$\therefore \lambda=\frac{\mathrm{c}}{\mathrm{v}}=\frac{3 \times 10^{8}}{3 \times 10^{6}}=100 \mathrm{~m}$

Now, $\mathrm{K}=\varepsilon _{\mathrm{r}}=4$

Speed of EM wave in the medium will be $v=\frac{\mathrm{c}}{\mathrm{n}}$

where $n$ is the refractive index of the medium.

Here $\left.\mathrm{n}=\frac{\mathrm{c}}{v}=\frac{1}{\sqrt{\mu _{0} \varepsilon _{0}}} \right\rvert, \frac{1}{\sqrt{\mu \varepsilon}}$

$ =\frac{\sqrt{\mu \varepsilon}}{\sqrt{\mu _{0} \varepsilon _{0}}}$

$ =\frac{\sqrt{\mu _{0} \mathrm{~K} \varepsilon _{0}}}{\sqrt{\mu _{0} \varepsilon _{0}}}\left(\because \mu=\mu _{0} \text { and } \in=\mathrm{K} \in _{0} \text { for the given medium }\right) $

$ =\sqrt{\mathrm{K}}$

$ =\sqrt{\mathrm{4}}$

$=2$

i.e. refractive index, $\mathrm{n}=2$

$\therefore$ Speed of EM waves in the dielectric is:

$ \mathrm{v}=\frac{\mathrm{c}}{2}=1.5 \times 10^{8} \mathrm{~ms}^{-1} $

But $\mathrm{v}=v \lambda^{\prime} \quad\left(\lambda=3 \times 10^{6} \mathrm{~Hz}\right)$

$\therefore \quad \lambda^{\prime}=\frac{\mathrm{v}}{\mathrm{v}}=\frac{1.5 \times 10^{8}}{3 \times 10^{6}} \mathrm{~m}$

$=50 \mathrm{~m}$

Average

EM Spectrum

45. A monochromatic source of power $0.66 \mathrm{~kW}$, and efficiency $100 \%$, emits $10^{20}$ photons is one second. These photons belong to the region of

(1) $x$ - rays

(2) uv-radiation

(3) visible light

(4) infrarred waves

Show Answer

Correct answer: (2)

Solution:

Given $\mathrm{P}=0.66 \mathrm{~kW}=0.66 \times 10^{3} \mathrm{~W}$

Since efficiency is $100 \%$ and source is monochromatic, all the photons have same energy $(E=h v)$ and the entire power is converted to photons energy. Number of photon emitted in 1 second $=10^{20}$

$\therefore$ Energy of one photon is $\mathrm{E}=\frac{\text { Total energy emitted is one second }}{\text { Number of photons emitted is one second }}$

$ =\frac{\mathrm{P}}{\mathrm{n}} $

$ =\frac{0.66 \times 10^{3}}{10^{20}}=0.66 \times 10^{-17} \mathrm{~J} $

Now, $\mathrm{E}=\mathrm{h} v=\frac{\mathrm{hc}}{\lambda}$

$ \begin{aligned} \therefore \quad \lambda & =\frac{\mathrm{hc}}{\mathrm{E}}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{0.66 \times 10^{-17}} \mathrm{~m} \\ & =30 \times 10^{-9} \mathrm{~m} \\ & =30 \mathrm{~nm} \quad \text { (or uv rays) } \end{aligned} $

Average

Intensity of EM Waves

46. A point source of electromagnetic radiation has an average power output of $135 \mathrm{~W}$. The maximum value of electric field, at a distance of $\mathbf{3}$ from the source, (in $\mathbf{V m}^{-1}$ ) is:

(1) $15 \sqrt{2}$

(2) 30

(3) $30 \sqrt{2}$

(4) 300

Show Answer

Correct answer: (2)

Solution:

Intensity at a distance of $\mathrm{r}=3 \mathrm{~m}$ from the source is:

$ \mathrm{I}=\frac{\mathrm{P}}{4 \pi \mathrm{r}^{2}}=\frac{135}{4 \pi \times 9} \frac{\mathrm{W}}{\mathrm{m}^{2}} $

Now, $\mathrm{I}=\frac{1}{2} \varepsilon _{0} \mathrm{cE} _{0}^{2}=\frac{1}{2} \cdot \frac{1}{4 \pi 9 \times 10^{9}} \times 3 \times 10^{8} \times \mathrm{E} _{0}^{2}$

$\therefore \frac{1}{2} \varepsilon _{0} \mathrm{cE} _{0}^{2}=\frac{\mathrm{P}}{4 \pi \mathrm{r}^{2}}$

or $\quad \frac{1}{2} \frac{1}{4 \pi \times 9 \times 10^{9}} \times 3 \times 10^{8} \times \mathrm{E} _{0}^{2}=\frac{135}{4 \pi \times 9}$

or $\mathrm{E} _{0}^{2}=900$

$\mathrm{E} _{0}=30 \mathrm{Vm}^{-1}$

Easy

EM Spectrum

47. The range of wavelengths allotted for commercial $F M$ radio is

(1) $0.313 \mathrm{~m}$ to $0.638 \mathrm{~m}$

(2) $0.1 \mathrm{~m}$ to $1.0 \mathrm{~m}$

(3) $2.78 \mathrm{~m}$ to $3.41 \mathrm{~m}$

(4) $277.8 \mathrm{~m}$ to $340.9 \mathrm{~m}$

Show Answer

Correct answer: (3)

Solution:

The range of frequencies alloted for FM radio is: $88 \mathrm{MHz}$ to $108 \mathrm{MHz}$.

Now this frequency range corresponds to a wavelength range of

$\frac{3 \times 10^{8}}{88 \times 10^{6}} \mathrm{~m}$ to $\frac{3 \times 10^{8}}{108 \times 10^{6}} \mathrm{~m}$

or $\quad 3.41 \mathrm{~m}$ to $2.78 \mathrm{~m}$

Correct answer: (3)

Average

Properties of EM Waves

48. The electric field vector $\mathbf{E} _{(r, t)}$ and the magnetic field vector $\mathbf{B} _{(r, t)}$ of a plane electromagnetic wave propagating in free-space are given by

$\mathbf{E} _{(r, t)}=\mathbf{E} _{0} \hat{\mathbf{i}} \cos (\omega t-k z)$

and $B _{(r, t)}=B _{0} \hat{j} \cos (\omega t-k z)$

where $B _{0}=\frac{E _{0}}{c}, k=\frac{2 \pi}{\lambda}$ is angular wave-number and $\omega=2 \pi v$ is angular frequency. The average value of the vector $\frac{1}{\mu _{0}}(E \times B)$, over a time period of the wave, is

(1) zero

(2) $\frac{1}{2} \varepsilon _{0} \mathrm{E} _{0}^{2} \hat{\mathrm{k}}$

(3) $\frac{1}{4} \varepsilon _{0} \mathrm{c} \mathrm{E} _{0}^{2} \hat{\mathrm{k}}$

(4) $\frac{1}{2} \varepsilon _{0} \mathrm{c} \mathrm{E} _{0}^{2} \hat{\mathrm{k}}$

Show Answer

Correct answer: (4)

Solution:

The vectors $\mathbf{E}$ and $\mathbf{B}$ are perpendicular to each other and are along $\hat{i}$ and $\hat{j}$ respectively, as the wave is progressing along the positive z-axis.

Therefore, $\mathbf{E} \times \mathbf{B}=\mathrm{EB} \hat{\mathrm{k}}$

$\mathrm{E}=\mathrm{E} _{\mathrm{rms}}$ and $\mathrm{B}=\mathrm{B} _{\mathrm{rms}}$ when averaged over a period of the wave. So this can be written as:

$$ \begin{aligned} \frac{1}{\mu _{0}}(\mathbf{E} \times \mathbf{B}) & =\frac{1}{\mu _{0}} E _{\mathrm{rms}} B _{\mathrm{rms}} \hat{\mathrm{k}} \\ & =\frac{1}{\mu _{0}} \frac{\mathrm{E} _{0}}{\sqrt{2}} \cdot \frac{\mathrm{B} _{0}}{\sqrt{2}} \cdot \hat{\mathrm{k}} \\ & =\frac{1}{\mu _{0}} \frac{1}{2} \cdot \mathrm{E} _{0} \cdot \frac{\mathrm{E} _{0}}{\mathrm{c}} \cdot \hat{\mathrm{k}} \\ & =\frac{1}{2} \frac{1}{\mu _{0}} \frac{\mathrm{E} _{0}^{2}}{\mathrm{c}} \cdot \hat{\mathrm{k}} \\ & =\frac{1}{2} \varepsilon _{0} \mathrm{c}^{2} \cdot \frac{\mathrm{E} _{0}^{2}}{\mathrm{c}} \cdot \hat{\mathrm{k}} \hspace{2cm} \left(\mathrm{c}^{2}=\frac{1}{\mu _{0} \varepsilon _{0}} \text { or } \frac{1}{\mu _{0}}=\mathrm{c}^{2} \varepsilon _{0}\right) \\ & =\frac{1}{2} \varepsilon _{0} \mathrm{cE} _{0}^{2} \hat{\mathrm{k}} \end{aligned} $$

Correct answer: (4)

Average

EM Spectrum

49. Monochromatic source, of power $4 \mathrm{~kW}$ and $100 \%$ efficiency produces $10^{20}$ photons every second. The radiations emitted by it belong to the part of the EM spectrum called:

(1) gamma rays

(2) $x$-rays

(3) microwaves

(4) infrared waves

Show Answer

Correct answer: (2)

Solution:

Power $\mathrm{P}=4000, \mathrm{~W}=4000 \mathrm{~J}$ per second.

$\mathrm{n}=10^{20}$ per second

$\therefore \mathrm{n}( \mathrm{h} v ) =\mathrm{P} \hspace {2cm} (\because$ efficiency is 100 $\%)$

or $\quad v=\frac{P}{n h}$

$ \begin{aligned} & =\frac{4000}{10^{20} \times 6.6 \times 10^{-34}} \mathrm{~Hz} \\ & =\frac{4}{6.6} \times 10^{17} \mathrm{~Hz} \\ & =0.61 \times 10^{17} \mathrm{~Hz} \end{aligned} $

This is in the range of frequencies for $\mathrm{x}$-rays.

Correct option is (2)

Difficult

Intensity EM Radiations

50. The maximum value of electric field, of the total radiant energy per unit ara (normal to the direction of incidence), recieved at a distance $R$, from the centre of a star of radius $r$, (whose outer surface radiates as a black body at temperature TK) can be given, in terms of Stefan’s consant $\sigma$, as;

(1) $\frac{\mathrm{rT}^{2}}{\mathrm{R}}\left(\frac{2 \sigma}{\varepsilon _{0} \mathrm{c}}\right)^{1 / 2}$

(2) $\frac{\mathrm{r}^{2} \mathrm{~T}^{4}}{\mathrm{R}^{2}}\left(\frac{2 \sigma}{\varepsilon _{0} \mathrm{c}}\right)^{1 / 2}$

(3) $\frac{\mathrm{rT}^{2}}{\mathrm{R}}\left(\frac{2 \sigma}{\varepsilon _{0} \mathrm{c}}\right)$

(4) $\frac{\mathrm{r}^{2} \mathrm{~T}^{4}}{\mathrm{cR}^{2}}\left(\frac{2 \sigma}{\varepsilon _{0}}\right)^{1 / 2}$

Show Answer

Correct answer: (1)

Solution:

Energy radiated by the surface of the star per second $=4 \pi \mathrm{r}^{2} . \sigma \mathrm{T}^{4}$

Energy recieved per unit normal area per second at a distance $ =\mathrm{R} = \mathrm{S} = \frac{4 \pi \mathrm{r}^{2} . \sigma \mathrm{T}^{4}}{4 \pi \mathrm{r}^{2}}$

But $\mathrm{S}=\mathrm{I}=\frac{1}{2} \varepsilon _{0} \mathrm{cE} _{0}^{2}$

$\therefore \frac{1}{2} \varepsilon _{0} \mathrm{cE} _{0}^{2}=\frac{\mathrm{r}^{2} \sigma \mathrm{T}^{4}}{\mathrm{R}^{2}}$

or $\mathrm{E} _{0}=\left(\frac{2 \mathrm{r}^{2} \sigma \mathrm{T}^{4}}{\varepsilon _{0} \mathrm{cR}^{2}}\right)^{1 / 2}$

$ =\frac{\mathrm{rT}^{2}}{\mathrm{R}}\left(\frac{2 \sigma}{\varepsilon _{0} \mathrm{c}}\right)^{1 / 2} $