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Which of the following wave cannot travel in vacuum?
(b) Infrasonic wave is not an electromagnetic wave. It is a mechanical wave which requires a material medium for propagation.
Energy of electromagnetic waves is due to their
(c) Electric and magnetic fields.
Explanation:
Electromagnetic waves are composed of oscillating electric and magnetic fields that travel perpendicular to each other. The energy of these waves is directly related to the strength of these fields. When these fields interact with matter, they can transfer energy to it.
Which of the following represents an infrared wavelength?
(a) \(10^{-4} \mathrm{~cm}\). Infrared wavelengths fall within the range of \(10^{-4}\) to \(10^{-3}\) centimeters.
Explanation:
Infrared wavelengths:
Infrared radiation has wavelengths longer than visible light, ranging from approximately 780 nanometers to 1 millimeter.
Which of the following has the shortest wavelength?
(b) \(\gamma\)-rays.
Explanation:
\(\gamma\)-rays:
These have the shortest wavelength among all electromagnetic radiation types, meaning they have the highest frequency and energy.
The frequency of visible light is of the order of
(a) The frequency of visible light ranges from approximately \(4 \times 10^{14} \mathrm{~Hz}\) to \(7.5 \times 10^{14} \mathrm{~Hz}\)
The structure of solids is investigated by using
(b) X-rays.
Explanation: X-rays are used to investigate the structure of solids because their wavelength is comparable to the interatomic distances in crystals. This allows X-rays to be diffracted by the crystal lattice, providing information about the arrangement of atoms within the solid.
Energy stored in electromagnetic oscillations is in the form of
(c) Both (a) and (b).
Explanation: Electromagnetic waves consist of oscillating electric and magnetic fields, so the energy stored in them is a combination of both electrical energy and magnetic energy.
Why other options are incorrect:
(a) Electrical energy:
While electromagnetic waves contain electrical energy, they also contain magnetic energy, so this option is incomplete.
(b) Magnetic energy:
Similar to (a), this option only describes one component of the energy stored in electromagnetic waves.
(d) None of these:
This is incorrect because electromagnetic waves clearly store energy in both electric and magnetic fields.
The range of wavelength of the visible light is
(b) \(4000 Å\) to \(7000 Å\).
Explanation:
Visible light encompasses a specific range of wavelengths within the electromagnetic spectrum. An Angstrom ( \(Å\) ) is a unit of length commonly used to describe wavelengths in this context. Visible light wavelengths typically fall between 4000 and 7000 Angstroms.
Which of the following electromagnetic waves have the longest wavelength?
(c) Radio waves.
Explanation:
Radio waves have the longest wavelengths among all electromagnetic waves listed, including heat waves (which are a type of infrared radiation), light waves, and microwaves.
Why other options are incorrect:
(a) Heat waves: Heat waves are considered a type of infrared radiation, which has a shorter wavelength than radio waves.
(b) Light waves: Visible light has a shorter wavelength than radio waves.
(d) Microwaves: While microwaves have longer wavelengths than visible light, they are still shorter than radio waves.
Radio waves diffract around building although light waves do not. The reason is that radio waves
(b) Radio waves diffract around building although light waves do not. The reason is that radio waves have longer wavelength than light.
Frequency of wave is \(6 \times 10^{15} \mathrm{~Hz}\). The wave is
(d) \(\text { The frequency range is } 10^{15}-10^{17} \text { for UV rays. Hence this wave is UV wave }\)
Explanation:
Radio waves: \(10^4 \mathrm{~Hz}\) to \(10^9 \mathrm{~Hz}\).
Microwaves: \(10^9 \mathrm{~Hz}\) to \(10^{12} \mathrm{~Hz}\).
Ultraviolet rays: \(10^{15} \mathrm{~Hz}\) to \(10^{17} \mathrm{~Hz}\).
X-rays: \(10^{17} \mathrm{~Hz}\) to \(10^{20} \mathrm{~Hz}\).
The frequency 1057 MHz of radiation arising from two close energy levels in hydrogen belongs to
(a)
\(
\begin{aligned}
& f=1057 \mathrm{MHz} \\
& \lambda=\frac{c}{f}=\frac{3 \times 10^8}{1057 \times 10^6} \mathrm{~m} \\
& =0.28 \mathrm{~m}=28 \mathrm{~cm} \text { (radiowaves) }
\end{aligned}
\)
The part of the spectrum of the electromagnetic radiation used to cook food is
(d) The correct answer is microwaves.
Explanation: Microwaves are the type of electromagnetic radiation used to cook food in a microwave oven.
Why other options are incorrect:
(a) Ultraviolet rays:
Ultraviolet radiation is harmful to living organisms and is associated with sunburn. It is not used for cooking.
(b) Cosmic rays:
Cosmic rays are high-energy radiation that originates from outer space. They are extremely powerful and have no use in cooking.
(c) X-rays:
X-rays are also highly penetrating radiation used in medical imaging and security scans. They would damage food rather than cook it.
Which of the following shows the greenhouse effect?
(b) Infrared radiations are reflected by lower clouds and keeps the earth warm. Hence, it shows greenhouse effect.
Which of the following waves are used in RADAR systems for aircraft navigation?
\(
\text { (d) } \lambda_{\text {Microwaves }}>\lambda_{\text {IR-rays }} \gg \lambda_{\text {UV-rays }}>\lambda_{\text {X-rays }}
\)
Hence, microwaves are less energetic electromagnetic waves, which are used in RADAR system for aircraft navigation.
In an electromagnetic wave, the average energy density associated with electric field is
(d) The energy density \(u_{\boldsymbol{E}}\) stored in an electric field is given by:
\(
u_E=\frac{1}{2} \varepsilon_0 E^2
\)
where \(\varepsilon_0\) is the permittivity of free space and \(E\) is the magnitude of the electric field.
Explanation: Energy stored in a capacitor:
\(
U_C=\frac{1}{2} C V^2
\)
Capacitance of a parallel-plate capacitor:
\(
C=\frac{\varepsilon_0 A}{d}
\)
Relationship between voltage and electric field: For a uniform electric field between the plates of a capacitor, the voltage is related to the electric field by:
\(
V=E d
\)
Substitute and simplify: Substitute the expressions for \(\boldsymbol{C}\) and \(\boldsymbol{V}\) into the energy formula:
\(
\begin{aligned}
& U_C=\frac{1}{2}\left(\frac{\varepsilon_0 A}{d}\right)(E d)^2 \\
& U_C=\frac{1}{2} \frac{\varepsilon_0 A}{d} E^2 d^2 \\
& U_C=\frac{1}{2} \varepsilon_0 E^2(A d)
\end{aligned}
\)
Energy density: The energy density, \(u_E\), is the energy per unit volume. The volume between the capacitor plates is \({A d}\). Divide the total energy by the volume to get the energy density:
\(
\begin{aligned}
& u_E=\frac{U_C}{A d}=\frac{\frac{1}{2} \varepsilon_0 E^2(A d)}{A d} \\
& u_E=\frac{1}{2} \varepsilon_0 E^2
\end{aligned}
\)
Total energy density of electromagnetic waves in vacuum is given by the relation
(d) The energy in EM waves is divided equally between the electric and magnetic fields.
The total energy per unit volume, \(u=u_e+u_m\)
\(
=\frac{1}{2} \varepsilon_0 E^2+\frac{1}{2} \frac{B^2}{\mu_0}
\)
For a medium with permittivity \(\varepsilon\) and permeability \(\mu\), the velocity of light is given by
(c) For a medium with permittivity \(\varepsilon\) and permeability \(\mu\), the velocity of light is given by \(c=\frac{1}{\sqrt{\mu \varepsilon}}\).
The speed of electromagnetic wave in vacuum
(c) Speed of electromagnetic waves in vacuum \(=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}=v \lambda\) \(=\) constant
As, we go from \(\gamma\)-rays to radio waves frequency decreases and wavelength increases thereby maintaining the product constant.
In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of \(2 \times 10^{10} \mathrm{~Hz}\) and amplitude \(48 \mathrm{Vm}^{-1}\). The wavelength of the wave is
(b) Speed, \(c={f} \lambda\)
\(\therefore\) Wavelength, \(\lambda=\frac{c}{f}=\frac{3 \times 10^8}{2 \times 10^{10}}=1.5 \times 10^{-2} \mathrm{~m}\)
The essential distinction between X-rays and \(\gamma\)-rays is that
(b) The wavelength of the \(\gamma\)-rays is shorter. However, the essential distinction between them is the nature of emission.
The sun delivers \(10^4 \mathrm{Wm}^{-2}\) of electromagnetic flux to the earth’s surface. The total power that is incident on a roof of dimensions \((10 \times 10) \mathrm{m}^2\) will be
\(
\text { (c) Total power }=10^4 \times(10 \times 10)=10^6 \mathrm{~W}
\)
The average electric field of electromagnetic waves in certain region of free space is \(9 \times 10^{-4} \mathrm{NC}^{-1}\). Then, the average magnetic field in the same region is of the order of
(b) Given, \(\quad E=9 \times 10^{-4} \mathrm{NC}^{-1}\)
We know that, \(c=\frac{E}{B}\)
\(
\Rightarrow \quad B=\frac{9 \times 10^{-4}}{3 \times 10^8}=3 \times 10^{-12} \mathrm{~T}
\)
The pressure exerted by an electromagnetic wave of intensity \(I \mathrm{Wm}^{-2}\) on a non-reflecting surface is \((c\) is the velocity of light)
(c) When a surface intercepts electromagnetic radiation, a force and a pressure are exerted on the surface. As the surface is non-reflecting, so it is completely absorbed and in such case, the force is
\(
F=\frac{I A}{c}
\)
The radiation pressure is the force per unit area,
\(
p=\frac{F}{A}=\frac{I}{c}
\)
The electric and the magnetic fields associated with an electromagnetic wave, propagating along the \(Z\)-axis, can be represented by
(d) The electromagnetic wave is propagating along the \(+z\) axis.
Since the electric and magnetic fields are perpendicular to each other and also perpendicular to the direction of propagation of wave.
Also, \(\vec{E} \times \vec{B}\) gives the direction of wave propagation.
\(
\vec{E}=E_0 \hat{i}, B=B_0 \hat{j}(\because \hat{i} \times \hat{j}=\hat{k})
\)
A parallel plate capacitor is charged to \(60 \mu \mathrm{C}\). Due to a radioactive source, the plate loses charge at the rate of \(1.8 \times 10^{-8} \mathrm{Cs}^{-1}\). The magnitude of displacement current is
\(
\begin{aligned}
&\text { (a) Displacement current is given by }\\
&I_d=\frac{d q}{d t}=1.8 \times 10^{-8} \mathrm{Cs}^{-1}
\end{aligned}
\)
Radiation of intensity \(1 \mathrm{~W} / \mathrm{m}^2\) are striking on a metal plate. The pressure on the plate is
(a) As metal is a reflecting surface and for reflecting surface, radiation pressure,
\(
p_r=\frac{2 I}{c}=\frac{2 \times 1}{3 \times 10^8}=0.66 \times 10^{-8} \mathrm{Nm}^{-2}
\)
An \(L-C\) resonant circuit contains a 200 pF capacitor and a \(200 \mu \mathrm{H}\) inductor. It is set into oscillation coupled to an antenna. The wavelength of the radiated electromagnetic waves is
(b) Frequency of \(L-C\) oscillations,
\(
\begin{aligned}
f=\frac{1}{2 \pi \sqrt{L C}} & =\frac{1}{2 \times 3.14 \sqrt{200 \times 10^{-6} \times 200 \times 10^{-12}}} \\
& =0.0796 \times 10^7 \mathrm{~Hz}
\end{aligned}
\)
So, wavelength, \(\lambda=\frac{c}{f}=\frac{3 \times 10^8}{0.0796 \times 10^7} \simeq 377 \mathrm{~m}\)
The magnetic field in a plane electromagnetic wave is given by \(B_y=2 \times 10^{-7} \sin \left(0.5 \times 10^3 x+1.5 \times 10^{11} t\right)\). This electromagnetic wave is
(c) We have, \(B_y=2 \times 10^{-7} \sin \left(0.5 \times 10^3 x+1.5 \times 10^{11} t\right)\)
Comparing with the standard equation, we get
\(
\begin{aligned}
B_y & =B_0 \sin (k x+\omega t) \\
k & =0.5 \times 10^3
\end{aligned}
\)
Wavelength, \(\lambda=\frac{2 \pi}{0.5 \times 10^3}=0.01256 \mathrm{~m} \quad\left(\because \lambda=\frac{2 \pi}{k}\right)\)
The wavelength range of microwaves is \(10^{-3} \mathrm{~m}\) to 0.1 m. The wavelength of this wave lies between \(10^{-3} \mathrm{~m}\) to 0.1 m, so the equation represents a microwave.
An electromagnetic wave going through vacuum is described by \(E=E_0 \sin (k x-\omega t)\), \(B=B_0 \sin (k x-\omega t)\). Which of the following equations is true?
(a) From \(k=\frac{2 \pi}{\lambda}\) and \(\omega=2 \pi f\), we get \(\frac{k}{\omega}=\frac{\frac{2 \pi}{\lambda}}{2 \pi f}=\frac{1}{f \lambda}=\frac{1}{c}\)
As, \(\frac{E_0}{B_0}=c \Rightarrow \frac{k}{\omega}=\frac{B_0}{E_0}\)
\(
\Rightarrow \quad E_0 k=B_0 \omega
\)
A plane electromagnetic wave propagating in the \(x\)-direction has a wavelength 6.0 mm. The electric field is in the \(y\)-direction and its maximum magnitude is \(33 \mathrm{Vm}^{-1}\). The equation for the electric field as a function of \(x\) and \(t\) is
\(
\begin{aligned}
&\text { (b) Angular frequency, }\\
&\begin{aligned}
\omega & =2 \pi f=\frac{2 \pi c}{\lambda} \quad(\because f=c / \lambda) \\
& =\frac{2 \pi \times 3 \times 10^8}{6 \times 10^{-3}}=\pi \times 10^{11} \mathrm{rads}^{-1}
\end{aligned}
\end{aligned}
\)
The equation for the electric field along \(Y\)-axis in the electromagnetic wave,
\(
E_y=E_0 \sin \omega\left(t-\frac{x}{c}\right)=33 \sin \pi \times 10^{11}(t-x / c)
\)
A perfectly reflecting mirror has an area of \(1 \mathrm{~cm}^2\). Light energy is allowed to fall on it for 1 h at the rate of \(10 \mathrm{Wcm}^{-2}\). The force that acts on the mirror is
(b) Let \(E=\) energy falling on the surface per second \(=10 \mathrm{~W} / \mathrm{cm}^2\)
Momentum of photons, \(p=\frac{h}{\lambda}=\frac{h}{(c / v)}=\frac{h v}{c}=\frac{E}{c}\)
On reflection, change in momentum per second
\(
=2 p=\frac{2 E}{c}=\frac{2 \times 10}{3 \times 10^8}=6.67 \times 10^{-8} \mathrm{~N}
\)
A plane electromagnetic wave is propagating along the \(z\)-direction. If the electric field component of this wave is in the direction \((\hat{\mathbf{i}}+\hat{\mathbf{j}})\), then which of the following is the direction of the magnetic field component?
(a) The magnetic field component is perpendicular to the direction of propagation and the direction of electric field.
Using vector algebra \(\mathbf{E} \times \mathbf{B}\) should be in \(z\)-direction.
\(
\therefore \quad(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \times(-\hat{\mathbf{i}}+\hat{\mathbf{j}})=2 \hat{\mathbf{k}}
\)
Magnitude of the electric and magnetic fields in an electromagnetic wave radiated by a 200 W bulb at a distance 2 m from it is (Assuming efficiency of bulb is \(5 \%\) and it behaves like a point source.)
(a) Effective power of the bulb, \(P=\frac{5}{100} \times 200=10 \mathrm{~W}\)
Intensity at distance \(r\) is given by
\(
\begin{aligned}
I & =\frac{P}{4 \pi r^2}=\frac{10}{4 \times 3.14 \times 2^2} \\
& =\frac{10}{16 \times 3.14} \\
& =0.199 \simeq 0.2 \mathrm{~W} / \mathrm{m}^2 \\
\text { Also, } \quad I & =\frac{1}{2} \varepsilon_0 E_0^2 c \\
\Rightarrow \quad E_0^2 & =\frac{2 I}{c \times \varepsilon_0}
\end{aligned}
\)
Magnitude of electric field, \(E_0=\sqrt{\frac{2 I}{c \times \varepsilon_0}}\)
\(
\begin{aligned}
& =\sqrt{\frac{2 \times 0.2}{3 \times 10^8 \times 8.85 \times 10^{-12}}} \\
& =12.27 \mathrm{~N} / \mathrm{C}
\end{aligned}
\)
\(\therefore\) Magnitude of magnetic field, \(B_0=\frac{E_0}{c}=\frac{12.27}{3 \times 10^8}\)
\(
\begin{aligned}
& =\frac{12.27}{3} \times 10^{-8} \mathrm{~T} \\
& =4.09 \times 10^{-8} \mathrm{~T}
\end{aligned}
\)
The electric vector vibration of an electromagnetic wave is given by \(E=\left(50 \mathrm{NC}^{-1}\right) \sin \omega\left(t-\frac{x}{c}\right)\). The intensity of the wave is [UP CPMT 2015]
\(
\begin{aligned}
&\text { (c) The intensity is, }\\
&\begin{aligned}
I & =\frac{1}{2} \varepsilon_0 E_0^2 c \\
& =\frac{1}{2}\left(8.85 \times 10^{-12}\right) \times(50)^2 \times 3 \times 10^8 \\
& =3.3 \mathrm{Wm}^{-2}
\end{aligned}
\end{aligned}
\)
The frequencies of X-rays, \(\gamma\)-rays and ultraviolet rays are respectively \(p, q\) and \(r\), then [Guj. CET 2015]
(b) Increasing order of wavelengths for the given EM waves are as given below
\(\gamma\)-rays < X-rays < Ultraviolet rays
Since, frequency \(\propto \frac{1}{\text { wavelength }}\)
\(\therefore\) Frequency order, \(\gamma\)-rays \(>\mathrm{X}\)-rays \(>\) Ultraviolet rays
\(
\begin{aligned}
& q>p>r \\
& q>p \\
& q>r \\
& p>r
\end{aligned}
\)
Which radiations are used in treatment of muscles ache? [Manipal 2015]
(b) Infrared rays are used for the treatment of muscles ache.
In electromagnetic wave, according to Maxwell, changing electric field gives [MHT CET 2014]
(d) According to Maxwell, time varying electric field produced displacement current.
\(
I_d=\varepsilon_0 \frac{d \phi_E}{d t}
\)
The speed of light in an isotropic medium depends on [Kerala CEE 2014]
(d) The speed of light in an isotropic medium depends on its wavelength, \({f}=\frac{c}{\lambda}\).
The energy of \(\gamma\)-ray photon is \(E_\gamma\) and that of an X -ray photon is \(E_X\). If the visible light photon has an energy of \(E_v\), then we can say that [WB JEE 2014]
(c)
\(
\begin{aligned}
&\begin{aligned}
& E_\gamma \geq 100 \mathrm{keV} \\
& E_X=100 \mathrm{eV} \text { to } 100 \mathrm{keV} \\
& E_v=2.48 \mathrm{eV}
\end{aligned}\\
&\text { So, we can say that }\\
&E_\gamma>E_X>E_v
\end{aligned}
\)
The speed of electromagnetic waves in vacuum is equal to [UK PMT 2014]
(c) According to Maxwell, it is found that the electromagnetic waves travel in free space (or vacuum) with a speed which is given by \(\quad c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}=3 \times 10^8 \mathrm{~ms}^{-1}\)
where, \(\mu_0\left(=4 \pi \times 10^{-7} \mathrm{~Wb} \mathrm{~A}^{-1} \mathrm{~m}^{-1}\right)\)
and \(\varepsilon_0\) \(\left(=8.85 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}\right)\) are absolute permeability and permittivity of the free space, respectively.
A plane electromagnetic wave of frequency 20 MHz travels through a space along \(x\)-direction. If the electric field vector at a certain point in space is \(6 \mathrm{Vm}^{-1}\), then what is the magnetic field vector at that point? [KCET 2014]
(a)
\(
\begin{aligned}
&\text { The magnetic field, } B_0=\frac{E_0}{c} \text {, where } c=3 \times 10^8 \mathrm{~ms}^{-1}\\
&B_0=\frac{6}{3 \times 10^8}=2 \times 10^{-8} \mathrm{~T}
\end{aligned}
\)
In electromagnetic spectrum, the frequencies of \(\gamma\)-rays and X-rays and ultraviolet rays are denoted by \(n_1, n_2\) and \(n_3\) respectively, then [MHT CET 2014]
(a) Frequency \(\left(n_3\right)\) range of ultraviolet rays \(\simeq 10^{17}-10^{14}\)
Frequency \(\left(n_2\right)\) range of X-rays \(\simeq 10^{20}-10^{17}\)
Frequency \(\left(n_1\right)\) range of \(\gamma[latex]-rays [latex]>10^{20}\)
\(
\therefore \quad n_1>n_2>n_3
\)
The electromagnetic waves detected using a thermopile and used in physical therapy are [Kerala CEE 2014]
(d) The electromagnetic waves detected using a thermopile and used in physical therapy are infrared radiations for treating muscular strain.
The wavelength of X-rays is in the range [Guj. CET 2014]
(b)
\(
\begin{aligned}
\text { X-rays wavelength range } & =1 \times 10^{-12} \mathrm{~m} \text { to } 1 \times 10^{-9} \mathrm{~m} \\
& =0.001 \mathrm{~nm} \text { to } 1 \mathrm{~nm}
\end{aligned}
\)
The wavelength of the short radio waves, microwaves, ultraviolet waves are \(\lambda_1, \lambda_2\) and \(\lambda_3\), respectively. Arrange them in decreasing order. [Guj. CET 2014]
(a) We know that
(i) Wavelength range of short radio waves, \(=1 \times 10^{-1} \mathrm{~m}\) to \(1 \times 10^4 \mathrm{~m}\)
(ii) Microwaves wavelength range \(=1 \times 10^{-3} \mathrm{~m}\) to \(3 \times 10^{-1} \mathrm{~m}\)
(iii) Ultraviolet waves wavelength range \(=1 \times 10^{-9} \mathrm{~m}\) to \(4 \times 10^{-7} \mathrm{~m}\)
So, decreasing order is \(\lambda_1, \lambda_2, \lambda_3\).
Which component of electromagnetic spectrum have maximum wavelength? [J & K CET 2013]
(a) Radio waves have maximum wavelength in the range of 0.1 m to \(10^4 \mathrm{~m}\).
The wave function (in SI unit) for a light wave is given as \(\psi(x, t)=10^3 \sin \pi\left(3 \times 10^6 x-9 \times 10^{14} t\right)\). The frequency of the wave is equal to [AMU 2012]
(a) The wave function
\(
\psi(x, t)=10^3 \sin \pi\left(3 \times 10^6 x-9 \times 10^{14} t\right)
\)
Here, angular frequency, \(\omega=9 \times 10^{14} \pi\)
\(
\Rightarrow \quad \begin{aligned}
\omega & =2 \pi f \\
f & =\frac{\omega}{2 \pi}=\frac{9 \times 10^{14} \pi}{2 \pi} \\
& =4.5 \times 10^{14} \mathrm{~Hz}
\end{aligned}
\)
The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is \(B_o=510 \mathrm{nT}\). What is the amplitude of the electric field part of the wave?
(a) Given \(B_o=510 \mathrm{nT}=510 \times 10^{-9} \mathrm{~T}\) \(c=3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}\)
For electromagnetic waves, \(c=\frac{E_o}{B_o}\) or \(E_o=c B_o\)
\(
=3 \times 10^8 \times 510 \times 10^{-9}=153 \mathrm{Vm}^{-1}
\)
Amplitude of electric field part of the wave is \(153 \mathrm{Vm}^{-1}\).
Suppose that the electric field amplitude of an electromagnetic wave is \(E_o=120 \mathrm{~N} / \mathrm{C}\) and that its frequency is \(v=50.0 \mathrm{MHz}\). Find expressions for \(E\) and \(B\).
(c)
\(
\begin{aligned}
&\text { Solution: Given } E_o=120 \mathrm{NC}^{-1}\\
&\begin{aligned}
& f=50 \mathrm{MHz}=50 \times 10^6 \mathrm{~Hz} \\
& c=3 \times 10^8 \mathrm{~ms}^{-1}
\end{aligned}
\end{aligned}
\)
For electromagnetic waves, we know that \(c=\frac{E_o}{B_o}\) or
\(
\begin{aligned}
B_o & =\frac{E_o}{c}=\frac{120}{3 \times 10^8}=40 \times 10^{-8} \mathrm{~T}=400 \times 10^{-9} \mathrm{~T} \\
& =400 \mathrm{nT}
\end{aligned}
\)
Also
\(
\begin{aligned}
\omega & =2 \pi f=2 \times 3.14 \times 50 \times 10^6 \\
& =3.14 \times 10^8 \mathrm{rad} \mathrm{~s}^{-1}
\end{aligned}
\)
Also \(\omega=k c\)
\(
k=\frac{\omega}{c}=\frac{3.14 \times 10^8 \mathrm{rad} \mathrm{~s}^{-1}}{3 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}}=1.05 \mathrm{rad} \mathrm{~m}^{-1}
\)
Now \(\quad c=f \lambda\)
\(
\lambda=\frac{c}{f}=\frac{3 \times 10^8}{50 \times 10^6}=6 \mathrm{~m}
\)
Expression for \(E\) and \(B\). If we suppose \(c\) along \(z\)-axis, \(E_o\) and \(B_o\) are along \(x\)-axis and \(y\)-axis.
Equation for \(E\) is
\(
\begin{aligned}
& E=E_o \sin (k x-\omega t) \text { along } x \text {-axis } \\
& \text { or } \quad \vec{E}=E_o \sin (k x-\omega t) \hat{i} \\
& =120 \sin \left(1.5 x-3.14 \times 10^8 \times t\right) \hat{i} \\
& \text { and } \quad B=B_o \sin (k x-\omega t) \text { along } y \text {-axis } \\
& \text { or } \quad \vec{B}=B_o \sin (k x-\omega t) \hat{j} \\
& =400 \mathrm{nT} \sin \left(1.5 x-3.14 \times 10^8 t\right) \hat{j} \text {. }
\end{aligned}
\)
In a plane em wave, the electric field oscillates sinusoidally at a frequency of \(2.0 \times 10^{10} \mathrm{~Hz}\) and amplitude \(48 \mathrm{~V} \mathrm{~m}^{-1}\).
(a) What is the wavelength of a wave?
(b) What is the amplitude of the oscillating magnetic field?
(c) What is the average energy density of the \(\vec{E}\) field and the average energy density of the \(\vec{B}\) field? \(\left[c=3 \times 10 \mathrm{~m} \mathrm{~s}^{-1}\right]\).
(a)
(a) \(\lambda=\frac{c}{f}=\frac{3 \times 10^8}{2 \times 10^{10}}=1.5 \times 10^{-2} \mathrm{~m}=1.5 \mathrm{~cm}\)
(b) \(B_o=\frac{E_o}{c}=\frac{48}{3 \times 10^8}=16 \times 10^{-8} \mathrm{~T}=-0.6 \mu \mathrm{~T}\)
(c) Energy density in \(E\) field, \(U_E=\frac{1}{2} \varepsilon_o E^2\)
Energy density in \(B[latex] field
[latex]
U_B=\frac{1}{2} \frac{1}{\mu_o} B^2
\)
\(
\text { Now consider } U_E=\frac{1}{2} \varepsilon_o(C B)^2 \quad\left[\because \frac{E}{B}=C\right]
\)
\(
=\frac{1}{2} \varepsilon_o C^2 B^2=\frac{1}{2} \varepsilon_o \frac{1}{\mu_o \varepsilon_o} B^2
\)
\(
=\frac{1}{2} \frac{1}{\mu_o} B^2 \quad\left[\because C=\frac{1}{\sqrt{\mu_o \varepsilon_o}}\right]
\)
\(
U_E=U_B
\)
Suppose that the electric field part of an electromagnetic wave in vacuum is
\(
E=\left\{(3.1 \mathrm{~N} / \mathrm{C}) \cos \left[(1.8 \mathrm{rad} / \mathrm{m}) y+\left(5.4 \times 10^6 \mathrm{rad} / \mathrm{s}\right) t\right]\right\} \hat{i}
\)
What is the amplitude of the magnetic field part of the wave?
(d)
\(
\begin{aligned}
&\text { We know that } c=\frac{E_o}{B_o}\\
&\begin{aligned}
\therefore B_o & =\frac{E_o}{c}=\frac{3.1}{3 \times 10^8}=1.03 \times 10^{-8} \mathrm{~T} \\
& =0.0103 \times 10^{-6} \mathrm{~T}=0.0103 \mu \mathrm{~T}
\end{aligned}
\end{aligned}
\)
Suppose that the electric field part of an electromagnetic wave in vacuum is
\(
E=\left\{(3.1 \mathrm{~N} / \mathrm{C}) \cos \left[(1.8 \mathrm{rad} / \mathrm{m}) y+\left(5.4 \times 10^6 \mathrm{rad} / \mathrm{s}\right) t\right]\right\} \hat{i}
\)
What is the expression for the magnetic field part of the wave?
(a) Expression of magnetic field part of the wave
\(
\begin{aligned}
& B=B_o \cos (k y+\omega t) \\
& =1.03 \times 10^{-8} \cos \left\{(1.8 \mathrm{rad} / \mathrm{m}) y+\left(5.4 \times 10^6 \mathrm{rad} / \mathrm{s}\right) t\right\}
\end{aligned}
\)
\(E\) is along \(\hat{i}\) and \(c\) is along \(-\hat{j}, c\) is the direction of
\(
\begin{aligned}
& \vec{E} \times \vec{B} \\
& -\hat{j}=\hat{i} \times ?
\end{aligned}
\)
Clearly “?” is in the direction of \(\hat{k}\) as \((k \times i=j)\) and
\(
(i \times k=-j)
\)
Thus \(\vec{B}\) is completely represented as
\(
\begin{aligned}
& \vec{B}=1.03 \times 10^{-8} \cos \left\{(1.8 \mathrm{rad} / \mathrm{m}) y+\left(5.4 \times 10^6\right.\right. \\
& \mathrm{rad} / \mathrm{s}) t\} \hat{k}
\end{aligned}
\)
About \(5 \%\) of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation
(a) at a distance of 1 m from the bulb?
(b) at a distance of 10 m ?
Assume that the radiation is emitted isotropically and neglect reflection.
(c)
(a) Power of bulb, \(P=100 \mathrm{~W}\)
Power is converted to visible radiation \(=5 \%[latex] [latex]=5 \mathrm{~J} / \mathrm{s}\).
Electric flux, \(\phi=5 \mathrm{~J} / \mathrm{s}, r=1 \mathrm{~m}\)
Using relation,
\(
\begin{aligned}
E & =\frac{\phi}{4 \pi r^2}=\frac{5}{4 \times 3.14 \times(1)^2} \\
& =0.4 \mathrm{~W} \mathrm{~m}^{-2}
\end{aligned}
\)
(b) \(r=10 \mathrm{~m}\)
Using relation, \(E=\frac{\phi}{4 \pi r^2}=\frac{5}{4 \times 3.14(10)^2}\)
The charge on a parallel plate capacitor is varying as \(q=q_o \sin 2 \pi n t\). the plates are very large and close together. Neglecting the edge effects, the displacement current through the capacitor is
(c) \(I=\frac{d q}{d t}=\frac{d}{d t}\left(q_0 \sin 2 \pi n t\right)=q_0 2 \pi n \cos 2 \pi n t\)
A plane electromagnetic wave, \(E_z=100 \cos \left(6 \times 10^8 t+\right.\) \(4 x) \mathrm{V} / \mathrm{m}\) propagates in a medium of dielectric constant
(b) Comparing the given equation with the equation of plane electromagnetic wave,
\(
E_z=E_o \cos (\omega t+k x)
\)
We have \(\omega=6 \times 10^8\) and \(k=4\)
Velocity of light in medium,
\(
\begin{aligned}
& v=\frac{\omega}{k}=\frac{6 \times 10^8}{4}=\frac{3}{2} \times 10^8 \mathrm{~m} / \mathrm{s} \\
& \therefore \text { Refractive Index, } \mu=\frac{c}{v}=\frac{3 \times 10^8}{\frac{3}{2} \times 10^8}=2
\end{aligned}
\)
The average energy density of an electromagnetic wave given by \(E=(50 \mathrm{~N} / \mathrm{C}) \sin (\omega t-k x)\) will be nearly:
(a)
\(
\begin{aligned}
&\text { Average energy density of electromagnetic wave, }\\
&\begin{aligned}
U_{\mathrm{av}} & =\frac{1}{2} \varepsilon_o E_o^2=\frac{1}{2} \times\left(8.85 \times 10^{-12}\right) \times(50)^2 \\
& \approx 10^{-8} \mathrm{~J} / \mathrm{m}^3
\end{aligned}
\end{aligned}
\)
A plane EM wave of wave intensity \(10 \mathrm{~W} / \mathrm{m}^2\) strikes a small mirror of area \(20 \mathrm{~cm}^2\), held perpendicular to the approaching wave. The radiation force on the mirror will be
(c) Radiation force \(=\) momentum transferred per sec by electromagnetic wave to the mirror
\(
\begin{aligned}
& =\frac{2 S_{\mathrm{av}} A}{c}=\frac{2 \times 10 \times 20 \times 10^{-4}}{3 \times 10^8} \\
& =1.33 \times 10^{-10} \mathrm{~N}
\end{aligned}
\)
The magnetic field in the plane electromagnetic field is given by:
\(
B_y=2 \times 10^{-7} \sin \left(0.5 \times 10^3 z+1.5 \times 10^{11} t\right) \mathrm{T}
\)
The expression for the electric field may be given by:
(d)
\(
B_y=2 \times 10^{-7} \sin \left(0.5 \times 10^3 z+1.5 \times 10^{11} t\right) \mathrm{T}
\)
The electric vector is perpendicular to B as will direction of propagation of electromagnetic wave.
Therefore, \(E_x\) has to be taken.
Further, \(E_o=B_o \times c\)
or \(=2 \times 10^{-7} \times 3 \times 10^8 \mathrm{~V} / \mathrm{m}\)
or \(E_o=2 \times 10^{-7} \times 3 \times 10^8=60 \mathrm{~V} / \mathrm{m}\)
\(\therefore\) The corresponding value of the electric field is,
\(
E_x=60 \sin \left(0.5 \times 10^3 z+1.5 \times 10^{11} t\right) \mathrm{V} / \mathrm{m}
\)
A point source of electromagnetic radiation has an average power output of 800 W. The maximum value of electric field at a distance of 4.0 m from the source is:
(d)
\(
\begin{aligned}
&\text { Intensity of electromagnetic wave is, }\\
&I=\frac{P_{\mathrm{av} .}}{2 \pi r^2}=\frac{E_o^2}{\mu_o c}
\end{aligned}
\)
\(
\begin{aligned}
E_o & =\sqrt{\frac{\mu_o c P_{\mathrm{av} .}}{2 \pi r^2}}=\sqrt{\frac{\left(4 \pi \times 10^{-7}\right)\left(3 \times 10^8\right) \times 800}{2 \pi \times(4)^2}} \\
& =54.77 \mathrm{~V} / \mathrm{m}
\end{aligned}
\)
A large parallel plate capacitor, whose plates have an area of \(1 \mathrm{~m}^2\) and are separated from each other by 1 mm, is being charged at a rate of \(25 \mathrm{~V} / \mathrm{s}\). If the dielectric between the plates has the dielectric constant 10 , then the displacement current at this instant is:
\(
\begin{aligned}
C & =\frac{\varepsilon_o k A}{d}=\frac{\left(8.85 \times 10^{-12}\right) \times 10 \times 1}{10^{-3}} \\
& =8.85 \times 10^{-8} \mathrm{~F} \\
I & =\frac{d}{d t}(C V)=C \frac{d v}{d t}=8.85 \times 10^{-8} \times 25 \\
& =2.2 \times 10^{-6} \mathrm{~A} \\
& =2.2 \mu \mathrm{~A}
\end{aligned}
\)
A parallel plate capacitor with plate area \(A\) and separation between the plates \(d\), is charged by a constant current \(I\). Consider a plane surface of area \(A / 2\) parallel to the plates and drawn simultaneously between the plates. The displacement current through this area is:
Charge on capacitor plates at time \(t\) is \(q=I t\) Electric field between the plates at this instant,
\(
E=\frac{q}{A \varepsilon_o}=\frac{I t}{A \varepsilon_o}
\)
Electric flux through the given area,
\(
\phi_E=\left(\frac{A}{2}\right) E=\frac{I t}{2 \varepsilon_o}
\)
\(
\begin{aligned}
&\text { Therefore, displacement current, }\\
&\begin{aligned}
I_d & =\varepsilon_o \frac{d \phi_E}{d t}=\varepsilon_o \frac{d}{d t}\left(\frac{I t}{2 \varepsilon_o}\right) \\
& =\frac{I}{2}
\end{aligned}
\end{aligned}
\)
A plane electromagnetic wave travelling along the \(X\)-direction has a wavelength of 3 mm . The variation in the electric field occurs in the \(Y\)-direction with an amplitude \(66 \mathrm{Vm}^{-1}\). The equation for the electric and magnetic fields as a function of \(x\) and \(t\) are respectively:
(d)
The equation of electric field occurring in \(Y\)-direction, \(E_y=66 \cos 2 \pi \times 10^{11}\left(t-\frac{x}{c}\right)\)
Therefore, for the magnetic field in \(Z\)-direction,
\(
\begin{aligned}
B_z & =\frac{E_y}{c}=\left(\frac{66}{3 \times 10^8}\right) \cos 2 \pi \times 10^{11}\left(t-\frac{x}{c}\right) \\
& =22 \times 10^{-8} \cos 2 \pi \times 10^{11}\left(t-\frac{x}{c}\right) \\
& =2.2 \times 10^{-7} \cos 2 \pi \times 10^{-11}\left(t-\frac{x}{c}\right)
\end{aligned}
\)
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