Case study (4 Marks)

Question 14 Marks

In a microwave oven, the food is kept in a plastic container and the microwave is directed towards the food. The food is cooked without melting or igniting the plastic container. Explain.

Answer

The natural frequency of water matches the frequency of microwave. This is the reason that food containing water gets cooked. The natural frequency of the plastic container does not match the frequency of microwave. So, the plastic container is not damaged.

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Question 24 Marks

A metal rod is placed along the axis of a solenoid carrying a high-freqμency alternating current. It is found that the rod gets heated. Explain why the rod gets heated.

Answer

The magnetic field along the axis of a solenoid carrying a high-frequency alternating current changes continuously. Due to the change in the magnetic field, e.m.f (or eddy current) is induced in the metal rod. There will be flow of charge due to the induced e.m.f. The direction of the induced e.m.f changes very frequently due to the high-frequency alternating current in the solenoid. Thus, the rod gets heated up due to the flow of charge in it.

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Question 34 Marks

Professor C.V Raman surprised his students by suspending freely a tiny light ball in a transparent vacuum chamber by shining a laser beam on it. Which property of EM waves was he exhibiting? Give one more example of this property.

Answer

The properties of an electromagnetic wave is same as other waves. Like other wave an electromagnetic wave also carries energy and momentum. Since, it carries momentum, an electromagnetic wave also exerts pressure called radiation pressure. This property of electromagnetic waves helped professor C V Raman surprised his students by suspending freely a tiny light ball in a transparent vacuum chamber by shining a laser beam on it.

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Question 44 Marks

Electrons oscillating in a circuit give rise to radiowaves. A transmitting antenna radiates most effectively the radiowaves of wavelength equal to the size of the antenna. The infrared waves incident on a substance set into oscillation all its electrons, atoms and molecules. This increases the internal energy and hence the temperature of the substance.

If $v_g, v_x$ and $v_m$ are the speeds of gamma rays, X-rays and microwaves respectively in vacuum, the

$v_g > v_x > v_m$
$v_g < v_x < v_m$
$v_g > v_x > v_m$
$v_g = v_x = v_m$

Which of the following wi II deflect in electric field?

X-rays.
$\gamma-\text{rays}.$
Cathode rays.
Ultraviolet rays.

$\gamma-\text{rays}$ are detected by:

Point contact diodes.
Thennopiles.
Ionization chamber.
Photocells.

The frequency of electromagnetic wave, which best suited to observe a particle ofradius $3 \times 10^{-4}cm$ is the order of,

$10^{15}Hz$
$10^{14} Hz$
$10^{13}Hz$
$10^{12}Hz$

We consider the radiation emitted by the human body. Which one of the following statements is true?

The radiation emitted is in the infrared region.
The radiation is emitted only during the day.
The radiation is emitted during the summers and absorbed during the winters.
The radiation emitted lies in the ultraviolet region and hence it is not visible.

Answer

(d) $v_g = v_x = v_m$

Explanation:
All electromagnetic waves travel in vacuum with the same speed.

Explanation:
Cathode rays (beamofelectrons) get deflected in an electric field.

Explanation:
$\gamma-\text{rays}$ are detected by ionization chamber.

(b) $10^{14}Hz$

Explanation:
Size of particle $=\lambda=\frac{\text{c}}{\upsilon}$
$\upsilon=\frac{\text{c}}{\lambda}=\frac{3\times10^10\text{cm}\ \text{s}^{-1}}{3\times10^{-4}\text{cm}}=3\times10^{14}\text{Hz}$

(a) The radiation emitted is in the infrared region.

Explanation:
Every body at a temperature T > 0 K emits radiation in the infrared region.

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Question 54 Marks

Radio waves are produced by the accelerated motion of charges in conducting wires. Microwaves are produced by special vacuum tubes. Infrared waves are produced by hot bodies and molecules also known as heat waves. UV rays are produced by special lamps and very hot bodies like Sun.

Solar radiation is:

Transverse electromagnetic wave.
Longitudinal electromagnetic waves.
Both longitudinal and transverse electromagnetic waves.
None of these.

What is the cause of greenhouse effect?

Infrared rays.
Ultraviolet rays
X-rays.
Radiowaves.

Biological importance of ozone layer is:

It stops ultraviolet rays.
It layer reduces greenhouse effect.
It reflects radiowaves.
None of these.

Ozone is found in.

Stratosphere.
Ionosphere.
Mesosphere.
Troposphere.

Earth's atmosphere is richest in.

Ultraviolet.
Infrared.
X-rays.
Microwave.

Answer

(a) Transverse electromagnetic wave.
(a) Infrared rays.

Explanation:

Greenhouse effect is due to infrared rays.

(a) It stops ultraviolet rays.

Explanation:

Ozone layer absorbs the harmful ultraviolet radiations coming from the sun.

(a) Stratosphere.

Explanation:

Ozone layer lies in stratosphere.

(b) Infrared.

Explanation:

Heatmosphere of earth is richest in infrared radiation.

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Question 64 Marks

A stationary charge produces only an electrostatic field while a charge in uniform motion produces a magnetic field, that does not change with time. An oscillating charge is an example of accelerating charge. It produces an oscillating magnetic field, which in turn produces an oscillating electric fields and so on. The oscillating electric and magnetic fields regenerate each other as a wave which propagates through space.
Magnetic field in a plane electromagnetic wave is given by $\vec{\text{B}}=\text{B}_0\sin(\text{kx}+\omega\text{t}) \hat{\text{j}}\text{T}.$

Expression for corresponding electric field will be (Where c is speed of light).

$\vec{\text{E}}=-\text{B}_0\text{c}\sin(\text{kx}+\omega\text{t}) \hat{\text{k}}\frac{\text{V}}{\text{m}}$
$\vec{\text{E}}=\text{B}_0\text{c}\sin(\text{kx}-\omega\text{t}) \hat{\text{k}}\frac{\text{V}}{\text{m}}$
$\vec{\text{E}}=\frac{\text{B}_0}{\text{c}}\sin(\text{kx}+\omega\text{t}) \hat{\text{k}}\frac{\text{V}}{\text{m}}$
$\vec{\text{E}}=\text{B}_0\text{c}\sin(\text{kx}+\omega\text{t}) \hat{\text{k}}\frac{\text{V}}{\text{m}}$

The electric field component ofa monochromatic radiation is given by $\vec{\text{E}} = 2\in_0\hat{\text{i}}\cos\text{kz}\cos\omega\text{t}.$ Its magnetic field $\vec{\text{B}}$ is then given by:

$\frac{2\in_0}{\text{c}}\hat{\text{j}}\cos\text{kz}\cos\omega\text{t}$
$\frac{2\in_0}{\text{c}}\hat{\text{j}}\sin\text{kz}\cos\omega\text{t}$
$\frac{2\in_0}{\text{c}}\hat{\text{j}}\sin\text{kz}\sin\omega\text{t}$
$-\frac{2\in_0}{\text{c}}\hat{\text{j}}\sin\text{kz}\sin\omega\text{t}$

A plane em wave of frequency 25MHz travels in a free space along x-direction. At a particular point in space and time, $\text{E}=(6.3\ \hat{\text{j}})\frac{\text{V}}{\text{m}}.$ What is magnetic field at that time?

$0.095\mu\text{T}$
$0.124\mu\text{T}$
$0.089\mu\text{T}$
$0.021\mu\text{T}$

A plane electromagnetic wave travelling along the x-direction has a wavelength of 3mm. The variation in the electric field occurs in they-direction with an amplitude $66Vm^1$. The equations for the electric and magnetic fields as a function of x and tare respectively.

$\text{E}_\text{y}=33\cos\pi\times10^{11}\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big),\\\text{B}_\text{z}=1.1\times10^{-7}\cos\pi\times10^{11}\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big)$
$\text{E}_\text{y}=11\cos2\pi\times10^{11}\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big),\\\text{B}_\text{y}=11\times10^{-7}\cos2\pi\times10^{11}\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big)$
$\text{E}_\text{x}=33\cos\pi\times10^{11}\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big),\\\text{B}_\text{x}=11\times10^{-7}\cos\pi\times10^{11}\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big)$
$\text{E}_\text{y}=66\cos2\pi\times10^{11}\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big),\\\text{B}_\text{z}=2.2\times10^{-7}\cos2\pi\times10^{11}\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big)$

A plane electromagnetic wave travels in free space along x-axis. At a particular point in space, the electric field along y-axis is $9.3Vm^{-1}.$ The magnetic induction (B) alongz-axis is:

$3.1 \times 10^{-8}T$
$3 \times 10^{-5}T$
$3 \times 10^{-6}T$
$9.3 \times 10^{-6}T$

Answer

(d) $\vec{\text{E}}=\text{B}_0\text{c}\sin(\text{Kx}+\omega\text{t}) \hat{\text{k}}\frac{\text{V}}{\text{m}}$

Explanation:
Given: $\vec{\text{B}}=\text{B}_0\text{c}\sin(\text{Kx}+\omega\text{t}) \hat{\text{k}}\frac{\text{V}}{\text{m}}$
The relation between electric and magnetic field is,
$\text{c}=\frac{\text{E}}{\text{B}}$ or E = cB
The electric 6 eld component is perpendicular to the direction of propagation and the direction of magnetic field. Therefore, the electric field component along z-axis is obtained as $\vec{\text{E}}=\text{cB}_0\sin(\text{kx}+\omega\text{t}) \hat{\text{k}}\frac{\text{V}}{\text{m}}.$

Explanation:
$\frac{\text{dE}}{\text{dz}}=-\frac{\text{dB}}{\text{dt}}$
$\frac{\text{dE}}{\text{dz}}=-2\text{E}_0\text{k}\sin\text{kz}\cos\omega\text{t}=-\frac{\text{dB}}{\text{dt}}$
${\text{dB}}=+2\text{E}_0\text{k}\sin\text{kz}\cos\omega\text{t}{\text{dt}}$
${\text{B}}=+2\text{E}_0\text{k}\sin\text{kz}\int\cos\omega\text{t}{\text{dt}}$
$=+2\text{E}_0\frac{\text{k}}{\omega}\sin\text{kz}\sin\omega\text{t}$
$\frac{\text{E}_0}{\text{B}_0}=\frac{\omega}{\text{k}}=\text{c}$
$\text{B}=\frac{2\text{E}_0}{\text{c}}\sin\text{kz}\sin\omega\text{t}$
$\therefore\text{B}=\frac{2\text{E}_0}{\text{c}}\sin\text{kz}\sin\omega\text{t}\hat{\text{j}}$
E is along y-direction and the wave propagates along x-axis.
$\therefore$ B should be in a direction perpendicular to both x and y-axis.

(d) $0.021\mu\text{T}$

Explanation:
Here, $\vec{\text{E}}=6.3\hat{\text{j}};\text{c}=3\times10^8\frac{\text{m}}{\text{s}}$
The magnitude of B is:
${\text{Bz}}=\frac{\text{E}}{\text{c}}=\frac{6.3}{3\times10^8}$
$=2.1\times10^8\ \text{T}=0.021\mu\text{T}$

(d) $\text{E}_\text{y}=66\cos2\pi\times10^{11}\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big),\text{B}_\text{z}=2.2\times10^{-7}\cos2\pi\times10^{11}\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big)$

Explanation:
Here: $\text{E}_0=66\text{Vm}^{-1},\text{E}_\text{y}=66\cos\omega\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big),$
$\lambda=3\text{mm}=3\times10^{-3}\text{m},\text{k}=\frac{2\pi}{\lambda}$
$\frac{\omega}{\text{k}}=\text{c}\Rightarrow\omega=\text{ck}=3\times10^8\times\frac{2\pi}{3\times10^{-3}}$
or $\omega=2\pi\times10^{11}$
$\therefore\text{E}_\text{y}=66\cos2\pi\times10^{11}\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big)$
${\text{Bz}}=\frac{\text{E}_\text{y}}{\text{c}}=\Big(\frac{66}{3\times10^8}\Big)\cos2\pi\times10^{11}\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big)$
$=2.2\times10^{-7}\cos2\pi\times10^{11}\Big(\text{t}-\frac{\text{x}}{\text{c}}\Big)$

(a) $3.1 \times 10^{-8}T$

Explanation:
At a particular point, $E = 9.3Vm^{-1}$
$\therefore$ Magnetic field at the same point $=\frac{9.3}{3\times10^8}$
$= 3.1 \times 10^{-8}T$

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Question 74 Marks

In an electromagnetic wave both the electric and magnetic fields are perpendicular to the direction of propagation, that is why electromagnetic waves are transverse in nature. Electromagnetic waves carry energy as they travel through space and this energy is shared equally by the electric and magnetic fields. Energy density of an electromagnetic waves is the energy in unit volume of the space through which the wave travels.

The electromagnetic waves propagated perpendicular to both $\vec{\text{E}}$ and $\vec{\text{B}}.$ The electromagnetic waves travel in the direction of.

$\vec{\text{E}}\times\vec{\text{B}}$
$\vec{\text{E}}\times\vec{\text{B}}$
$\vec{\text{B}}\times\vec{\text{E}}$
$\vec{\text{B}}\times\vec{\text{E}}$

Fundamental particle in an electromagnetic wave is:

Photon
Electron
Phonon
Proton

Electromagnetic waves are transverse in nature is evident by:

Polarisation
Interference
Reflection
Diffraction

For a wave propagating in a medium, identify the property that is independent of the others.

Velocity
Wavelength
Frequency
All these depend on each other.

The electric and magnetic fields of an electromagnetic waves are:

In opposite phase and perpendicular to each other.
In opposite phase and parallel to each other.
In phase and perpendicular to each other.
In phase and parallel to each other.

Answer

(b) $\vec{\text{E}}\times\vec{\text{B}}$

Explanation:

Electromagnetic waves propagate in the direction of $\vec{\text{E}}\times\vec{\text{B}}.$

(a) Photon

Explanation:

Photon is the fundamental particle in an electromagnetic wave.

(a) Polarisation

Explanation:

Polarisation establishes the wave nature of electromagnetic waves.

Explanation:

Frequency u remains unchanged when a wave propagates from one medium to another. Both wavelength and velocity get changed.

Explanation:

The electric and magnetic fields of an electromagnetic wave are in phase and perpendicular to each other.

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Question 84 Marks

Maxwell showed that the speed of an electromagnetic wave depends on the penneability and pennittivity of the medium through which it travels. The speed of an electromagnetic wave in free space is given by $\text{c}=\frac{1}{\sqrt{\mu_0\in_0}}.$ The fact led Maxwell to predict that light is an electromagnetic wave. The emergence of the speed of light from purely electromagnetic considerations is the crowning achievement of Maxwell's electromagnetic theory. The speed of an electromagnetic wave in any medium of permeability $\mu$ and pennittivity $\in$ will be $\frac{\text{c}}{\sqrt{\text{K}\mu_\text{r}}}$ where K is the dielectric constant of the medium and $\mu,$ is the relative permeability.

The dimensions of $\frac{1}{2}\in_0\text{E}^2$ ($\in:$ pennittivity of free space; E = electric field) is:

$MLT^{-1}$
$ML^2T^{-2}$
$ML^{-1}T^{-2}$
$ML^2T^{-1}$

Let $[\in_0]$ denote the dimensional formula of the permittivity of the vacuum. UM = mass, L = length, T = time and A = electric current, then

$[\in_0]=\text{M}^{-1}\text{L}^{-3}\text{T}^2\text{A}$
$[\in_0]=\text{M}^{-1}\text{L}^{-3}\text{T}^4\text{A}^2$
$[\in_0]=\text{MLT}^{-2}\text{A}^{-2}$
$[\in_0]=\text{ML}^{2}\text{A}^{-1}$

An electromagnetic wave offrequency 3MHz passes from vacuum into adielectricmedium with permittivity $\in=4.$ Then

Wavelength and frequency both remain unchanged.
Wavelength is doubled and the frequency remains unchanged.
Wavelength is doubled and the frequency becomes half.
Wavelength is halved and the frequency remains unchanged.

Which of the following are not electromagnetic waves?

Cosmic rays
$\gamma-\text{rays}$
$\beta-\text{rays}$
X-rays

The electromagnetic waves travel with,

The same speed in all media.
The speed oflight $c = 3 \times 10^8ms^{-1}$ in free space.
The speed oflight $c = 3 \times 10^8ms^{-1}$ in solid medium
The speed of light $c = 3 \times 10^8ms^{-1}$ in fluid medium.

Answer

Explanation:
$\frac{1}{2}\in_0\text{E}^2=\text{energy density}=\frac{\text{Energy}}{\text{Volume}}$
$\therefore\big[\frac{1}{2}\in_0\text{E}^2\big]=\frac{\text{ML}^2\text{T}^{-2}}{\text{L}^3}=[\text{ML}^{-1}\text{T}^{-2}]$

(b) $[\in_0]=\text{M}^{-1}\text{L}^{-3}\text{T}^4\text{A}^2$

Explanation:
As $\in_0=\frac{\text{q}_1\text{q}_2}{4\pi\text{FR}^2}$ (from Coulomb's law)
$\in_0=\frac{\text{C}_2}{\text{MLT}^{-2}\text{L}^{2}}=\text{M}^{-1}\text{L}^{-3}\text{T}^4\text{A}^2$

(d) Wavelength is halved and the frequency remains unchanged.

Explanation:
The frequency of the electromagnetic wave remains same when it passes from one medium to another.
Refractive index of the medium, $\text{n}=\sqrt{\frac{\in}{\in_0}}=\sqrt{\frac{4}{1}}=2$
Wavelength of the electromagnetic wave in the medium,
$\lambda_\text{med}=\frac{\lambda}{\text{n}}=\frac{\lambda}{2}$

(b) $\gamma-\text{rays}$

Explanation:
$\beta-\text{rays}$ consists of electrons which are not electromagnetic in nature.

(b) The speed oflight $c = 3 \times 10^8ms^{-1}$ in free space.

Explanation:
The velocity of electromagnetic waves in free space (vacuum) is equal to velocity of light in vacuum.
$(i.e., 3 \times 10^8ms^{-1})$.

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Question 94 Marks

All the known radiations from a big family of electromagnetic waves which stretch over a large range of wavelengths. Electromagnetic wave include radio waves, microwaves, visible light waves, infrared rays, UV rays, X-rays and gamma rays. The orderly distribution of the electromagnetic waves in accordance with their wavelength or frequency into distinct groups having widely differing properties is electromagnetic spectrum.

Which wavelength of the Sun is used finally as electric energy?

Radio waves.
Nfrared waves.
Visible light.
Microwaves.

Which of the following electromagnetic radiations have the longest wavelength?

X-rays.
$\gamma-\text{rays}.$
Microwaves.
Radiowaves.

Which one of the following is not electromagnetic in nature?

X-rays.
Gamma rays.
Cathode rays.
Infrared rays.

Which of the following has minimum wavelength?

X-rays.
Ultraviolet rays.
$\gamma-\text{rays}.$
Cosmic rays.

The decreasing order of wavelength of infrared, microwave, ultraviolet and gamma rays is:

Microwave, infrared, ultraviolet, gamma rays.
Gamma rays, ultraviolet, infrared, microwave.
Microwave, gamma rays, infrared, ultraviolet.
Infrared, microwave, ultraviolet, gamma rays.

Answer

(b) Nfrared waves.

Explanation:

Infrared rays can be converted into electric energy as in solar cell.

(d) Radiowaves.

Explanation:

Radiowaves have longest wavelength.

Explanation:

Cathode rays are invisible fast moving streams ofelectrons emitted by the cathode of a discharge tube which is maintained at a pressure of about 0.01 mm of mercury.

Explanation:

$\gamma-\text{rays}$ have minimum wavelength.

(a) Microwave, infrared, ultraviolet, gamma rays.

Explanation:

$\lambda_\text{micro}>\lambda_\text{infra}>\lambda_\text{ultra}>\lambda_\text{gamma}$

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Question 104 Marks

An electromagnetic wave transports linear momentum as it travels through space. If an electromagnetic wave transfers a total energy U to a surface in time t, then total linear momentum delivered to the surface is $\text{p}=\frac{\text{U}}{\text{c}}.$ When an electromagnetic wave falls on a surface, it exerts pressure on the surface. ln 1903, the American scientists Nichols and Hull succeeded in measuring radiation pressures of visible light where other had failed, by making a detailed empirical analysis of the ubiquitous gas heating and ballistic effects.

The pressure exerted by an electromagnetic wave of intensity I $(Wm^{-2})$ on a non-reflecting surface is (c is the velocity of light).

$\text{Ic}$
$\text{Ic}^2$
$\frac{\text{I}}{\text{c}}$
$\frac{\text{I}}{\text{c}^2}$

Light with an energy flux of $18\frac{\text{W}}{\text{cm}^2}$ falls on a non-reflecting surface at normal incidence. The pressure exerted on the surface is:

$3\frac{\text{N}}{\text{m}^2}$
$2\times10^{-4}\frac{\text{N}}{\text{m}^2}$
$6\frac{\text{N}}{\text{m}^2}$
$6\times10^{-4}\frac{\text{N}}{\text{m}^2}$

Radiation of intensity $0.5Wm^{-2}$ are striking a metal plate. The pressure on the plate is:

$0.166 \times 10^{-8}Nm^{-2}$
$0.212 \times 10^{-8}Nm^{-2}$
$0.132 \times 10^{-8}Nm^{-2}$
$0.083 \times 10^{-8}Nm^{-2}$

A point source of electromagnetic radiation has an average power out-put of 1500W. The maximum value of electric field at a distance of 3m from this source $($in $Vm^{-1})$ is:

$500$
$100$
$\frac{500}{3}$
$\frac{250}{3}$

The radiation pressure of the visible light is of the order of,

$10^{-2}\frac{\text{N}}{\text{m}^2}$
$10^{-4}\frac{\text{N}}{\text{m}}$
$10^{-6}\frac{\text{N}}{\text{m}^2}$
$10^{-8}\text{N}$

Answer

Explanation:
Pressure exerted by an electromagnetic radiation, $\text{P}=\frac{\text{I}}{\text{c}}.$

(d) $6\times10^{-4}\frac{\text{N}}{\text{m}^2}$

Explanation:
$\text{p}_\text{rad}=\frac{\text{Energy flux}}{\text{Speed of light}}=\frac{18\frac{\text{W}}{\text{cm}^2}}{3\times8^8\frac{\text{m}}{\text{s}}}$
$=\frac{18\times10^4\frac{\text{W}}{\text{cm}^2}}{3\times8^8\frac{\text{m}}{\text{s}}}=6\times10^{-4}\frac{\text{N}}{\text{m}^2}$

(a) $0.166 \times 10^{-8}Nm^{-2}$

Explanation:
$\text{p}=\frac{\text{I}}{\text{c}}=\frac{0.5}{3\times10^8}=0.166\times10^{-8}\text{Nm}^{-2}$

(b) $100$

Explanation:
Intensity of EM wave is given by $\text{I}=\frac{\text{P}}{4\pi\text{R}^2}$
$\text{V}_\text{av}=\frac{1}{2}\in_0\text{E}_0^2\times\text{c}$
$\Rightarrow\text{E}_0=\sqrt{\frac{\text{P}}{2\pi\text{R}^2\in_0\text{c}}}=\sqrt{\frac{1500}{2\times3.14(3)^2\times8.85\times10^{-12}\times3\times18^8}}$
$=\sqrt{10,000}=100\text{Vm}^{-1}$

Explanation:
The radiation pressure of visible light $=7\times10^{-6}\frac{\text{N}}{\text{m}^2}$

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Physics Case study (4 Marks)

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