12th Standard Syllabus & Materials
12th Standard
TN 12th Computer Applications தரவுதள மேலாண்மை அமைப்பு - ஓர் அறிமுகம் Sample Question Papers Study Material - QB365 Set A
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Published on: 27/11/2019
Electromagnetic Waves
Download Tamil Nadu 12th Standard Physics question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.
Questions + Answers key
Take MCQ Physics Test1.
If the velocity of light in a medium is 2.25 x 108 ms-1 then the refractive index of the medium will be _______________.
1.5
0.5
1.33
1.73
2.
Which of the following E.M waves has the longest wavelength?
Radio waves
IR
X-rays
visible
3.
Which one of them is used to produce a propagating electromagnetic wave?
an accelerating charge
a charge moving at constant velocity
a stationary charge
an uncharged particle
4.
During the propagation of electromagnetic waves in a medium __________________.
electric energy density is double of the magnetic energy density
electric energy density is half of the magnetic energy density
electric energy density is equal to the magnetic energy density
both electric and magnetic energy densities are zero
5.
The electric and the magnetic fields, associated with an electromagnetic wave, propagating along negative X axis can be represented by _____.
\(\vec { E } ={ E }_{ 0 }\hat { i } \) and \(\vec { B } ={ B }_{ 0 }\hat { k } \)
\(\vec { E } ={ E }_{ 0 }\hat { k } \) and \(\vec { B } ={ B }_{ 0 }\hat { j } \)
\(\vec { E } ={ E }_{ 0 }\hat { i } \) and \(\vec { B } ={ B }_{ 0 }\hat { j } \)
\(\vec { E } ={ E }_{ 0 }\hat { j } \) and \(\vec { B } ={ B }_{ 0 }\hat { i } \)
6.
What is the orgin of displacement currtent?
7.
Which part of Electromagnetic is absorbed from sunlight by ozone layer?
(i) Write its source and
(ii) mention its uses.
8.
Write the uses of Micro waves.
9.
Consider a parallel plate capacitor whose plates are closely spaced. Let R be the radius of the plates and the current in the wire connected to the plates is 5 A, calculate the displacement current through the surface passing between the plates by directly calculating the rate of change of flux of electric field through the surface.
10.
What is meant by Fraunhofer lines?
11.
What are electromagnetic waves?
12.
Identify the following Electromagnetic radiations
(a) 109 Hz
(b) 1011 Hz. Give one application of each.
13.
Identify the Electromagnetic waves whose wavelength vary as
(a) 10-12 m to 10-8 m
(b) 10-4 m and write their uses.
14.
Let an electromagnetic wave propagate along the x-direction, the magnetic field oscillates at a frequency of 1010 Hz and has an amplitude of 10−5 T, acting along the y-direction. Then, compute the wavelength of the wave. Also write down the expression for electric field in this case.
15.
In an electric circuit, there is a capacitor of reactance 100 Ω connected across the source of 220 V, find the displacement current.
16.
The magnetic field amplitude of an Electromagnetic wave is 1.6 x 10-7 T. If the frequency is 30 MHz. determine electric field, any velocity K and λ.
17.
Discuss the source of electromagnetic waves.
18.
19.
Explain the Maxwell’s modification of Ampere’s circuital law.
1.
(c)
1.33
2.
(a)
Radio waves
3.
(a)
an accelerating charge
4.
(c)
electric energy density is equal to the magnetic energy density
5.
\( { E } ={ E }_{ 0 }\hat { k } \) and \({ B } ={ B }_{ 0 }\hat { j } \)
6.
Displacement of does not arise due to motion of charge carries but it arises due to time variation of electric flux.
7.
UV light is absorbed by the ozone layer
(i) Source: Sun, arc and ionized gases.
(ii) Uses: To destroy bacteria, sterilizing the surgical instruments, burglar alarm etc
8.
It is used in radar systems for aircraft navigation, speed of the vehicle, microwave oven for cooking, and very long-distance wireless communication through satellites.
9.
Area of the capacitor = A
Radius = R
Current in the wire connected to the plates I = 5 A
The electric field, between the plates of a parallel plate capacitor,
\(E=\frac{\sigma}{\varepsilon_0} \)
\(E=\frac{Q}{A \varepsilon_0}\)
Q is the charge accumulated at the positive plate.
The flux of this field, \(\phi_E=\frac{Q}{A \varepsilon_0} \times A=\frac{Q}{\varepsilon_0}\)
Displacement current \(i_d=\varepsilon_0 \frac{d \phi_E}{d t}\)
\(=\varepsilon_0 \frac{d}{d t}\left(\frac{Q}{\varepsilon_0}\right)=i_c\)
\(\therefore \mathrm{i}_{\mathrm{d}}=5 \mathrm{~A} \quad\left(\because\right.\) The current through the capacitor ic = 5 A)
Displacement current = 5 A
10.
When the spectrum obtained from the Sun is examined, it consists of large number of dark lines (line absorption spectrum). These dark lines in the solar spectrum are known as Fraunhofer lines.
11.
An electromagnetic waves are the waves that are radiated by an accelerated charge which propagates through space as coupled electric and magnetic fields, oscillating perpendicular to each other and to the direction of propagation of the wave.
12.
(a) 109 Hz - Radiowaves
Application: For radio & television communication.
(b) 1011 Hz - microwaves
Application: used in Radar for aircraft navigation, (microwave oven for cooking).
13.
(a) X - rays are used as diagnostic tool in medicine. X-rays are used extensively in studying structures of inner atomic electron shells and crystal structures. It is used in detecting fractures, diseased organs, formation of bones and stones, observing the progress of healing bones. Further, in a finished metal product, it is used to detect faults, cracks, flaws and holes.
(b) Radio are produced by oscillators in electric circuits. It obeys reflection and diffraction. It is used in radio and communication system and cellphones.
14.
Amplitude of magnetic field B = 10-5 T
Frequency, f = 1010 HZ
(i) Wavelength of the wave,
\({\lambda}=\frac{c}{f} =\frac{3 \times 10^8}{ 10^{10}} \)
= 3 x 108 - 10
Wavelength = 3 x 10-2 m
(ii) Electric field E(x,t)\(\hat i\)
Angular frequency \(\omega =2 \pi f \)
\(\omega=2 \times 3.14 \times 10^{10}=6.28 \times 10^{10} rads^{-1}\)
\(k=\frac{2 \pi}{\lambda} =\frac{2 \times 3.14}{3 \times 10^{-2}} \)
\(=\frac{6.28}{3 \times 10^{-2}}=\frac{628}{3}=2.09 \times 10^{2} \)
k = 2.09 x 102
(iii) Eo = BoC
Eo = 10-5 x 3 x 108
Eo = 3 x 103 V m-1
The required expression for electric field is
\(\vec{E}(x, t)=E_o \sin \left(\frac{2 \pi}{\lambda} x-2 \pi f t\right) \hat{i} N C^{-1} \)
\(\vec{E}(x , t)=3 \times 10^{3} \sin \left(2.09 \times 10^{2} \mathrm{x}-6.28 \times 10^{10} \mathrm{t}\right) \hat(-{k}) N C^{-1} \)
15.
Since displacement current = conduction current
\({ I }_{ d }=\frac { V }{ { X }_{ C } } =\frac { 220 }{ 100 } =2.2A\)
16.
Given: The amplitude of magnetic field of an Electromagnetic wave B = 1.6 x 10-7 T
To find:
The amplitude of electric field of an Electromagnetic wave E = ?
frequency ૪ = 30 Mhz = 30 x 106 Hz.
To find: Angle velocity ω =?
Wavelength of Electromagnetic wave λ = ?
(i) Ampere of electric field E = ?
\(\frac { E }{ B } =C\Rightarrow E=C.B\Rightarrow 3\times { 10 }^{ 8 }\times 1.6\times { 10 }^{ -7 }\)
E = 48Vm-1.
(ii) Angle velocity, ω = 2π૪
ω = 2 x 3.14 x 30 x 106
ω = 1.885 x 108 rad /s.
(iii) Wavelength of Electromagnetic wave, λ = \(\frac{C}{\gamma}\)
\(\gamma=\frac{3\times 10^8}{30\times 10^6}\) = 10m
λ = 10m
17.
(i) Any stationary source charge produces only electric field. When the charge moves with uniform velocity, it produces steady current which gives rise to magnetic field (not time dependent, only space· dependent) around the conductor in which charge flows.
(ii) If the charged particle accelerates, it produces magnetic field in addition to electric field. Both electric and magnetic fields are time varying fields. Since the electromagnetic waves are transverse waves, the direction of propagation of electromagnetic waves is perpendicular to the plane containing electric and magnetic field vectors.
(iii) Any oscillatory motion is also an accelerating motion, so, when the charge oscillates (oscillating molecular dipole) about their mean position as shown in Figure, it produces electromagnetic waves.
(iv) Suppose the electromagnetic field in free space propagates along z-direction, and if the electric field vector points along x-axis then the magnetic field vector will be mutually perpendicular to both electric field and the direction of wave propogation. Thus,
Ex = Eo sin (kz - ωt)
By = Bo sin(Kz - ωt)
Where, Eo and Bo are amplitude of the oscillating electric and magnetic field, k is a wave number, ω is the angular frequency of the wave and \(\hat { k } \) (unit vector, here it is called propagation vector) denotes the direction of propagation of electromagnetic wave.
(vi) Note that both electric field and magnetic field oscillate with a frequency (frequency of electromagnetic wave) which is equal to the frequency of the source (here, oscillating charge is the source for the production of electromagnetic waves). In free space or in vacuum, the ratio between Eo and Bo is equal to the speed of electromagnetic wave, which is equal to speed of light c.
\(c=\frac { { E }_{ 0 } }{ { B }_{ 0 } } \)
In any medium, the ratio of Eo and Bo is equal to the speed of electromagnetic wave in that medium. Thus,
Further, the energy of electromagnetic waves comes from the energy of the oscillating charge.
18.
19.
(i) We have stated Ampere's law as \(\oint \vec{B} \cdot \overrightarrow{d l}=\mu_oi\)
(ii) Where, i is the electric current crossing a surface bounded by a closed curve and the line integral of \(\vec{B}\) is calculated along that closed curve. This equation is valid only when the electric field at the surface does not change with time.
(iii) Maxwell strongly believed that when the time varying magnetic field produces an electric field, the time varying electric field must produce a magnetic field.
(iv) To understand how a varying electric field produces magnetic field, let us consider a situation of charging a parallel plate capacitor.
(v) Let ic be the conduction current. To calculate the magnetic field at P (fig. 1 ) an amperian loop. S1 is drawn. Applying Ampere circuital law for the surface S1, we get
\(\oint \vec{B} \cdot \overrightarrow{d l}=\mu_0 i_c\) Where, \(\mu_0\) is permeability of free space.
(vi) Applying the same for the surface S2, we get \(\oint \vec{B} \cdot \overrightarrow{d l}=0.\)
Because the surface S2 nowhere touches the wire carrying conduction current. Therefore for the point P at one surface (S1) it has some value and at another surface (S2) it has zero value.
(vii) So, Maxwell believed that there must be a current associated with the changing electric field in between the capacitor and he called that current as displacement current.
(viii) Applying Gauss law to the electric flux between the plates of the capacitor \(\phi_E=\oint \vec{E} \cdot \overrightarrow{\mathrm{dA}}=E A=\frac{q}{\varepsilon_0}\) where, A is the area of the plate.
The change in electric flux is \(\frac{d \phi_F}{d t}=\frac{1}{\varepsilon_0} \frac{d q}{d t} (or) \frac{\mathrm{dq}}{\mathrm{dt}}=\mathrm{i}_{\mathrm{d}}=\varepsilon_0 \frac{\mathrm{d} \phi_{\mathrm{E}}}{\mathrm{dt}}\), where id is the displacement current.
(ix) The displacement current can be defined as the current which comes into play in the region in which the electric field and electric flux are changing with time.
(x) So, Maxwell modified Ampere's law \(\oint_{l} \vec{B} \cdot d \vec{l}=\mu_{0} i_c+\mu_{0}-i_d\) which means the total current enclosed by the surface is sum of conduction current and displacement current.
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