Ans: 1
Radius of each circular plate, r= 12 cm = 0.12 m
Distance between the plates, d= 5 cm = 0.05 m
Charging current, I= 0.15 A
Permittivity of free space,
= 8.85 × 10−12C2N−1m−2
(a) Capacitance between the two plates is given by the relation,
C
Where,
A= Area of each plate
Charge on each plate, q = CV
Where,
V = Potential difference across the plates
Differentiation on both sides with respect to time (t) gives:
Therefore, the change in potential difference between the plates is 1.87 ×109V/s.
(b) The displacement current across the plates is the same as the conduction current. Hence, the displacement current, idis 0.15 A.
(c) Yes
Kirchhoff’s first rule is valid at each plate of the capacitor provided that we take the sum of conduction and displacement for current.
Ans: 2
Radius of each circular plate, R= 6.0 cm = 0.06 m
Capacitance of a parallel plate capacitor, C= 100 pF = 100 × 10−12F
Supply voltage, V= 230 V
Angular frequency, ω= 300 rad s−1
(a) Rms value of conduction current, I
Where,
XC= Capacitive reactance
∴I = V × ωC
= 230 × 300 × 100 × 10−12
= 6.9 × 10−6 A
= 6.9 μA
Hence, the rms value of conduction current is 6.9 μA.
(b) Yes, conduction current is equal to displacement current.
(c) Magnetic field is given as:
B
Where,
μ0= Free space permeability
I0= Maximum value of current =
r= Distance between the plates from the axis = 3.0 cm = 0.03 m
∴B
= 1.63 × 10−11 T
Hence, the magnetic field at that point is 1.63 × 10−11T.
Ans: 3
The speed of light (3 × 108m/s) in a vacuum is the same for all wavelengths. It is independent of the wavelength in the vacuum.
Ans: 4
The electromagnetic wave travels in a vacuum along the z-direction. The electric field (E) and the magnetic field (H) are in the x-yplane. They are mutually perpendicular.
Frequency of the wave, ν= 30 MHz = 30 × 106s−1
Speed of light in a vacuum, c = 3 × 108 m/s
Wavelength of a wave is given as:
Ans: 5
A radio can tune to minimum frequency, ν1= 7.5 MHz= 7.5 × 106Hz
Maximum frequency, ν2= 12 MHz = 12 × 106Hz
Speed of light, c= 3 × 108m/s
Corresponding wavelength for ν1can be calculated as:
Corresponding wavelength for ν2can be calculated as:
Thus, the wavelength band of the radio is 40 m to 25 m.
Ans: 6
The frequency of an electromagnetic wave produced by the oscillator is the same as that of a charged particle oscillating about its mean position i.e., 109Hz.
Ans: 7
Amplitude of magnetic field of an electromagnetic wave in a vacuum,
B0= 510 nT = 510 × 10−9T
Speed of light in a vacuum, c = 3 × 108 m/s
Amplitude of electric field of the electromagnetic wave is given by the relation,
E = cB0
= 3 × 108 × 510 × 10−9 = 153 N/C
Therefore, the electric field part of the wave is 153 N/C.
Ans: 8
Electric field amplitude, E0 = 120 N/C
Frequency of source, ν= 50.0 MHz = 50 × 106 Hz
Speed of light, c= 3 × 108 m/s
(a) Magnitude of magnetic field strength is given as:
Angular frequency of source is given as:
ω= 2πν
= 2π × 50 × 106
= 3.14 × 108 rad/s
Propagation constant is given as:
Wavelength of wave is given as:
(b) Suppose the wave is propagating in the positive x direction. Then, the electric field vector will be in the positive y direction and the magnetic field vector will be in the positive zdirection. This is because all three vectors are mutually perpendicular.
Equation of electric field vector is given as:
And, magnetic field vector is given as:
Ans: 9
Energy of a photon is given as:
Where,
h = Planck’s constant = 6.6 × 10−34 Js
c = Speed of light = 3 × 108 m/s
λ = Wavelength of radiation
The given table lists the photon energies for different parts of an electromagnetic spectrum for differentλ.
Ans: 10
Frequency of the electromagnetic wave, ν= 2.0 × 1010 Hz
Electric field amplitude, E0 = 48 V m−1
Speed of light, c= 3 × 108 m/s
(a) Wavelength of a wave is given as:
(b) Magnetic field strength is given as:
(c) Energy density of the electric field is given as:
And, energy density of the magnetic field is given as:
Where,
∈0= Permittivity of free space
μ0= Permeability of free space
We have the relation connecting E and B as:
E= cB… (1)
Where,
…(2)
Putting equation (2) in equation (1), we get
Squaring both sides, we get
Ans: 11
(a) From the given electric field vector, it can be inferred that the electric field is directed along the negative xdirection. Hence, the direction of motion is along the negative ydirection i.e.,
.
(b) It is given that,
The general equation for the electric field vector in the positive xdirection can be written as:
On comparing equations (1) and (2), we get
Electric field amplitude, E0 = 3.1 N/C
Angular frequency, ω = 5.4 × 108 rad/s
Wave number, k = 1.8 rad/m
Wavelength,
= 3.490 m
(c) Frequency of wave is given as:
(d) Magnetic field strength is given as:
Where,
c= Speed of light = 3 × 108 m/s
(e) On observing the given vector field, it can be observed that the magnetic field vector is directed along the negative z direction. Hence, the general equation for the magnetic field vector is written as:
Ans: 12
Power rating of bulb, P= 100 W
It is given that about 5% of its power is converted into visible radiation.
Power of visible radiation,
Hence, the power of visible radiation is 5W.
(a) Distance of a point from the bulb, d = 1 m
Hence, intensity of radiation at that point is given as:
(b) Distance of a point from the bulb, d1 = 10 m
Hence, intensity of radiation at that point is given as:
Ans: 13
A body at a particular temperature produces a continous spectrum of wavelengths. In case of a black body, the wavelength corresponding to maximum intensity of radiation is given according to Planck’s law. It can be given by the relation,
Where,
λm= maximum wavelength
T = temperature
Thus, the temperature for different wavelengths can be obtained as:
For λm= 10−4 cm;
For λm= 5 ×10−5 cm;
For λm= 10−6 cm;
and so on.
The numbers obtained tell us that temperature ranges are required for obtaining radiations in different parts of an electromagnetic spectrum. As the wavelength decreases, the corresponding temperature increases.
Ans: 14
(a) Radio waves; it belongs to the short wavelength end of the electromagnetic spectrum.
(b) Radio waves; it belongs to the short wavelength end.
(c) Temperature, T = 2.7 °K
λmis given by Planck’s law as:
This wavelength corresponds to microwaves.
(d) This is the yellow light of the visible spectrum.
(e) Transition energy is given by the relation,
E= hν
Where,
h= Planck’s constant = 6.6 × 10−34Js
ν= Frequency of radiation
Energy,E = 14.4 K eV
This corresponds to X-rays.
Ans: 15
(a) Long distance radio broadcasts use shortwave bands because only these bands can be refracted by the ionosphere.
(b) It is necessary to use satellites for long distance TV transmissions because television signals are of high frequencies and high energies. Thus, these signals are not reflected by the ionosphere. Hence, satellites are helpful in reflecting TV signals. Also, they help in long distance TV transmissions.
(c) With reference to X-ray astronomy, X-rays are absorbed by the atmosphere. However, visible and radio waves can penetrate it. Hence, optical and radio telescopes are built on the ground, while X-ray astronomy is possible only with the help of satellites orbiting the Earth.
(d) The small ozone layer on the top of the atmosphere is crucial for human survival because it absorbs harmful ultraviolet radiations present in sunlight and prevents it from reaching the Earth’s surface.
(e) In theabsenceof an atmosphere, there would be no greenhouse effect on the surface of the Earth. As a result, the temperature of the Earth would decrease rapidly, making it chilly and difficult for human survival.
(f) A global nuclear war on the surface of the Earth would have disastrous consequences. Post-nuclear war, the Earth will experience severe winter as the war will produce clouds of smoke that would cover maximum parts of the sky, thereby preventing solar light form reaching the atmosphere. Also, it will lead to the depletion of the ozone layer.
Radius of each circular plate, r= 12 cm = 0.12 m
Distance between the plates, d= 5 cm = 0.05 m
Charging current, I= 0.15 A
Permittivity of free space,
= 8.85 × 10−12C2N−1m−2(a) Capacitance between the two plates is given by the relation,
C
Where,
A= Area of each plate
Charge on each plate, q = CV
Where,
V = Potential difference across the plates
Differentiation on both sides with respect to time (t) gives:
Therefore, the change in potential difference between the plates is 1.87 ×109V/s.
(b) The displacement current across the plates is the same as the conduction current. Hence, the displacement current, idis 0.15 A.
(c) Yes
Kirchhoff’s first rule is valid at each plate of the capacitor provided that we take the sum of conduction and displacement for current.
Ans: 2
Radius of each circular plate, R= 6.0 cm = 0.06 m
Capacitance of a parallel plate capacitor, C= 100 pF = 100 × 10−12F
Supply voltage, V= 230 V
Angular frequency, ω= 300 rad s−1
(a) Rms value of conduction current, I
Where,
XC= Capacitive reactance
∴I = V × ωC
= 230 × 300 × 100 × 10−12
= 6.9 × 10−6 A
= 6.9 μA
Hence, the rms value of conduction current is 6.9 μA.
(b) Yes, conduction current is equal to displacement current.
(c) Magnetic field is given as:
B
Where,
μ0= Free space permeability
I0= Maximum value of current =
r= Distance between the plates from the axis = 3.0 cm = 0.03 m
∴B
= 1.63 × 10−11 T
Hence, the magnetic field at that point is 1.63 × 10−11T.
Ans: 3
The speed of light (3 × 108m/s) in a vacuum is the same for all wavelengths. It is independent of the wavelength in the vacuum.
Ans: 4
The electromagnetic wave travels in a vacuum along the z-direction. The electric field (E) and the magnetic field (H) are in the x-yplane. They are mutually perpendicular.
Frequency of the wave, ν= 30 MHz = 30 × 106s−1
Speed of light in a vacuum, c = 3 × 108 m/s
Wavelength of a wave is given as:
Ans: 5
A radio can tune to minimum frequency, ν1= 7.5 MHz= 7.5 × 106Hz
Maximum frequency, ν2= 12 MHz = 12 × 106Hz
Speed of light, c= 3 × 108m/s
Corresponding wavelength for ν1can be calculated as:
Corresponding wavelength for ν2can be calculated as:
Thus, the wavelength band of the radio is 40 m to 25 m.
Ans: 6
The frequency of an electromagnetic wave produced by the oscillator is the same as that of a charged particle oscillating about its mean position i.e., 109Hz.
Ans: 7
Amplitude of magnetic field of an electromagnetic wave in a vacuum,
B0= 510 nT = 510 × 10−9T
Speed of light in a vacuum, c = 3 × 108 m/s
Amplitude of electric field of the electromagnetic wave is given by the relation,
E = cB0
= 3 × 108 × 510 × 10−9 = 153 N/C
Therefore, the electric field part of the wave is 153 N/C.
Ans: 8
Electric field amplitude, E0 = 120 N/C
Frequency of source, ν= 50.0 MHz = 50 × 106 Hz
Speed of light, c= 3 × 108 m/s
(a) Magnitude of magnetic field strength is given as:
Angular frequency of source is given as:
ω= 2πν
= 2π × 50 × 106
= 3.14 × 108 rad/s
Propagation constant is given as:
Wavelength of wave is given as:
(b) Suppose the wave is propagating in the positive x direction. Then, the electric field vector will be in the positive y direction and the magnetic field vector will be in the positive zdirection. This is because all three vectors are mutually perpendicular.
Equation of electric field vector is given as:
And, magnetic field vector is given as:
Ans: 9
Energy of a photon is given as:
Where,
h = Planck’s constant = 6.6 × 10−34 Js
c = Speed of light = 3 × 108 m/s
λ = Wavelength of radiation
The given table lists the photon energies for different parts of an electromagnetic spectrum for differentλ.
λ (m) 103 1 10−3 10−6 10−8 10−10 10−12 E (eV) 12.375 × 10−10 12.375 × 10−7 12.375 × 10−4 12.375 × 10−1 12.375 × 101 12.375 × 103 12.375 × 105
Ans: 10
Frequency of the electromagnetic wave, ν= 2.0 × 1010 Hz
Electric field amplitude, E0 = 48 V m−1
Speed of light, c= 3 × 108 m/s
(a) Wavelength of a wave is given as:
(b) Magnetic field strength is given as:
(c) Energy density of the electric field is given as:
And, energy density of the magnetic field is given as:
Where,
∈0= Permittivity of free space
μ0= Permeability of free space
We have the relation connecting E and B as:
E= cB… (1)
Where,
…(2)Putting equation (2) in equation (1), we get
Squaring both sides, we get
Ans: 11
(a) From the given electric field vector, it can be inferred that the electric field is directed along the negative xdirection. Hence, the direction of motion is along the negative ydirection i.e.,
.(b) It is given that,
The general equation for the electric field vector in the positive xdirection can be written as:
On comparing equations (1) and (2), we get
Electric field amplitude, E0 = 3.1 N/C
Angular frequency, ω = 5.4 × 108 rad/s
Wave number, k = 1.8 rad/m
Wavelength,
= 3.490 m(c) Frequency of wave is given as:
(d) Magnetic field strength is given as:
Where,
c= Speed of light = 3 × 108 m/s
(e) On observing the given vector field, it can be observed that the magnetic field vector is directed along the negative z direction. Hence, the general equation for the magnetic field vector is written as:
Ans: 12
Power rating of bulb, P= 100 W
It is given that about 5% of its power is converted into visible radiation.
Power of visible radiation,Hence, the power of visible radiation is 5W.
(a) Distance of a point from the bulb, d = 1 m
Hence, intensity of radiation at that point is given as:
(b) Distance of a point from the bulb, d1 = 10 m
Hence, intensity of radiation at that point is given as:
Ans: 13
A body at a particular temperature produces a continous spectrum of wavelengths. In case of a black body, the wavelength corresponding to maximum intensity of radiation is given according to Planck’s law. It can be given by the relation,
Where,
λm= maximum wavelength
T = temperature
Thus, the temperature for different wavelengths can be obtained as:
For λm= 10−4 cm;
For λm= 5 ×10−5 cm;
For λm= 10−6 cm;
and so on.The numbers obtained tell us that temperature ranges are required for obtaining radiations in different parts of an electromagnetic spectrum. As the wavelength decreases, the corresponding temperature increases.
Ans: 14
(a) Radio waves; it belongs to the short wavelength end of the electromagnetic spectrum.
(b) Radio waves; it belongs to the short wavelength end.
(c) Temperature, T = 2.7 °K
λmis given by Planck’s law as:
This wavelength corresponds to microwaves.
(d) This is the yellow light of the visible spectrum.
(e) Transition energy is given by the relation,
E= hν
Where,
h= Planck’s constant = 6.6 × 10−34Js
ν= Frequency of radiation
Energy,E = 14.4 K eV
This corresponds to X-rays.
Ans: 15
(a) Long distance radio broadcasts use shortwave bands because only these bands can be refracted by the ionosphere.
(b) It is necessary to use satellites for long distance TV transmissions because television signals are of high frequencies and high energies. Thus, these signals are not reflected by the ionosphere. Hence, satellites are helpful in reflecting TV signals. Also, they help in long distance TV transmissions.
(c) With reference to X-ray astronomy, X-rays are absorbed by the atmosphere. However, visible and radio waves can penetrate it. Hence, optical and radio telescopes are built on the ground, while X-ray astronomy is possible only with the help of satellites orbiting the Earth.
(d) The small ozone layer on the top of the atmosphere is crucial for human survival because it absorbs harmful ultraviolet radiations present in sunlight and prevents it from reaching the Earth’s surface.
(e) In theabsenceof an atmosphere, there would be no greenhouse effect on the surface of the Earth. As a result, the temperature of the Earth would decrease rapidly, making it chilly and difficult for human survival.
(f) A global nuclear war on the surface of the Earth would have disastrous consequences. Post-nuclear war, the Earth will experience severe winter as the war will produce clouds of smoke that would cover maximum parts of the sky, thereby preventing solar light form reaching the atmosphere. Also, it will lead to the depletion of the ozone layer.
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