
Chapter 40 – Electromagnetic Waves
Page No 338:
Question 1:
In a microwave oven, the food is kept in a plastic container and the microwaves 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.
Question 2:
A metal rod is placed along the axis of a solenoid carrying a highfrequency 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 highfrequency 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 highfrequency alternating current in the solenoid. Thus, the rod gets heated up due to the flow of charge in it.
Question 3:
Can an electromagnetic wave be deflected by an electric field or a magnetic field?
Answer:
No, an electromagnetic wave cannot be deflected by an electric field or a magnetic field. This is because according to Maxwell’s theory, an electromagnetic wave does not interact with the static electric field and magnetic field. Even if we consider the particle nature of the wave, the photon is electrically neutral. So, it is not affected by the static magnetic and electric fields.
Question 4:
A wire carries an alternating current i = i_{0} sin ωt. Is there an electric field in the vicinity of the wire?
Answer:
When an alternating current passes through a conductor, the changing magnetic field create a changing electric field outside it. An electromagnetic field is radiated from the surface of the conductor. There is a timevarying electric field outside the conductor. Hence, there is a timevarying electric field in the vicinity of the wire.
Question 5:
A capacitor is connected to an alternatingcurrent source. Is there a magnetic field between the plates?
Answer:
When an alternatingcurrent source is connected to a capacitor, the electric field between the plates of the capacitor keeps on changing with the applied voltage. Due to the changing electric field, a magnetic field exists in between the plates of the capacitor.
Question 6:
Can an electromagnetic wave be polarised?
Answer:
An electromagnetic wave is a transverse wave; thus, it can be polarised. An unpolarised wave consists of many independent waves, whose planes of vibrations of electric and magnetic fields are randomly oriented. They are polarised by restricting the vibrations of the electric field vector or magnetic field vector in one direction only.
Question 7:
A plane electromagnetic wave is passing through a region. Consider (a) electric field (b) magnetic field (c) electrical energy in a small volume and (d) magnetic energy in a small volume. Construct the pairs of the quantities that oscillate with equal frequencies.
Answer:
Let the electromagnetic wave be propagating in the zdirection. The vibrations of the electric and magnetic fields are given by:
E_{x}= E_{0} sin (kz – ωt)
B_{y}= B_{0} sin (kz – ωt)
Let the volume of the region be V.
The angular frequency of the vibrations of the electric and magnetic fields are same and are equal to ω. Therefore, their frequency,
f=ω2π, is same.
The electrical energy in the region,
U_{E} =
12∈0E2×VIt can be written as:
UE=12∈0(E02sin2(kzωt)×VUE=12∈0E02×1cos2(kzωt)2×VUE=14∈0E02×1cos2(kzωt)×VThe magnetic energy in the region,
UB=B22μ0×VUB=B02sin2(kzωt)2μ0×V⇒UB=B021cos(2kz2ωt)4μ0×VThe angular frequency of the electric and magnetic energies is same and is equal to 2ω.
Therefore, their frequency,
f’=2ω2π=2f, will be same.
Thus, the electric and magnetic fields have same frequencies and the electrical and magnetic energies will have same frequencies.
Question 1:
A magnetic field can be produced by
(a) a moving charge
(b) a changing electric field
(c) Neither of them
(d) Both of them
Answer:
(d) Both of them
According to AmpereMaxwell’s Law, a magnetic field is produced due to the conduction current in a conductor and the displacement current. The conduction current is actually the motion of the charge. The displacement current is due to the changing electric field. The displacement current is given by
id=ε0dΦEdt (
∵Ï•_{E} is the electric flux).
Thus, the magnetic field is produced by the moving charge as well as the changing electric field.
Question 2:
A compass needle is placed in the gap of a parallel plate capacitor. The capacitor is connected to a battery through a resistance. The compass needle
(a) does not deflect
(b) deflects for a very short time and then comes back to the original position
(c) deflects and remains deflected as long as the battery is connected
(d) deflects and gradually comes to the original position in a time that is large compared to the time constant
Answer:
(d) deflects and gradually comes to the original position in a time that is large compared to the time constant
The compass needle deflects due to the presence of the magnetic field. Inside the capacitor, a magnetic field is produced when there is a changing electric field inside it. As the capacitor is connected across the battery, the charge on its plates at a certain time t is given by:
Q = CV (1
–e
t/RC),
where
Q = charge developed on the plates of the capacitor
R = resistance of the resistor connected in series with the capacitor
â€‹C = capacitance of the capacitor
V = potential difference of the battery
The time constant of the capacitor is given, τ = RC
The capacitor keeps on charging up to the time τ. The development of charge on the plates will be gradual after t = RC. The change in electric field will be up to the time the charge is developing on the plates of the capacitor. Thus, the compass needle â€‹deflects and gradually comes to the original position in a time that is large compared to the time constant.
Question 3:
Dimensions of 1/(µ_{0}Ïµ_{0}) is
(a) L/T
(b) T/L
(c) L^{2}/T^{2}
(d) T^{2}/L^{2}.
Answer:
(c) L^{2}/T^{2}
The speed of light,
C =1μ0∈0.
The dimensions of
1μ0∈0 are of velocity, i.e. L/T .
â€‹Therefore,
1∈0μ0 will have dimensions L^{2}/T^{2}â€‹.
Question 4:
Electromagnetic waves are produced by
(a) a static charge
(b) a moving charge
(c) an accelerating charge
(d) chargeless particles
Answer:
(c) an accelerating charge
A static charge produces an electrostatic field. A moving charge produces a magnetic field. Electromagnetic waves are produced by an accelerating charge.
Question 5:
An electromagnetic wave going through vacuum is described by
E = E_{0} sin (kx − ωt); B = B_{0} sin (kx − ωt).
Then,
(a) E_{0} k = B_{0} ω
(b) E_{0} B_{0} = ωk
(c) E_{0} ω = B_{0} k
(d) None of these
Answer:
(a) E_{0} k = B_{0} ω
The relation between E_{0} and B_{0}_{ }is given by
E0B0=c …(1)
Here, c = speed of the electromagnetic wave
The relation between â€‹ω (the angular frequency) and k (wave number):
ωk=c …(2)
Therefore, from (1) and (2), we get:
E0B0= ωk=c
⇒E0k=B0ω
Page No 339:
Question 6:
An electric field
E →and a magnetic field
B →exist in a region. The fields are not perpendicular to each other.
(a) This is not possible.
(b) No electromagnetic wave is passing through the region.
(c) An electromagnetic wave may be passing through the region.
(d) An electromagnetic wave is certainly passing through the region.
Answer:
(c) An electromagnetic wave may be passing through the region.
For an electromagnetic wave,electric field ,magnetic field and direction of propagation are mutually perpendicular to each other.We can have a region in which electric and magnetic fields are applied at an angle with each other.In transmission lines Different modes exist. In transverse electric (TE) modeno electric field exist in the direction of propagation. These are sometimes called H modes because there is only a magnetic field along the direction of propagation (H is the conventional symbol for magnetic field).
Question 7:
Consider the following two statements regarding a linearly polarised, plane electromagnetic wave:
(A) The electric field and the magnetic field have equal average values.
(B) The electric energy and the magnetic energy have equal average values.
(a) Both A and B are true.
(b) A is false but B is true.
(c) B is false but A is true
(d) Both A and B are false.
Answer:
(a) Both A and B are true.
For a linearly polarised, plane electromagnetic wave
E = E0sinω(txc)B = B0sinω(txc)The average value of either E or B over a cycle is zero ( average of sin(
θ) over a cycle is zero).
Also the electric energy density (u_{E}) and magnetic energy density (u_{B})are equal.
uE = 12∈0E2 = B22μ0 = uBEnergy can be found out by integrating energy density over the entire volume of full space.
As the energy of the electromagnetic wave is equally shared between electric and magnetic field so their average values will also be equal.
Question 8:
A free electron is placed in the path of a plane electromagnetic wave. The electron will start moving
(a) along the electric field
(b) along the magnetic field
(c) along the direction of propagation of the wave
(d) in a plane containing the magnetic field and the direction of propagation
Answer:
(a) along the electric field
As the electron is at rest initially, only the electric field will exert force on it. There will be no magnetic force on the electron in the stating. Hence, the electron will start moving along the electric field.
Question 9:
A plane electromagnetic wave is incident on a material surface. The wave delivers momentum p and energy E.
(a) p = 0, E ≠ 0
(b) p ≠ 0, E = 0
(c) p ≠ 0, E ≠ 0
(d) p = 0, E = 0
Answer:
(c) p ≠ 0, E ≠ 0.
When an electromagnetic wave strikes a material surface, it transports the momentum, as well as the energy, to the surface. The striking electromagnetic wave exerts pressure on the surface. The total energy transferred to the surface by the electromagnetic wave is given by E = pc. Therefore, â€‹p ≠ 0, E ≠ 0.
Question 1:
An electromagnetic wave going through vacuum is described by
E = E_{0} sin (kx − ωt).
Which of the following is/are independent of the wavelength?
(a) k
(b) ω
(c) k/ω
(d) kω
Answer:
(c) k/ω
The given quantities can be expressed as:
k is given by
k=2πλω is given by
ω=2πνc=νλ⇒ω=2πcλk/ω is given by
kω=2π/λ2πc/λ=1ckω is given by
k×ω=2πλ×2πcλ=4π2cλ2Thus, â€‹k/ω is independent of the wavelength.
Question 2:
Displacement current goes through the gap between the plates of a capacitor when the charge of the capacitor
(a) increases
(b) decreases
(c) does not change
(d) is zero
Answer:
(a) increases
(b) decreases
Displacement current inside a capacitor,
id=ε0dϕEdt, where
Ï•_{E}_{ }is the electric flux inside the capacitor.
Up to the time the electric flux changes, there will be a displacement current. This is possible when the charge on a capacitor changes. Therefore, the displacement current goes through the gap between the plates of a capacitor when the charge of the capacitor or electric field increases or decreases inside the capacitor.
Question 3:
Speed of electromagnetic waves is the same
(a) for all wavelengths
(b) in all media
(c) for all intensities
(d) for all frequencies
Answer:
(c) for all intensities
For any given medium, the speed (c) of an electromagnetic wave is given by
c = νλ,
where
ν = â€‹frequency of the electromagnetic wave
λ = â€‹wavelength of the electromagnetic wave
As the frequency and wavelength are changed, the speed of the electromagnetic wave changes. So, the speed of an electromagnetic wave is not same for â€‹all wavelengths andâ€‹ all frequencies in any medium. The velocity of an electromagnetic wave changes with change in medium. Also, the speed of an electromagnetic wave is same for all the intensities in any medium.
Question 4:
Which of the following have zero average value in a plane electromagnetic wave?
(a) Electric field
(b) Magnetic field
(c) Electric energy
(d) Magnetic energy
Answer:
(a) Electric field
(b) Magnetic field
In a plane electromagnetic wave, the electric and the magnetic fields oscillate sinusoidally. For an electromagnetic wave propagating in the zdirection, the electric and magnetic fields are given by:
E_{x} = E_{0} sin (kz – ωt)
B_{y} = B_{0} sin (kz – ωt)
These are sinusoidal functions. Therefore, for a fixed value of z, the average value of the electric and magnetic fields are zero.
Question 5:
The energy contained in a small volume through which an electromagnetic wave is passing oscillates with
(a) zero frequency
(b) the frequency of the wave
(c) half the frequency of the wave
(d) double the frequency of the wave
Answer:
(d) double the frequency of the wave
The energy per unit volume of an electromagnetic wave,
u=12∈0E2+B22μ0The energy of the given volume can be calculated by multiplying the volume with the above expression.
U=u×V=12∈0E2+B22μ0×V …(1)
Let the direction of propagation of the electromagnetic wave be along the zaxis. Then, the electric and magnetic fields at a particular point are given by
E_{x}= E_{0} sin (kz – ωt)
B_{y}= B_{0} sin (kz – ωt)
Substituting the values of electric and magnetic fields in (1), we get:
U=12∈0(E02sin2(kzωt)+B02sin2(kzωt)2μ0×V⇒U=∈0E02(1cos(2kz2ωt))4+B02(1cos(2kz2ωt))4μ0×VFrom the above expression, it can be easily understood that the energy of the electric and magnetic fields have angular frequency 2ω. Thus, the frequency of the energy of the electromagnetic wave will also be double.
Question 1:
Show that the dimensions of the displacement current
ε0dφEdtare that of an electric current.
Answer:
Displacement current,
ID=∈0dφedtElectric flux,
ϕe =EA
[ϕe] = [E][A]=[14π∈0qr2][A]Also, [∈0] = [M1L3T4A2]⇒[ϕe] = [M1L3T4A2][AT][L2][L2]=[ML3T3A1]Displacement current,
ID= [∈0] [ϕe] [T1]ID= [M1L3T4A2][ML3T3A1][T1]ID= [A][I_{D}]=[current]
Question 2:
A point charge is moving along a straight line with a constant velocity v. Consider a small area A perpendicular to the motion of the charge. Calculate the displacement current through the area when its distance from the charge is x. The value of x is not large, so that the electric field at any instant is essentially given by Coulomb’s law.
Figure
Answer:
From Coulomb’s law:
Electric field strength,
E= kqx2Electric flux,
ϕE=EAϕE=kqAx2Displacement current = I_{d}
Id = ∈0dϕEdt Id =∈0ddtkqAx2 Id=∈0kqAddtx2 Id= ∈014π∈0×q×A×(2)x3×dxdt Id=qAv2πx3
Question 3:
A parallelplate capacitor of platearea A and plate separation d is joined to a battery of emf ε and internal resistance R at t = 0. Consider a plane surface of area A/2, parallel to the plates and situated symmetrically between them. Find the displacement current through this surface as a function of time.
Answer:
Electric field strength for a parallel plate capacitor,
E=Q∈0AElectric flux linked with the area,
ϕE=EA=Q∈0A×A2=Q2∈0Displacement current,
Id=∈0dϕEdt = ∈0ddtQ∈02Id=12dQdt …(i)Charge on the capacitor as a function of time during charging,
Q=εC1et/RCPutting this in equation (i), we get:
Id=12εCddt1et/RCId=12εCet/RC×1RC C = A∈0d⇒Id=ε2R×etdε0AR
Question 4:
Consider the situation of the previous problem. Define displacement resistance R_{d} = V/i_{d} of the space between the plates, where V is the potential difference between the plates and i_{d} is the displacement current. Show that R_{d} varies with time as
Rd=R(et/τ1).
Answer:
Electric field strength for a parallel plate capacitor =
E=Q∈0A
Electric flux, ϕ=E.A=Q∈0A.A=Q∈0Displacement current, id=∈0dϕEdt = ∈0ddtQ∈0id=dQdtAlso, Q = CVid=ddtE0Cet/RCid=E0C1RCet/RCDisplacement resistance, Rd = E0id R⇒Rd=E0E0Ret/RCR⇒Rd=Ret/RCR⇒Rd=R(et/RC1)
Question 5:
Using B = µ_{0} H, find the ratio E_{0}/H_{0} for a plane electromagnetic wave propagating through vacuum. Show that it has the dimensions of electric resistance. This ratio is a universal constant called the impedance of free space.
Answer:
Given, B = µ_{0}H
For vacuum we can rewrite this equation as,
B_{0} = µ_{0}H_{0} …(i)
Relation between magnetic field and electric field for vacuum is given as,
B_{0} = µ_{0}
∈0cE_{0} …(ii)
From equation (i) by (ii),
μ0H0=μ0∈0cE0⇒E0H0=1∈0c⇒E0H0=18.85×1012×3×108⇒E0H0≈377 Ω
Dimension of 1∈0c=1[LT1][M1L3T4A2]=1M1L2T3A2=M4L2T3A2=[R]
Question 6:
The sunlight reaching Earth has maximum electric field of 810 Vm^{−1}. What is the maximum magnetic field in this light?
Answer:
Given:
Electric field amplitude, E_{0} = 810 V/m
Maximum value of magnetic field = Magnetic field amplitude = B_{0} = ?
We know:
Speed of a wave =
EBFor electromagnetic waves, speed = speed of light
B_{0} = µ_{0} ε_{0} cE_{0}
Putting the values in the above relation, we get:
B0= 4π×107×8.85×1012×3×108×810B0= 4×3.14×8.85×3×81×1010B0= 27010.9×1010B0= 2.7×106 T=2.7 μT
Question 7:
The magnetic field in a plane electromagnetic wave is given by
B = (200 µT) sin [(4.0 × 10^{15}s^{−1})(t−x/c)].
Find the maximum electric field and the average energy density corresponding to the electric field.
Answer:
Maximum value of a magnetic field, B_{0} = 200
μT
The speed of an electromagnetic wave is c.
So, maximum value of electric field,
E0=cB0
E0=c×B0=200×106×3×108E0=6×104 NC1(b) Average energy density of a magnetic field,
Uav=12μ0B02=(200×106)22×4π×107Uav=4×1088π×107=120πUav=0.0159≈0.016 J/m3For an electromagnetic wave, energy is shared equally between the electric and magnetic fields.
Hence, energy density of the electric field will be equal to the energy density of the magnetic field.
Question 8:
A laser beam has intensity 2.5 × 10^{14} W m^{−2}. Find amplitudes of electric and magnetic fields in the beam.
Answer:
Given:
Intensity, I = 2.5 × 10^{14} W/m^{2}
We know:
I=12∈0E02c
⇒E02=2I∈0c⇒ E0=2I∈0c⇒E0=2×2.5×10148.85×1012×3×108⇒E0=0.4339×109⇒E0=4.33×108 N/CMaximum value of magnetic field,
B0=E0cB0=4.33×1083×108⇒B0=1.43 T
Question 9:
The intensity of the sunlight reaching Earth is 1380 W m^{−2}. Assume this light to be a plane, monochromatic wave. Find the amplitudes of electric and magnetic fields in this wave.
Answer:
Given:
I =1380 W/m2I=12∈0E02c⇒E0=2I∈0c⇒ E0=2×13808.83×1012×3×108⇒E0=103.95×104⇒E0=10.195×102⇒E0=1.02×103 N/CSince E0=B0c, B0=E0c=1.02×1023×1028⇒B0=3.398×106⇒B0=3.4×106 T
HC Verma Solutions for Class 11 Physics – Part 1
 Chapter 1 – Introduction to Physics
 Chapter 2 – Physics and Mathematics
 Chapter 3 – Rest and Motion: Kinematics
 Chapter 4 – The Forces
 Chapter 5 – Newton’s Laws of Motion
 Chapter 6 – Friction
 Chapter 7 – Circular Motion
 Chapter 8 – Work and Energy
 Chapter 9 – Center of Mass, Linear Momentum, Collision
 Chapter 10 – Rotational Mechanics
 Chapter 11 – Gravitation
 Chapter 12 – Simple Harmonic Motion
 Chapter 13 – Fluid Mechanics
 Chapter 14 – Some Mechanical Properties of Matter
 Chapter 15 – Wave Motion and Wave on a String
 Chapter 16 – Sound Wave
 Chapter 17 – Light Waves
 Chapter 18 – Geometrical Optics
 Chapter 19 – Optical Instruments
 Chapter 20 – Dispersion and Spectra
 Chapter 21 – Speed of Light
 Chapter 22 – Photometry
HC Verma Solutions for Class 12 Physics – Part 2
 Chapter 23 – Heat and Temperature
 Chapter 24 – Kinetic Theory of Gases
 Chapter 25 – Calorimetry
 Chapter 26 – Laws of Thermodynamics
 Chapter 27 – Specific Heat Capacities of Gases
 Chapter 28 – Heat Transfer
 Chapter 29 – Electric Field and Potential
 Chapter 30 – Gauss’s Law
 Chapter 31 – Capacitors
 Chapter 32 – Electric Current in Conductors
 Chapter 33 – Thermal and Chemical Effects of Electric Current
 Chapter 34 – Magnetic Field
 Chapter 35 – Magnetic Field due to a Current
 Chapter 36 – Permanent Magnets
 Chapter 37 – Magnetic Properties of Matter
 Chapter 38 – Electromagnetic Induction
 Chapter 39 – Alternating Current
 Chapter 40 – Electromagnetic Waves
 Chapter 41 – Electric Current through Gases
 Chapter 42 – Photoelectric Effect and Wave Particle Duality
 Chapter 43 – Bohr’s Model and Physics of the Atom
 Chapter 44 – Xrays
 Chapter 45 – Semiconductors and Semiconductor Devices
 Chapter 46 – The Nucleus
 Chapter 47 – The Special Theory of Relativity