HC Verma ii Solutions for Class 12 Science Physics Chapter 18 Electromagnetic Waves are provided here with simple step-by-step explanations. These solutions for Electromagnetic Waves are extremely popular among class 12 Science students for Physics Electromagnetic Waves Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the HC Verma ii Book of class 12 Science Physics Chapter 18 are provided here for you for free. You will also love the ad-free experience on Meritnation’s HC Verma ii Solutions. All HC Verma ii Solutions for class 12 Science Physics are prepared by experts and are 100% accurate.

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.

Page No 338:

Question 2:

A metal rod is placed along the axis of a solenoid carrying a high-frequency 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.

Page No 338:

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.

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Question 4:

A wire carries an alternating current i = i0 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 time-varying electric field outside the conductor. Hence, there is a time-varying electric field in the vicinity of the wire.

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Question 5:

A capacitor is connected to an alternating-current source. Is there a magnetic field between the plates?

Answer:

When an alternating-current 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.

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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.

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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 z-direction. The vibrations of the electric and magnetic fields are given by:
Ex= E0 sin (kz – ωt)
By= B0 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,
UE = 120E2×V
It can be written as:
UE=120(E02sin2(kz-ωt)×VUE=120E02×1-cos2(kz-ωt)2×VUE=140E02×1-cos2(kz-ωt)×V
The magnetic energy in the region,
UB=B22μ0×VUB=B02sin2(kz-ωt)2μ0×VUB=B021-cos(2kz-2ωt)4μ0×V

The 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.

Page No 338:

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 Ampere-Maxwell'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 ( ϕE is the electric flux).
Thus, the magnetic field is produced by the  moving charge as well as the changing electric field.

Page No 338:

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 
​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 ​deflects and gradually comes to the original position in a time that is large compared to the time constant.

Page No 338:

Question 3:

Dimensions of 1/(µ0ϵ0) is
(a) L/T
(b) T/L
(c) L2/T2
(d) T2/L2.

Answer:

(c) L2/T2

The speed of light, C =1μ00.
The dimensions of  1μ00 are of velocity, i.e. L/T .
​Therefore, 10μ0 will have dimensions L2/T2​.

Page No 338:

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.

Page No 338:

Question 5:

An electromagnetic wave going through vacuum is described by
E = E0 sin (kx − ωt); B = B0 sin (kx − ωt).
Then,
(a) E0 k = B0 ω
(b) E0 B0 = ωk
(c) E0 ω = B0 k
(d) None of these

Answer:

(a) E0 k = B0 ω

The relation between E0  and B0 is given by E0B0=c  ...(1)
Here, c = speed of the electromagnetic wave
The relation between ​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) mode-no 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).

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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(t-xc)B = B0sinω(t-xc)
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 (uE) and magnetic energy density (uB)are equal.
uE = 120E2 = B22μ0 = uB
Energy 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.

Page No 339:

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.

Page No 339:

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, ​p ≠ 0, E ≠ 0.

Page No 339:

Question 1:

An electromagnetic wave going through vacuum is described by
E = E0 sin (kx − ωt).
Which of the following is/are independent of the wavelength?
(a) k
(b) ω
(c) k
(d) kω

Answer:

(c) k

The given quantities can be expressed as:
k is given by
k=2πλ
ω is given by
ω=2πνc=νλω=2πcλ
k/ω is given by 
kω=2π/λ2πc/λ=1c

kω is given by

k×ω=2πλ×2πcλ=4π2cλ2
Thus, ​k/ω is independent of the wavelength.

Page No 339:

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
ϕE 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.

Page No 339:

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 = ​frequency of the electromagnetic wave
λ = ​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 ​all wavelengths and​ 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.

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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 z-direction, the electric and magnetic fields are given by:
Ex = E0 sin (kz – ωt)
By = B0 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.

Page No 339:

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=120E2+B22μ0
The energy of the given volume can be calculated by multiplying the volume with the above expression.

U=u×V=120E2+B22μ0×V  ...(1)
Let the direction of propagation of the electromagnetic wave be along the z-axis. Then, the electric and magnetic fields at a particular point are given by
ExE0 sin (kz – ωt)
ByB0 sin (kz – ωt)

Substituting the values of electric and magnetic fields in (1), we get:

U=120(E02sin2(kz-ωt)+B02sin2(kz-ωt)2μ0×VU=0E02(1-cos(2kz-2ωt))4+B02(1-cos(2kz-2ωt))4μ0×V
From 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.

Page No 339:

Question 1:

Show that the dimensions of the displacement current ε0dφEdt are that of an electric current.

Answer:

Displacement current,
ID=0dφedt
Electric flux,
 ϕe =EA
[ϕe] = [E][A]=[14π0qr2][A]Also, [0] = [M-1L-3T4A2][ϕe] = [M1L3T-4A-2][AT][L-2][L2]=[ML3T-3A-1]

Displacement current,
ID= [0] [ϕe] [T-1]ID= [M-1L-3T4A2][ML3T-3A-1][T-1]ID= [A]
[ID]=[current]

Page No 339:

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= kqx2 
Electric flux,
ϕE=EAϕE=kqAx2
Displacement current = Id
 Id = 0dϕEdt Id =0ddtkqAx2 Id=0kqAddtx-2 Id= 014π0×q×A×(-2)x-3×dxdt Id=qAv2πx3

Page No 339:

Question 3:

A parallel-plate capacitor of plate-area 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=Q0A
Electric flux linked with the area,
ϕE=EA=Q0A×A2=Q20
Displacement current,
Id=0dϕEdt = 0ddtQ02Id=12dQdt   ...(i)

Charge on the capacitor as a function of time during charging,
Q=εC1-e-t/RC

Putting this in equation (i), we get:
Id=12εCddt1-e-t/RCId=12εC-e-t/RC×-1RC C = A0dId=ε2R×e-tdε0AR

Page No 339:

Question 4:

Consider the situation of the previous problem. Define displacement resistance Rd = V/id of the space between the plates, where V is the potential difference between the plates and id is the displacement current. Show that Rd varies with time as Rd=R(et/τ-1).

Answer:

Electric field strength for a parallel plate capacitor =  E=Q0A
Electric flux, ϕ=E.A=Q0A.A=Q0Displacement current, id=0dϕEdt = 0ddtQ0id=dQdtAlso, Q = CVid=ddtE0Ce-t/RCid=E0C-1RCe-t/RCDisplacement resistance, Rd = E0id -RRd=E0E0Re-t/RC-RRd=Ret/RC-RRd=R(et/RC-1)

Page No 339:

Question 5:

Using B = µ0 H, find the ratio E0/H0 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 = µ0H

For vacuum we can rewrite this equation as,
B0 = µ0H0   ...(i)

Relation between magnetic field and electric field for vacuum is given as,
B0 = µ00cE0   ...(ii)

From equation (i) by (ii),
μ0H0=μ00cE0E0H0=10cE0H0=18.85×10-12×3×108E0H0377 Ω
Dimension of 10c=1[LT-1][M-1L-3T4A2]=1M-1L-2T3A2=M4L2T-3A-2=[R]

Page No 339:

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, E0 = 810 V/m
Maximum value of magnetic field = Magnetic field amplitude = B0 = ?
We know:
Speed of a wave =EB
For electromagnetic waves, speed = speed of light
B0 = µ0 ε0 cE0
Putting the values in the above relation, we get:
B0= 4π×10-7×8.85×10-12×3×108×810B0= 4×3.14×8.85×3×81×10-10B0= 27010.9×10-10B0= 2.7×10-6 T=2.7 μT

Page No 339:

Question 7:

The magnetic field in a plane electromagnetic wave is given by
B = (200 µT) sin [(4.0 × 1015s−1)(tx/c)].
Find the maximum electric field and the average energy density corresponding to the electric field.

Answer:

Maximum value of a magnetic field, B0 = 200 μT
The speed of an electromagnetic wave is c.

So, maximum value of electric field,
E0=cB0
E0=c×B0=200×10-6×3×108E0=6×104 NC-1

(b) Average energy density of a magnetic field,
Uav=12μ0B02=(200×10-6)22×4π×10-7Uav=4×10-88π×10-7=120πUav=0.01590.016 J/m3

For 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.

Page No 339:

Question 8:

A laser beam has intensity 2.5 × 1014 W m−2. Find amplitudes of electric and magnetic fields in the beam.

Answer:

Given:
Intensity, I = 2.5 × 1014 W/m2
We know:
I=120E02c
E02=2I0c E0=2I0cE0=2×2.5×10148.85×10-12×3×108E0=0.4339×109E0=4.33×108 N/C

Maximum value of magnetic field,
B0=E0cB0=4.33×1083×108B0=1.43 T

Page No 339:

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=120E02cE0=2I0c E0=2×13808.83×10-12×3×108E0=103.95×104E0=10.195×102E0=1.02×103 N/CSince E0=B0c, B0=E0c=1.02×1023×1028B0=3.398×10-6B0=3.4×10-6 T



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