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HC Verma ii Solutions for Class 12-science Physics Chapter 18 - Electromagnetic Waves

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:

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 338:

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.

Page No 338:

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 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

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]



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