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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.

#### 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

(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 ${i}_{\mathrm{d}}={\epsilon }_{0}\frac{\mathrm{d}{\Phi }_{E}}{\mathrm{d}t}$ ($\because$ ϕE 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

(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.

#### Question 3:

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

(c) L2/T2

The speed of light, .
The dimensions of  $\frac{1}{\sqrt{{\mu }_{0}{\in }_{0}}}$ are of velocity, i.e. L/T .
​Therefore, $\frac{1}{{\in }_{0}{\mu }_{0}}$ will have dimensions L2/T2​.

#### Question 4:

Electromagnetic waves are produced by
(a) a static charge
(b) a moving charge
(c) an accelerating charge
(d) chargeless particles

(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 = 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

(a) E0 k = B0 ω

The relation between E0  and B0 is given by $\frac{{E}_{0}}{{B}_{0}}=c$  ...(1)
Here, c = speed of the electromagnetic wave
The relation between ​ω (the angular frequency) and k (wave number):
$\frac{\omega }{k}=c$   ...(2)
Therefore, from (1) and (2), we get:
$\frac{{E}_{0}}{{B}_{0}}=$$\frac{\omega }{k}=c$
$⇒{E}_{0}k={B}_{0}\omega$

#### Question 6:

Can an electromagnetic wave be polarised?

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.

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, $f=\frac{\omega }{2\pi }$, is same.
The electrical energy in the region,
UE = $\left(\frac{1}{2}{\in }_{0}{E}^{2}\right)×V$
It can be written as:
${U}_{\mathrm{E}}=\left(\frac{1}{2}{\in }_{0}\left({{E}_{0}}^{2}{\mathrm{sin}}^{2}\left(kz-\omega t\right)\right)×V\phantom{\rule{0ex}{0ex}}{U}_{E}=\left(\frac{1}{2}{\in }_{0}{{E}_{0}}^{2}×\frac{\left(1-\mathrm{cos}2\left(kz-\omega t\right)\right)}{2}\right)×V\phantom{\rule{0ex}{0ex}}{U}_{E}=\left(\frac{1}{4}{\in }_{0}{{E}_{0}}^{2}×\left(1-\mathrm{cos}2\left(kz-\omega t\right)\right)\right)×V\phantom{\rule{0ex}{0ex}}$
The magnetic energy in the region,
${U}_{\mathrm{B}}=\left(\frac{{B}^{2}}{2{\mu }_{0}}\right)×V\phantom{\rule{0ex}{0ex}}{U}_{\mathrm{B}}=\left(\frac{{{B}_{0}}^{2}{\mathrm{sin}}^{2}\left(kz-\omega t\right)}{2{\mu }_{0}}\right)×V\phantom{\rule{0ex}{0ex}}⇒{U}_{\mathrm{B}}=\left(\frac{{{B}_{0}}^{2}\left(1-\mathrm{cos}\left(2kz-2\omega t\right)\right)}{4{\mu }_{0}}\right)×V$

The angular frequency of the electric and magnetic energies is same and is equal to 2ω.
Therefore, their frequency, $f\text{'}=\frac{2\omega }{2\pi }=2f\phantom{\rule{0ex}{0ex}}$, will be same.

Thus, the electric and magnetic fields have same frequencies and the electrical and magnetic energies will have same frequencies.

#### Question 6:

The sunlight reaching Earth has maximum electric field of 810 Vm−1. What is the maximum magnetic field in this light?

Given:
Electric field amplitude, E0 = 810 V/m
Maximum value of magnetic field = Magnetic field amplitude = B0 = ?
We know:
Speed of a wave =$\frac{E}{B}$
For electromagnetic waves, speed = speed of light
B0 = µ0 ε0 cE0
Putting the values in the above relation, we get:

#### 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.

Maximum value of a magnetic field, B0 = 200 $\mathrm{\mu }$T
The speed of an electromagnetic wave is c.

So, maximum value of electric field,
${E}_{0}=c{B}_{0}$

(b) Average energy density of a magnetic field,

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.

#### Question 8:

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

Given:
Intensity, I = 2.5 × 1014 W/m2
We know:
$I=\frac{1}{2}{\in }_{0}{{E}_{0}}^{2}c$

Maximum value of magnetic field,

#### 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.

Given:

#### Question 1:

Show that the dimensions of the displacement current ${\mathrm{\epsilon }}_{0}\frac{d{\phi }_{E}}{dt}$ are that of an electric current.

Displacement current,
${I}_{\mathrm{D}}=\frac{{\in }_{0}d{\phi }_{\mathrm{e}}}{dt}$
Electric flux,

Displacement current,

[ID]=[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

From Coulomb's law:
Electric field strength,

Electric flux,
${\varphi }_{\mathrm{E}}=EA\phantom{\rule{0ex}{0ex}}{\varphi }_{\mathrm{E}}=\frac{kqA}{{x}^{2}}$
Displacement current = Id

#### 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.

Electric field strength for a parallel plate capacitor,
$E=\frac{Q}{{\in }_{0}A}$
Electric flux linked with the area,
${\varphi }_{E}\mathit{=}EA\mathit{=}\frac{Q}{{\mathit{\in }}_{\mathit{0}}A}\mathit{×}\frac{A}{\mathit{2}}=\frac{Q}{2{\in }_{0}}$
Displacement current,

Charge on the capacitor as a function of time during charging,
$Q=\epsilon C\left[1-{e}^{\mathit{-}t\mathit{/}RC}\right]$

Putting this in equation (i), we get:

#### 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 ${R}_{d}=R\left({e}^{t/\tau }-1\right)$.

Electric field strength for a parallel plate capacitor =  $E=\frac{Q}{{\in }_{0}A}$

#### 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.

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 = µ0${\in }_{0}$cE0   ...(ii)

From equation (i) by (ii),

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