Electromagnetic Waves Test

Result

Electromagnetic Waves Test - 4

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

If c is the speed of electromagnetic waves in vacuum, then their speed in a medium of dielectric constant K and relative permeability μ_r will be

A
v =

B
v = c

C
v =

D
v =

Solution
Question 2
1 / -0

A plane electromagnetic wave of frequency 25 MHz travels in free space along x-direction. At a particular point in space and time, = 6.3 Vm^-1. What is the value of at that point?

A
(4.1 × 10^-11) T

B
(2.9 × 10^-9) T

C
(3.1 × 10^-8) T

D
(2.1 × 10^-8) T

Solution

Here, E = 6.3 Vm^-1, C = 3 × 10⁸ ms^-1
Using B = , we get
B = = 2.1 × 10^-8T
Since wave is travelling along x-direction and is along y-direction, therefore must be along z-direction.
=
Question 3
1 / -0

Velocity of electromagnetic waves in a medium depends upon

A
thermal properties of the medium

B
mechanical and electrical properties of the medium

C
electrical and magnetic properties of the medium

D
mechanical and magnetic properties of the medium

Solution

Velocity of light in space = , where ε is the permittivity of free space and μ is the permeability of free space denoting electric and magnetic properties, respectively.
Question 4
1 / -0

An electromagnetic wave travels along the z-direction. Which of the following pairs of space and time-varying E and B fields would generate such a wave?

A
E_xB_y

B
E_yB_x

C
E_zB_x

D
E_xB_x

Solution

As the pointing vector gives the direction of energy flow, i.e. direction of propagation , direction of is aong z-axis. Form the equation, we can conclude that the direction of the electric field is along x-axis and direction of the magnetic field is along y-axis.
Question 5
1 / -0

The direction of electromagnetic waves is given by

A

B

C

D
EB

Solution

As the pointing vector gives the direction of the energy flow, which is in the direction of propagation.
Question 6
1 / -0

A laser emits a beam of light of 2 mm diameter. If the power of the beam is 10 mW, then find its intensity.

A
3.183 × 10³ Wm^-2

B
3.183 × 10⁷ Wm^-2

C
3.183 × 10⁶ Wm^-2

D
3.183 × 10⁵ Wm^-2

Solution

Here, diameter of beam = 2 mm
= 2 × 10^-3 m
Area of the beam ==
= 3.142 × 1 × 10^-6
= 3.142 × 10^-6 m²
Power, P = 10 mW = 10 × 10^-3 W
Intensity, I =
= 3.183 × 10³ Wm^-2
Question 7
1 / -0

If u_e and u_m respectively are the electric and magnetic field densities in an electromagnetic (EM) wave and 'c' is the speed of light, then

A
u_e = u_m

B
u_e = cu_m

C

D
None of these

Solution
Question 8
1 / -0

In the absence of atmosphere around Earth, the average surface temperature of Earth

A
will increase

B
will decrease

C
will remain the same

D
cannot be decided

Solution

In the presence of the atmosphere, heat radiations reflected by the earth are further reflected back to the earth by the atmosphere, which increases the temperature. In the absence of the atmosphere, no such reflection back to earth will take place. Hence the heat radiations will be reflected back and temperature of the earth will decrease.
Question 9
1 / -0

The Poynting vector gives the

A
magnitude and direction of the energy flow in an EM wave

B

C

D

Solution

The Poynting vector gives the magnitude and direction of the energy flow in an EM wave. In physics, the Poynting vector represents the directional energy flux (the energy transfer per unit area per unit time) of an electromagnetic field. The SI unit of the Poynting vector is the watt per square metre (W/m²). It is named after its discoverer, John Henry Poynting, who first derived it in 1884.
Question 10
1 / -0

The dielectric constant of air is 1.006. The speed of an electromagnetic wave travelling in air is a × 10⁸ms^–1, where a is about

A
3

B
3.88

C
2.5

D
3.2

E
2.8

Solution

For an electromagnetic wave (in vacuum),
Velocity c =
Air acts almost as vacuum, hence
a = 3 (approx.)
Question 11
1 / -0

Which of the following electromagnetic waves emitted by the sun are responsible for heating Earth's atmosphere due to greenhouse effect?

A
Visible light

B
Infrared radiation

C
Ultraviolet rays

D
-rays

Solution

Infrared radiations emitted by the sun are responsible for heating Earth's atmosphere due to greenhouse effect. The correct choice is (2).
Question 12
1 / -0

The pressure exerted by an electromagnetic wave of intensity I (W/m²) on a non-reflecting surface is
(where c is the velocity of light)

A
Ic

B
Ic²

C
I/c

D
I/c²

Solution

As we know,
Pressure = Force/Area
Energy of the incident beam is given by E = pc
For a non-reflecting surface, final momentum of the incident beam = 0

Hence, change in momentum of the beam =
Rate of change of momentum in time Δt =
Force exerted =

Pressure exerted =

Here, I is the intensity of the light, which is defined as energy per unit area per second.
Question 13
1 / -0

If Earth receives 20 cal min^-1 cm^-2 of solar energy, then what are the respective amplitudes of electric and magnetic fields of radiations? (Use SI units)

A
4.24 and 3.08

B
3248.4 and 8.616

C
8.616 and 3248.4

D
3 and 4.42

Solution

Pointing vector,
= EH sin 90^o
= EH
Solar energy = 20 cal min^-1 cm^-2
= joules m^-2 sec^_1
EH = 14000 ......... (1)
Also, = 377 ohm ............ (2)
Multiplying (1) and (2), we get
E² = 14,000 x 377
E = = 2297.38 volt/m
H = = 6.093 A-m
Amplitudes of E and H are
E₀ = E = 3248.49
H₀ = H= 8.616
Question 14
1 / -0

At what speed (m/s) does an electron have a de Broglie wavelength of 1.0 nm?

A
5.6 × 105

B
1.2 × 10⁵

C
3.4 × 10⁶

D
7.3 × 10⁵

Solution

v = 6.6 × 10^-34/ ( 9.1 × 10^-31 × 1 × 10^-9)
= 7.3 × 10⁵m/s
Question 15
1 / -0

Which of the following electromagnetic waves are used in telecommunication?

A
Radio waves

B
Visible radiations

C
Ultraviolet rays

D
Microwaves

Solution

Microwaves are used in telecommunication.
So, option 4 is correct.
Radio waves - These are used for television and radio programmes.
Visible radiation - These are used in photography.
Ultraviolet rays - These can be used to kill microbes. Hospitals use UV lamps to sterilise surgical equipment and the air in operating theatres.