Electromagnetic Waves Test

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Electromagnetic Waves Test - 24

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following cannot be polarized ?

A
Ultraviolet rays

B
Ultrasonic waves

C
X-rays

D
Radiowaves

Solution

All the longitudinal waves like sound etc cannot be polarized because the motion of the particles is already in one dimension that is the direction of propagation of wave.
Thus all the transverse waves like electromagnetic waves can be polarized.

Thus, (B) Ultrasonic waves being sound waves having frequency greater than 20 kHz but being longitudinal in nature cannot be polarized
Question 2
1 / -0

The electric field for a plane , electomagnetic wave travelling in the +y direction is shown in figure.If the electric field of the wave $$\overline{E}$$ is in the Z direction , then the $$\overline{B}$$ field is

A
In the x direction and in phase with the $$\overline{E}$$ field

B
In the x direction and out of phase with the $$\overline{E}$$ field

C
In the z direction and in phase with the $$\overline{E}$$ field

D
In the z direction and one fourth of a cycle out of phase with the $$\overline{E}$$ field

Solution

The wave equation for a plane electric wave traveling in the x direction in space is
$$\dfrac {\partial^2 E}{\partial y^2} = \dfrac{1}{c^2} \dfrac {\partial^2 E}{\partial t^2}$$
with the same form applying to the magnetic field wave in a plane perpendicular the electric field. Both the electric field and the magnetic field are perpendicular to the direction of travel y.
Question 3
1 / -0

An electromagnetic wave passing through vacuum is described by the equations. $$E=E_0\sin(kx-\omega t)$$ and $$B=B_0\sin(kx-\omega t)$$. Then.

A
$$E_0k=B_0\omega$$

B
$$E_0\omega =B_0k$$

C
$$E_0B_0=\omega k$$

D
$$E_0=B_0$$

Solution

Equations of an electromagnetic wave $$E=E_0\sin(kx-\omega t)$$ and $$B=B_0\sin(kx -\omega t)$$
The relation between electric field $$(E_0)$$ and magnetic field $$(B_0)$$ is $$E_0=cB_0$$ ........(i)
Velocity of light $$c=\displaystyle v\lambda =\displaystyle\frac{\omega}{2\pi}\lambda =\frac{\omega}{k}$$ (where $$k=2\pi /\lambda$$).
Substituting this value of $$c$$ in equation (i), we get
$$E_0=\displaystyle\frac{\omega}{k}B_0$$
or $$E_0k=B_0\omega$$.
Question 4
1 / -0

Radiation pressure on any surface :

A
is dependent on wavelength of the light used

B
is dependent on nature of surface and intensity of light used

C
is dependent on frequency and nature of surface

D
depends on the nature of source from which light is coming and on nature of surface on which it is falling.

Solution

Radiation pressure is given by $$P_R = \dfrac{(1+\alpha) I}{c}$$
where $$\alpha$$ is the coefficient of reflection of the surface.
For completely reflecting surface $$\alpha =1$$
For completely absorbing surface $$\alpha = 0$$
So, radiation pressure depends on the nature of surface on which the light is falling but independent of wavelength of light falling.
Question 5
1 / -0

An em wave is propagating in a medium with a velocity $$\overrightarrow{V}=V\hat{i}$$. The instantaneous oscillating electric field of this em wave is along $$+$$y axis. Then the direction of oscillating magnetic field of the em wave will be along.

A
$$-$$y direction

B
$$-$$z direction

C
$$-$$x direction

D
$$+$$z direction

Solution

Propagation of wave is in X direction $$ = \hat i$$
Oscillation of electric field is in y direction $$ = \hat j$$
SO direction of magnetic field is given by $$ = \hat i \times \hat j = \hat k $$
So +z direction is correct answer option 4
Question 6
1 / -0

Choose the correct answer from the alternatives given.
An electromagnetic wave of frequency $$\nu= 3\ MHz$$ passes from vacuum into a dielectric medium with permittivity $$\varepsilon= 4$$. Then

A
wavelength and frequency both become half.

B
wavelength is doubled and frequency remains unchanged.

C
wavelength and frequency both remain unchanged.

D
wavelength is halved and frequency remains unchanged.

Solution

Given : frequency $$v =3 MHz=3\times10^6Hz$$, relative permitivity $$\varepsilon_r = 4$$
Here the frequency of electromagnetic wave remains unchanged but the wavelength of electromagnetic wave changes when it passes from one medium to another.
The refractive index is the square root of permeability and permittivity product.
For formula,
$$c=\dfrac 1{\sqrt {\mu_0\varepsilon_0}}\\\implies c\propto \dfrac1{\sqrt{\varepsilon_0}}$$
Similarly,
$$v\propto\dfrac1{\sqrt{\varepsilon}}$$
Therefore,
$$\dfrac cv=\sqrt{\dfrac {\varepsilon}{\varepsilon_0}}=\sqrt{\dfrac 41}=2........(i)$$
But
$$\dfrac cv=\dfrac {\nu\lambda}{\nu\lambda'}\\\implies \dfrac cv=\dfrac{\lambda}{\lambda'}\\\implies 2=\dfrac{\lambda}{\lambda'}\\\implies \lambda'=\dfrac \lambda2$$
Hence wavelength is halved.
Question 7
1 / -0

Choose the correct answer from the alternatives given.
A plane electromagnetic wave of frequency $$25 MHz$$ travels in free space along $$X$$-direction. At a particular point in space and time, electric field $$\vec E=6.3\ \hat j\ V/m$$. What is $$B$$ at this point.

A
$$1.2 \, \times \, 10^{-6} \, T$$

B
$$1.2 \, \times \, 10^{-8} \, T$$

C
$$2.1 \, \times \, 10^{-6} \, T$$

D
$$2.1 \, \times \, 10^{-8} \, T$$

Solution

Given: The frequency of the electromagnetic wave is $$25\ MHz$$.
The electric field at the particular point is $$6.3\hat j\ V/m$$

To find: The magnetic field at that point.

The magnetic field of the electromagnetic wave at a point is given by:
$$B = \dfrac{E}{c}\\= \dfrac{6.3}{3 \times 10^8}\\ \Rightarrow2.1 \times 10^{-8} T$$

So, option $$(D)$$ is correct.
Question 8
1 / -0

Choose the correct answer from the alternatives given.
A radio can tune to any station in 7.5 MHz to 12 MHz band. The corresponding wavelength band is

A
40 m to 25 m

B
30 m to 25 m

C
25 m to 10 m

D
10 m to 5 m

Solution

$$Here, \, \upsilon_1 \, =\, 7.5 \, MHz, \, \upsilon_2 \, =\,12 \, MHz$$
$$\therefore \, \lambda_1 = \dfrac{c}{\upsilon_1} = \dfrac{3 \times 10^8}{7.5 \times 10^6} = 40m$$
$$and \, \lambda_2 = \dfrac{c}{\upsilon_2} = \dfrac{3 \times 10^8}{12 \times 10^6} = 25m$$
Question 9
1 / -0

Generation, propagation and detection of electromagnetic waves is the basis of

A
lasers

B
reactors

C
radio and television

D
computer

Solution

The communication and broadcasting following the base on generation, propagation, and detection of electromagnetic waves.

The electromagnetic spectrum describes a different range of electromagnetic waves. These EM waves are a special type of wave that can travel without a medium.

Electromagnetic waves are named like this due to the fact that they have both an electric and a magnetic component. In a vacuum, EM waves always travel at the same speed i.e. the speed of light. So, other EM waves besides light are infrared, ultraviolet, radio waves, and microwaves. Therefore radio and television both are based on EM wave properties. Other options like lasers, reactors, and computers are not guided by EM waves.

Thus the correct option is C.
Question 10
1 / -0

Choose the correct answer from the alternatives given.
Electromagnetic wave consists of periodically oscillating electric and magnetic vectors

A
in mutually perpendicular planes but vibrating with a phase difference of $$\pi$$

B
in mutually perpendicular planes but vibrating with a phase difference of $$\dfrac{\pi}{2}$$

C
in randomly oriented planes but vibrating in phase

D
in mutually perpendicular planes but vibrating in phase

Solution

The Electromagnetic wave consists of the two mutually perpendicular electric and magnetic fields present in the medium and also both being perpendicular to the direction of propagation of the wave. The two fields are in same phase as they obtain their peaks at the same instant.

An electromagnetic wave consists of periodically oscillating electric and the magnetic vector in mutually perpendicular planes but vibrating in phase.

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Electromagnetic Waves Test - 24

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Electromagnetic Waves Test - 24

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SHARING IS CARING

Question 1

Ultraviolet rays

Ultrasonic waves

X-rays

Radiowaves

Solution

Question 2

In the x direction and in phase with the $$\overline{E}$$ field

In the x direction and out of phase with the $$\overline{E}$$ field

In the z direction and in phase with the $$\overline{E}$$ field

In the z direction and one fourth of a cycle out of phase with the $$\overline{E}$$ field

Solution

Question 3

$$E_0k=B_0\omega$$

$$E_0\omega =B_0k$$

$$E_0B_0=\omega k$$

$$E_0=B_0$$

Solution

Question 4

is dependent on wavelength of the light used

is dependent on nature of surface and intensity of light used

is dependent on frequency and nature of surface

depends on the nature of source from which light is coming and on nature of surface on which it is falling.

Solution

Question 5

$$-$$y direction

$$-$$z direction

$$-$$x direction

$$+$$z direction

Solution

Question 6

wavelength and frequency both become half.

wavelength is doubled and frequency remains unchanged.

wavelength and frequency both remain unchanged.

wavelength is halved and frequency remains unchanged.

Solution

Question 7

$$1.2 \, \times \, 10^{-6} \, T$$

$$1.2 \, \times \, 10^{-8} \, T$$

$$2.1 \, \times \, 10^{-6} \, T$$

$$2.1 \, \times \, 10^{-8} \, T$$

Solution

Question 8

40 m to 25 m

30 m to 25 m

25 m to 10 m

10 m to 5 m

Solution

Question 9

lasers

reactors

radio and television

computer

Solution

Question 10

in mutually perpendicular planes but vibrating with a phase difference of $$\pi$$

in mutually perpendicular planes but vibrating with a phase difference of $$\dfrac{\pi}{2}$$

in randomly oriented planes but vibrating in phase

in mutually perpendicular planes but vibrating in phase

Solution

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