Electromagnetic Waves Test

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Electromagnetic Waves Test - 22

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Weekly Quiz Competition

Question 1
1 / -0

The concept of displacement current introduced by Maxwell removes asymmetry between

A
Gauss's law in electrostatics and Gauss's law in magnetism

B
Faraday's law and Ampere' law

C
Gauss's law and Ampere's law

D
Gauss's law and Faraday's law

Solution

Maxwell added the concept of displacement current in AMpere Circuit Law which governs the conduction in the wire of conduction current. After the deviation compass between capacitor Maxwell thought of magnetic lines which would be the result of varying current known as displacement current. So by continuing the displacement in Amperes law, Maxwell was able to show the result of Amperes conduction in circuit moving electrons and also the result of faraday generation of ME waves.
Question 2
1 / -0

A. Current flow inside the capacitor due to accumulation of charges on the capacitor walls is called displacement current.
B. Current due to the flow of electrons due to some potential difference is called as conduction current.
C. Displacement current came into existence when Maxwell observed that if a magnetic compass is placed between the capacitors the needle gets deflected which signifies presence of magnetic fields which would possibly caused due to some changing current.
D. Displacement current change the actual motion of electric charges.
Which of the above statement(s) is/are correct?

A
A and B only

B
A, B and C only

C
C only

D
A,B,C and D

Solution

Displacement current is a quantity appearing in Maxwell's equations that is defined in terms of the rate of change of electric displacement field. Displacement current came into existence when Maxwell observed that if a magnetic compass is placed between the capacitors the needle gets deflected which signifies presence of magnetic fields which would possibly caused due to some changing current. Current flow inside the capacitor due to accumulation of charges on the capacitor walls is called displacement current.However it is not an electric current of moving charges, but a 'time-varying electric field'. Hence option D is incorrect.
Contrarily, current due to the flow of electrons due to some potential difference is called as conduction current.
Question 3
1 / -0

According to Maxwell's hypothesis, changing of electric filed give rise to

A
magnetic field

B
pressure gradient

C
charge

D
voltage

Solution

According to Maxwell's hypothesis, changing of electric field gives rise to Magnetic field.
We know that $$F=qE,$$, where $$F$$ is force and $$E$$ is electric field.
We can relate magnetic field and force by $$F=qvB$$, where $$v$$ is velocity and $$B$$ is the magnetic field.
Therefore we can obtain magnetic field by changing electric field.
Therefore option $$A$$ is correct.
Question 4
1 / -0

Beyond which frequency, the ionosphere bands any incident electromagnetic radiation but do not reflect it back towards the earth?

A
$$50 MHz$$

B
$$40 MHz$$

C
$$30 MHz$$

D
$$20 MHz$$

Solution

The ionosphere can reflect electromagnetic waves of frequency less than $$40 MHz$$ but not of frequency more than $$40 MHz$$.
Question 5
1 / -0

If $${\epsilon}_0$$ and $${\mu}_0$$ represent the permittivity and permeability of vaccum and $$\epsilon$$ and $$\mu$$ represent the permittivity and permeability of medium, then refractive index of the medium is given by

A
$$\sqrt{\dfrac{{\epsilon}_0{\mu}_0}{\epsilon \mu}}$$

B
$$\sqrt{\dfrac{\epsilon \mu}{{\epsilon}_0{\mu}_0}}$$

C
$$\sqrt{\dfrac{{\epsilon}_0{\mu}_0}{\epsilon}}$$

D
$$\sqrt{\dfrac{{\epsilon}_0{\mu}_0}{\mu}}$$

Solution

Given,
$$\epsilon_o= $$ permittivity of free space and $$\mu_o=$$ permeablilty of vaccum and,
$$\epsilon=$$ permittivity of material and $$\mu=$$permeability of meadium.
$$\eta=$$refractive index of the material.
So, The relation between permittivity, permability and the refractive index of the medium is,
Refractive index of a medium $$\eta=\sqrt{\dfrac{\mu \epsilon}{{\mu}_0{\epsilon}_0}}$$
Question 6
1 / -0

Choose the correct answer from the alternatives given.
The amplitude of an electromagnetic wave in vaccum is doubled with no other changes made to the wave. As a result of this doubling of the amplitude, which of the following statement is correct?

A
The frequency of the wave changes only

B
The wave length of the wave changes only

C
The speed of the wave propagation changes only

D
None of the above is correct

Solution

As we know, velocity of electromagnetic wave,
$$c=\dfrac{1}{\sqrt{{\mu}_0{\epsilon}_0}}=3\times 10^8 m/s$$ which is constant
So It is independent of amplitude of electromagnetic wave, frequency and wavelength of electromagnetic wave.
so none of the above is correct.
Question 7
1 / -0

Choose the correct answer from alternatives given.
If $$\vec { E } $$ and $$\vec { B } $$ represent electric and magnetic field vectors of an electromagnetic wave, the direction of propagation of the wave is along

A
$$\vec { B } \times \vec { E } $$

B
$$\vec { B } $$

C
$$\vec { E } $$

D
$$\vec { E } \times \vec { B } $$

Solution

Electromagnet waves have electric field as well as magnetic field which are perpendicular to each other and the electromagnetic waves propagate in a direction which is perpendicular to both the fields.
Thus the propagation vector of EM waves $$\vec{k} =\vec{E} \times \vec{B}$$
Question 8
1 / -0

According to Maxwell's equation, the velocity of light in any medium is expressed as

A
$$\cfrac { 1 }{ \sqrt { { \mu }_{ 0 }{ \varepsilon }_{ 0 } } } $$

B
$$\cfrac { c }{ \sqrt { \mu \varepsilon } } $$

C
$$\cfrac { \mu}{ \sqrt { \varepsilon } } $$

D
$$\sqrt { \cfrac { { \mu }_{ 0 } }{ \varepsilon } } $$

Solution

Aim: Finding a wave equation from Maxwell's equation
Lets start with
$$\nabla \times \vec{E} = -\dfrac{\partial\vec{B}}{\partial t}$$

Now we take a partial derivative of both sides with respect to time. The curl operator has no part with respect to time.
$$\nabla \times \dfrac{\partial\vec{E}}{\partial t} = -\dfrac{\partial^2\vec{B}}{\partial t^2}$$

Again from the other equation
$$\nabla \times \vec{B} = \mu_0\epsilon_0\dfrac{\partial \vec{E}}{\partial t}$$
Solving this for $$\dfrac{\partial\vec{E}}{\partial t}$$ and plugging into the previous expression to get
$$\nabla \times \dfrac{(\nabla \times \vec{B})}{\mu_0\epsilon_0} = -\dfrac{\partial^2 \vec{B}}{\partial t^2}$$
or $$\dfrac{1}{\mu_0 \epsilon_0}\left(\nabla(\nabla \cdot \vec{B}) - \nabla^2\vec{B}\right) = -\dfrac{\partial^2 \vec{B}}{\partial t^2}$$
Now divergence of the magnetic field is zero.
So rearranging we get, $$\dfrac{-1}{\mu_0 \epsilon_0}\nabla^2\vec{B} + \dfrac{\partial^2 \vec{B}}{\partial t^2} = 0$$ is the required wave equation, one solution of which is,
$$\vec{B} = B_0 e^{i (\vec{x}\cdot\vec{k} - \omega t) }$$
which s epresents a plane wave traveling in the direction of the vector $$\vec{k}$$ and freuency $$\omega$$ and phase velocity $$v =\dfrac{ \omega}{|\vec{k}|}$$
In order to be a solution, this equation needs to have
$$\dfrac{\omega^2}{k^2} = \dfrac{1}{\mu_0\epsilon_0}$$
or $$v =\dfrac{ 1}{\sqrt{\mu_0\epsilon_0}}$$
So electromagnetic signals in a vacuum travel at speed $$v = \dfrac{1}{\sqrt{\mu_0\epsilon_0}}$$
Question 9
1 / -0

A plane electromagnetic wave travels in vacuum along $$\hat { k }$$ direction, where $$ \hat { i } ,\hat { j }$$ and $$ \hat { k } $$ are unit vectors along the $$x, y$$ and $$z$$ directions. The direction along which the electric and the magnetic field vectors point may be respectively

A
$$\hat { i }$$ and $$ \hat { j } $$

B
$$\hat { i }$$ and $$ -\hat { j } $$

C
$$\hat { j }$$ and $$ \hat { i } $$

D
$$\hat { k }$$ and $$ \hat { i } $$

Solution

Electromagnetic wave is a transverse wave that means the electric and magnetic field associated to it will not only be perpendicular to each other but will also be perpendicular to the direction in which the wave travels.
So, if waves travel along $$\hat k$$ direction then the electric and the magnetic field will be along $$\hat i$$ and $$\hat j$$ directions.
Question 10
1 / -0

In Thomson's experiment to measure $${e}/{m}$$ of electron, the electric and the magnetic fields are

A
In the same direction

B
In the opposite direction

C
At an angle of $${45}^{o}$$ with each other

D
Perpendicular to each other

Solution

The experimental set up of Thomson's experiment is shown in figure,
According to this figure, the electric field is applied between two horizontal parallel plates, this field is directed in downward direction. The cross in figure shows the magnetic field is directed inside the paper. Therefore, both the fields are perpendicular to each other.
Hence option D is correct.

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Electromagnetic Waves Test - 22

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Electromagnetic Waves Test - 22

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Question 1

Gauss's law in electrostatics and Gauss's law in magnetism

Faraday's law and Ampere' law

Gauss's law and Ampere's law

Gauss's law and Faraday's law

Solution

Question 2

A and B only

A, B and C only

C only

A,B,C and D

Solution

Question 3

magnetic field

pressure gradient

charge

voltage

Solution

Question 4

$$50 MHz$$

$$40 MHz$$

$$30 MHz$$

$$20 MHz$$

Solution

Question 5

$$\sqrt{\dfrac{{\epsilon}_0{\mu}_0}{\epsilon \mu}}$$

$$\sqrt{\dfrac{\epsilon \mu}{{\epsilon}_0{\mu}_0}}$$

$$\sqrt{\dfrac{{\epsilon}_0{\mu}_0}{\epsilon}}$$

$$\sqrt{\dfrac{{\epsilon}_0{\mu}_0}{\mu}}$$

Solution

Question 6

The frequency of the wave changes only

The wave length of the wave changes only

The speed of the wave propagation changes only

None of the above is correct

Solution

Question 7

$$\vec { B } \times \vec { E } $$

$$\vec { B } $$

$$\vec { E } $$

$$\vec { E } \times \vec { B } $$

Solution

Question 8

$$\cfrac { 1 }{ \sqrt { { \mu }_{ 0 }{ \varepsilon }_{ 0 } } } $$

$$\cfrac { c }{ \sqrt { \mu \varepsilon } } $$

$$\cfrac { \mu}{ \sqrt { \varepsilon } } $$

$$\sqrt { \cfrac { { \mu }_{ 0 } }{ \varepsilon } } $$

Solution

Question 9

$$\hat { i }$$ and $$ \hat { j } $$

$$\hat { i }$$ and $$ -\hat { j } $$

$$\hat { j }$$ and $$ \hat { i } $$

$$\hat { k }$$ and $$ \hat { i } $$

Solution

Question 10

In the same direction

In the opposite direction

At an angle of $${45}^{o}$$ with each other

Perpendicular to each other

Solution

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