Electromagnetic Waves Test

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Electromagnetic Waves Test - 26

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Weekly Quiz Competition

Question 1
1 / -0

An electromagnetic wave is propagating along x-axis. At x = 1 m and t = 10 s, its electric vector |$$\overset{-}{E}| = 6 V/m$$ then the magnitude of its magnetic vector is:

A
$$2 \, \times \, 10^{-8} \, T$$

B
$$3 \, \times \, 10^{-7} \, T$$

C
$$6 \, \times \, 10^{-8} \, T$$

D
$$5 \, \times \, 10^{-7} \, T$$

Solution

Electric and magnetic compounds of an electromagnetic field are related by
$$E = CB$$
$$B = \dfrac{E}{C}$$
$$B = \dfrac{6}{3 \times 10^8}$$ (when $$E$$ is given)
$$B = 2 \times 10^{-8} T$$
Question 2
1 / -0

The frequency of electromagnetic wave which is best suitable to observe particle of radius $$3 \times 10^{-4}$$ cm is of the order of

A
$$10^{15}$$ Hz

B
$$10^{14}$$ Hz

C
$$10^{13}$$ Hz

D
$$10^{12}$$ Hz

Solution

The wavelength of radiation used should be less than the size of the particle
Size of particle $$=3 \times 10^{-4}cm/s=3\times 10^{-6}m/s$$
$$\text{Size > } \lambda=\dfrac{c}{υ}$$

$$3×10^{−6}>\dfrac{3×10^8}{υ}$$ or $$υ>10^{14}$$ Hz

In the above option higher frequency is $$10^{15}$$ Hz
Option A
Question 3
1 / -0

Maxwell in his famous equations of electromagnetism, introduced the concept of

A
ac current

B
displacement current

C
impedance

D
reactance

Solution

Maxwell's equations are:
1. $$\nabla .E=\rho / \epsilon$$
2. $$\nabla.B=0$$
3. $$\nabla \times E= -\dfrac{dB}{dt}$$
4. $$\nabla \times B= \mu_0J+ \dfrac{1}{c^2} \dfrac{dE}{dt}$$
so, considering the last eqn. written,
$$\nabla \times B=\mu_0 J$$ is the Ampere's eqn.
so, Maxwell modified the Ampere's eqn. and introduced the concept of displacement current.
So, displacement current =$$\dfrac{1}{c^2} \dfrac{dE}{dt}$$

Hence the correct option is $$(B)$$
Question 4
1 / -0

A charged particle oscillates about its mean equilibrium position with a frequency of $${10}^{9}$$ Hz. The frequency of electromagnetic waves produced by the oscillator is

A
$${10}^{6}$$ Hz

B
$${10}^{7}$$ Hz

C
$${10}^{8}$$ Hz

D
$${10}^{9}$$ Hz

Solution

The frequency of the electromagnetic wave is same as that of oscillating charged particle about its equilibrium position, which is $$10^9 Hz$$.
Question 5
1 / -0

The ratio of amplitude of a magnetic field to the amplitude of electric field for an electromagnetic wave propagating in vacuum is equal to

A
The speed of light in vacuum

B
Reciprocal of speed of light in vacuum

C
The ration of magnetic permeability to the electric susceptibility of vacuum

D
Unity
Solution
Hint: Apply the relation between electric and magnetic field.

Explanation :-
Both electric field and magnetic field are vector quantities.
They both are components of electromagnetic waves.
We know that,
$$C=\dfrac{E_0}{B_0}$$

$$\dfrac{B_0}{E_0}=\dfrac{1}{C}$$

Hence, the ratio of amplitude of magnetic field to the amplitude of electric field for an electromagnetic wave in a vacuum is equal to the reciprocal of speed of light.

The correct answer is option (B).
Question 6
1 / -0

An electromagnetic wave propagating along north has its electric field vector upwards. Its magnetic field vector point towards.

A
North

B
East

C
West

D
Downwards
Solution
Hint: Apply Right - hand thumb rule.

Explanation :-
Electric field and magnetic field vectors for an electromagnetic wave are cross - field vectors.
So, the direction of an electromagnetic wave is given by the product of electric field vector and magnetic field vector.
According to the question, electric field vector is directed upwards and EM wave is directed towards North. So, according to the right - hand thumb rule, the magnetic field vector points towards the East.
Hence, the direction of magnetic field is towards East.
The correct answer is option (B).
Question 7
1 / -0

The amplitude of the magnetic field of a harmonic electromagnetic wave in vacuum is $${B}_{0}$$ = 510 nT. The amplitude of the electric field part of the wave is

A
120 N $${C}^{-1}$$

B
163 N $${C}^{-1}$$

C
510 N $${C}^{-1}$$

D
153 N $${C}^{-1}$$

Solution

Given,
$$B_0=510nT$$
$$c=3\times 10^8m/s$$
The magnitude of electric field is given by
$$E_0=B_0 c$$
$$E_0=510\times 10^{-9}\times 3\times 10^8$$
$$E_0=153NC^{-1}$$
The correct option is D.
Question 8
1 / -0

Which of the following doesnt represent Maxwells equation?

A
$$\oint { \overrightarrow { E } .\bar { dA } = } \dfrac { q }{ { \varepsilon }_{ 0 } } $$

B
$$\oint { \overrightarrow { B } .\bar { dA } = } 0$$

C
$$\oint { \overrightarrow { E } .\bar { dl } =\dfrac { -dB }{ dt } } $$

D
$$\oint { \overrightarrow { B } .\bar { dl } = } { \mu }_{ 0 }{ I }_{ C }+{ \mu }_{ 0 }{ \varepsilon }_{ 0 }\dfrac { d{ \phi }_{ E } }{ dt } $$

Solution

Following are Maxwell's equations :

Therefore, the equations $$(A), (B)$$ and $$(D)$$ represent the correct form of Maxwell's equations whereas the equation $$(C)$$ shows the rate of change of magnetic field $$\vec B$$ but actually it should be the rate of change of magnetic flux $$\phi$$. So, equation $$(C)$$ is wrong.

So, option $$(C)$$ is correct.
Question 9
1 / -0

The electric field associated with an electromagnetic wave in vacuum is given by $$|\overrightarrow { E } |= 40\ cos (kz -6\times{10}^{8}t )$$, where $$E$$, $$z$$ and $$t$$ are in volt per meter, meter and second respectively. The value of wave vector $$k$$ is:

A
$$2\ {m}^{-1}$$

B
$$0.5\ {m}^{-1}$$

C
$$3\ {m}^{-1}$$

D
$$6\ {m}^{-1}$$

Solution

Given: The electric field associated with an electromagnetic wave in vacuum is given by $$|\vec E|=40 \cos(kz−6\times 10^8t)$$

To find: Value of wave vector $$k$$

Solution:
We know electromagnetic wave eqution is
$$|\vec E|=E_0\cos(kz-\omega t)$$

And given equation is
$$|\vec E|=40 \cos(kz−6\times 10^8t)$$

By comparing these two, we get
$$\omega=6\times10^8$$ and
$$E_0=40$$

We also know,
Speed of electromagnetic wave is given by:
$$v=\dfrac \omega k$$
where v is the speed of the light.

Hence,
$$k=\dfrac \omega v\\\implies k=\dfrac {6\times 10^8}{3\times 10^8}\\\implies k=2m^{-1}$$

Option $$(A)$$ is correct.
Question 10
1 / -0

An electromagnetic waves can be produced, when charge is

A
Moving with constant velocity

B
Moving in a circular orbit

C
Falling in an electric field

D
both (b) and (c)

Solution

An accelerated charge is the source of electromagnetic waves (EMWs). When the charge is in a circular motion, the direction of its velocity continuously changes and thus it is in accelerated motion and produces EMWs.
A charge falling in an electric field is accelerated by the electric force and thus produces EMWs.

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Electromagnetic Waves Test - 26

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Electromagnetic Waves Test - 26

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Question 1

$$2 \, \times \, 10^{-8} \, T$$

$$3 \, \times \, 10^{-7} \, T$$

$$6 \, \times \, 10^{-8} \, T$$

$$5 \, \times \, 10^{-7} \, T$$

Solution

Question 2

$$10^{15}$$ Hz

$$10^{14}$$ Hz

$$10^{13}$$ Hz

$$10^{12}$$ Hz

Solution

Question 3

ac current

displacement current

impedance

reactance

Solution

Question 4

$${10}^{6}$$ Hz

$${10}^{7}$$ Hz

$${10}^{8}$$ Hz

$${10}^{9}$$ Hz

Solution

Question 5

The speed of light in vacuum

Reciprocal of speed of light in vacuum

The ration of magnetic permeability to the electric susceptibility of vacuum

Unity

Solution

Question 6

North

East

West

Downwards

Solution

Question 7

120 N $${C}^{-1}$$

163 N $${C}^{-1}$$

510 N $${C}^{-1}$$

153 N $${C}^{-1}$$

Solution

Question 8

$$\oint { \overrightarrow { E } .\bar { dA } = } \dfrac { q }{ { \varepsilon }_{ 0 } } $$

$$\oint { \overrightarrow { B } .\bar { dA } = } 0$$

$$\oint { \overrightarrow { E } .\bar { dl } =\dfrac { -dB }{ dt } } $$

$$\oint { \overrightarrow { B } .\bar { dl } = } { \mu }_{ 0 }{ I }_{ C }+{ \mu }_{ 0 }{ \varepsilon }_{ 0 }\dfrac { d{ \phi }_{ E } }{ dt } $$

Solution

Question 9

$$2\ {m}^{-1}$$

$$0.5\ {m}^{-1}$$

$$3\ {m}^{-1}$$

$$6\ {m}^{-1}$$

Solution

Question 10

Moving with constant velocity

Moving in a circular orbit

Falling in an electric field

both (b) and (c)

Solution

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