Electromagnetic Waves Test

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Electromagnetic Waves Test - 17

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Weekly Quiz Competition

Question 1
1 / -0

Calculate the wavelength of electromagnetic waves of frequency $$300 \,MHz.$$

A
$$3\, m$$

B
$$2,m$$

C
$$30\,m$$

D
$$1\,m$$

Solution
Question 2
1 / -0

Which of the following relations is correct for electromagnetic waves:

A
$$\cfrac { { \partial }^{ 2 }E }{ \partial { x }^{ 2 } } =\cfrac { 1 }{ { c }^{ 2 } } \cfrac { { \partial }^{ 2 }E }{ \partial { t }^{ 2 } } $$

B
$$\cfrac { { \partial }^{ 2 }E }{ \partial { x }^{ 2 } } ={ \left( \cfrac { \partial E }{ \partial t } \right) }^{ 2 }$$

C
$$\cfrac { { \partial }^{ 2 }E }{ \partial { x }^{ 2 } } =\cfrac { { c }^{ 2 }{ \partial }^{ 2 }E }{ \partial { t }^{ 2 } } $$

D
$$\cfrac { { \partial }^{ 2 }E }{ \partial { x }^{ 2 } } =c\cfrac { { \partial }^{ 2 }E }{ \partial { t }^{ 2 } } $$

Solution

The equation of electromagnetic wave propagating along x direction in vaccum is: $$\dfrac{ \partial ^2 E}{ \partial x^2}=\dfrac{1}{c^2}\dfrac{ \partial ^2 E}{ \partial t^2}$$
Question 3
1 / -0

The electric field part of an electromagnetic wave in a medium is represented by
$$E_x=0$$
$$E_y=2.5\dfrac{N}{C}\cos\left[\left(2\pi \times10^6\dfrac{rad}{s}\right)t-\left(\pi \times10^{-2}\dfrac{rad}{m}\right)x\right]$$;
$$E_z=0$$. The wave is :

A
Moving along $$x$$ direction with frequency $$2\pi \times 10^6\ Hz$$ and wavelength $$200\ m$$

B
Moving along $$y$$ direction with frequency $$2\pi\times 10^6\ Hz$$ and wavelength $$200\ m$$

C
Moving along $$x$$ direction with frequency $$10^6\ Hz$$ and wavelength $$100\ m$$

D
Moving along $$x$$ direction with frequency $$10^6\ Hz$$ and wavelength $$200\ m$$

Solution

As the coefficient of x is negative, it is moving along +ve x-axis and equating the equation
             $$E_y=2.5\cos[(2\pi \times10^6)t-(\pi \times10^{-2})x]$$
          with $$y=A\; \cos(\omega t-kx)$$
      $$\omega=2\pi\times106$$
           $$\Rightarrow f=\dfrac{\omega}{2\pi}=10^6Hz$$
   $$k=\pi\times10^{-2}$$
      $$\Rightarrow\lambda=\dfrac{2\pi}{k}$$
   $$=\dfrac{2\pi}{\pi\times10 ^{-2}}=200m$$
Question 4
1 / -0

Wavelength of, monochromatic light is $$5000 \mathring A$$. It's wave number is:

A
$$5000 \times 10^{-10} m^{-1}$$

B
$$2 \times 10^{5} m^{-1}$$

C
$$2 \times 10^{6} m^{-1}$$

D
$$2 \times 10^{4} m^{-1}$$

Solution

$$\text{Wave number}=\dfrac{1}{\text{wavelength}}$$
$$=\dfrac{1}{5000\times 10^{-10}}$$
$$=2\times 10^{6} m^{-1}$$
Question 5
1 / -0

The electric field of an electromagnetic wave in free space is given by $$\vec{E} =10cos (10^7 t + kx) \hat j V/m$$ where t and x are in seconds and metres respectively. It can be iferred that
(i) the wavelength $$\lambda$$ is 188.4 m.
(ii) the wave number k is 0.33 rad/m
(iii) the wave amplitude is 10 V/m
(iv) the wave is propagating along +x direction
Which one of the following pairs of statements is correct ?

A
(iii) and (iv)

B
(i) and (ii)

C
(ii) and (iii)

D
(i) and (iii)

Solution

D. (i) and (iii)

The electric field of an electromagnetic wave in free space is given by

$$E= 10 Cos(10^7t+kx)j V/m$$

The wave magnitude is $$10 V/m$$

The wavelength of electomagnetic wave in free space,

$$\lambda=$$ $$\dfrac{2\pi c}{w}$$

$$w=10^7$$

$$c=3\times10^8 m/s$$

$$\lambda=\dfrac{2\times 3.14\times3\times10^8}{10^7}m$$

$$\lambda=188.4m$$

The wave propagating along $$-ve$$ direction.

The wavenumber, $$K= \dfrac{2\pi}{\lambda}$$

$$K= \dfrac{2\times3.14}{188.4}$$

$$K=0.033 rad/m$$
Question 6
1 / -0

Two waves having same velocity enter electric and magnetic fields respectively. If $$\lambda _{1}$$and$$\lambda _{2}$$ are their wavelengths as they move in the fields, then

A
$$\lambda _{1},\lambda _{2}$$ are constants

B
$$\lambda _{1}$$ and $$\lambda _{2}$$ are variable

C
$$\lambda _{1}$$ is variable $$\lambda _{2}$$ is constant

D
$$\lambda _{2}$$ is variable, $$\lambda _{1}$$ is constant

Solution

Velocity of a wave is given by:
$$v = \cfrac{E}{B}$$
Hence wave velocity change in both the cases.

Frequency of the wave remains the same.
Using $$v = f\lambda$$, it can be concluded that both $$\lambda_1$$ and $$\lambda_2$$ are variable.
Question 7
1 / -0

A monochromatic light of wavelength 589 nm is incident from air onto water surface. The refractive index of water is 1.33. What is the wavelength and speed of refracted light?

A
$$589 nm, 2.256\times 10^8 ms^{-1}$$

B
$$443 nm, 2.256\times 10^8 ms^{-1}$$

C
$$632 nm, 3\times 10^8 ms^{-1}$$

D
$$726 nm, 3\times 10^8 ms^{-1}$$

Solution

On refraction, $$\lambda $$ of light changes but frequency remains unchanged.
$$\therefore \nu=5. 1\times 10^{14} Hz$$
$$v_w=\dfrac {c}{\mu}=\dfrac {3\times 10^8}{1. 33}=2.256\times 10^8 ms^{-1}$$.
Also,
$$\lambda_w=\dfrac {v_w}{\nu}=\dfrac {2.256\times 10^8}{5.1\times 10^{14}}=443\times 10^{-9}\ m=443\ nm$$
Question 8
1 / -0

A standing em wave frequency $$2.2\times { 10 }^{ 10 }Hz$$ is produced in a certain material nodal planes of magnetic field are $$3.5mm$$ apart. Find wavelength and speed of the wave in this material.

A
$$2.81\times { 10 }^{ 8 }m{ s }^{ -1 }$$

B
$$1.79\times { 10 }^{ 8 }m{ s }^{ -1 }$$

C
$$3.08\times { 10 }^{ 8 }m{ s }^{ -1 }$$

D
$$1.54\times { 10 }^{ 8 }m{ s }^{ -1 }$$

Solution

Wave length of original wave is equal to twice the wavelength of the standing waves.
$$\lambda= 2\lambda_{st}= 2\times 3.5 mm = 7 mm$$
Given: $$\nu= 2.2 \times 10^{10} Hz$$
Thus velocity, $$v= \nu \lambda= 2.2\times 10^{10} \times 7 \times 10^{-3} $$
$$\implies v= 1.54 \times 10^8 m/s$$
Question 9
1 / -0

A radio wave of frequency $$90MHz$$ (FM) enter into a ferrite rod. If $${ \varepsilon }_{ r }={ 10 }^{ 3 }$$ and $${{\mu}_{r}}=10$$, then the velocity and wavelength radio wave in ferrite rod is:

A
$$3\times { 10 }^{ 6 }{ ms }^{ -1 },3.33\times { 10 }^{ -2 }m$$

B
$$3\times { 10 }^{ 5 }{ ms }^{ -1 },3.33\times { 10 }^{ -1 }m$$

C
$$3\times { 10 }^{ 6 }{ ms }^{ -1 },3.33\times { 10 }^{ -3 }m$$

D
$$3\times { 10 }^{ 5 }{ ms }^{ -1 },3.33\times { 10 }^{ -2 }m$$

Solution

Velocity of radio wave in ferrite rod: $$v= \dfrac{c}{\sqrt{ \epsilon _r \mu _r}}= \dfrac{3 \times 10^8}{10^2}= 3 \times 10^6m/s$$
Wavelength $$\lambda = \dfrac{3 \times 10^6}{90 \times 10^6}=0.333 \times 10^{-1}m= 3.33 \times 10^{-2}m$$
Question 10
1 / -0

A parallel plate capacitor consists of two circular plates each of radius $$12cm$$ and separated by $$5.0mm$$. The capacitor is being charged by an external source. The charging current is constant and is equal to $$0.15A$$.
The displacement current is:

A
$$15A$$

B
$$1.5A$$

C
$$0.15A$$

D
$$0.015A$$

Solution

According to Maxwell's hypothesis, a displacement current will flow through a capacitor when the potential difference across its plates is varying. Thus a varying electric field will exist between the plates and this displacement current is same in magnitude to the current flowing in outer circuit.
Here, the current in the outer circuit is 0.15 A. Thus $$0.15 A$$ will be the displacement current.

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Electromagnetic Waves Test - 17

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Electromagnetic Waves Test - 17

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Question 1

$$3\, m$$

$$2,m$$

$$30\,m$$

$$1\,m$$

Solution

Question 2

$$\cfrac { { \partial }^{ 2 }E }{ \partial { x }^{ 2 } } =\cfrac { 1 }{ { c }^{ 2 } } \cfrac { { \partial }^{ 2 }E }{ \partial { t }^{ 2 } } $$

$$\cfrac { { \partial }^{ 2 }E }{ \partial { x }^{ 2 } } ={ \left( \cfrac { \partial E }{ \partial t } \right) }^{ 2 }$$

$$\cfrac { { \partial }^{ 2 }E }{ \partial { x }^{ 2 } } =\cfrac { { c }^{ 2 }{ \partial }^{ 2 }E }{ \partial { t }^{ 2 } } $$

$$\cfrac { { \partial }^{ 2 }E }{ \partial { x }^{ 2 } } =c\cfrac { { \partial }^{ 2 }E }{ \partial { t }^{ 2 } } $$

Solution

Question 3

Moving along $$x$$ direction with frequency $$2\pi \times 10^6\ Hz$$ and wavelength $$200\ m$$

Moving along $$y$$ direction with frequency $$2\pi\times 10^6\ Hz$$ and wavelength $$200\ m$$

Moving along $$x$$ direction with frequency $$10^6\ Hz$$ and wavelength $$100\ m$$

Moving along $$x$$ direction with frequency $$10^6\ Hz$$ and wavelength $$200\ m$$

Solution

Question 4

$$5000 \times 10^{-10} m^{-1}$$

$$2 \times 10^{5} m^{-1}$$

$$2 \times 10^{6} m^{-1}$$

$$2 \times 10^{4} m^{-1}$$

Solution

Question 5

(iii) and (iv)

(i) and (ii)

(ii) and (iii)

(i) and (iii)

Solution

Question 6

$$\lambda _{1},\lambda _{2}$$ are constants

$$\lambda _{1}$$ and $$\lambda _{2}$$ are variable

$$\lambda _{1}$$ is variable $$\lambda _{2}$$ is constant

$$\lambda _{2}$$ is variable, $$\lambda _{1}$$ is constant

Solution

Question 7

$$589 nm, 2.256\times 10^8 ms^{-1}$$

$$443 nm, 2.256\times 10^8 ms^{-1}$$

$$632 nm, 3\times 10^8 ms^{-1}$$

$$726 nm, 3\times 10^8 ms^{-1}$$

Solution

Question 8

$$2.81\times { 10 }^{ 8 }m{ s }^{ -1 }$$

$$1.79\times { 10 }^{ 8 }m{ s }^{ -1 }$$

$$3.08\times { 10 }^{ 8 }m{ s }^{ -1 }$$

$$1.54\times { 10 }^{ 8 }m{ s }^{ -1 }$$

Solution

Question 9

$$3\times { 10 }^{ 6 }{ ms }^{ -1 },3.33\times { 10 }^{ -2 }m$$

$$3\times { 10 }^{ 5 }{ ms }^{ -1 },3.33\times { 10 }^{ -1 }m$$

$$3\times { 10 }^{ 6 }{ ms }^{ -1 },3.33\times { 10 }^{ -3 }m$$

$$3\times { 10 }^{ 5 }{ ms }^{ -1 },3.33\times { 10 }^{ -2 }m$$

Solution

Question 10

$$15A$$

$$1.5A$$

$$0.15A$$

$$0.015A$$

Solution

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