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Electromagnetic Waves

JEE Physics · 127 questions · Page 12 of 13 · Click an option or "Show Solution" to reveal answer

Q111
A plane electromagnetic wave of frequency 50 MHz travels in free space along the positive x-direction. At a particular point in space and time, E→=6.3j^ V/m.\overrightarrow E = 6.3\widehat j\,V/m. The corresponding magnetic field B,→\overrightarrow {B,} at that point will be :
A 18.9 × \times 10−-8 k^\widehat kT
B 2.1 × \times 10−-8 k^\widehat kT
C 6.3 × \times 10−-8 k^\widehat kT
D 18.9 × \times 108 k^\widehat kT
Correct Answer
Option B
Solution

Given,

E→=6.3j^\overrightarrow E = 6.3\widehat j

∴\therefore

∣E→∣=(6.3)2=6.3\left| {\overrightarrow E } \right| = \sqrt {{{\left( {6.3} \right)}^2}} = 6.3

∴\therefore

∣B→∣=∣E→∣C\left| {\overrightarrow B } \right| = {{\left| {\overrightarrow E } \right|} \over C}
=6.33×108= {{6.3} \over {3 \times {{10}^8}}}
=2.1×10−8 = 2.1 \times {10^{ - 8}}\,

T Now we have to find the direction of

B→\overrightarrow B

. We know,

E^×B^=C^\widehat E \times \widehat B = \widehat C

and given

E→\overrightarrow E

is in y-direction and wave moving in positive x-direction. ∴\therefore

J^×B^=i^\widehat J \times \widehat B = \widehat i

⇒\Rightarrow

B^=K^\widehat B = \widehat K

∴\therefore

B→=∣B→∣B^\overrightarrow B = \left| {\overrightarrow B } \right|\widehat B
=2.1×10−8 K^ T= 2.1 \times {10^{ - 8}}\,\widehat K\,T
Q112
If the magnetic field in a plane electromagnetic wave is given by B→\overrightarrow B = 3 × \times 10-8 sin(1.6 × \times 103x + 48 × \times 1010t)j^\widehat j T, then what will be expression for electric field ?
A E→\overrightarrow E = (9sin(1.6 × \times 103x + 48 × \times 1010t)k^\widehat k V/m)
B E→\overrightarrow E = (60sin(1.6 × \times 103x + 48 × \times 1010t)k^\widehat k V/m)
C E→\overrightarrow E = (3 × \times 10-8 sin(1.6 × \times 103x + 48 × \times 1010t)i^\widehat i V/m)
D E→\overrightarrow E = (3 × \times 10-8 sin(1.6 × \times 103x + 48 × \times 1010t)j^\widehat j V/m)
Correct Answer
Option A
Solution

Given

B→\overrightarrow B

= 3 ×\times 10-8 sin(1.6 ×\times 103x + 48 ×\times 1010t)

j^\widehat j

T We know,

E0B0=c{{{E_0}} \over {{B_0}}} = c

⇒\Rightarrow E0 = (3 ×\times 10-8) ×\times (3 ×\times 10-8) = 9 V/m ∴\therefore

E→\overrightarrow E

= (9sin(1.6 ×\times 103x + 48 ×\times 1010t)

k^\widehat k

V/m)

Q113
The electric field in an electromagnetic wave is given as E→=20sin⁡ω(t−xc)j→NC−1\overrightarrow{\mathrm{E}}=20 \sin \omega\left(\mathrm{t}-\dfrac{x}{\mathrm{c}}\right) \overrightarrow{\mathrm{j}} \mathrm{NC}^{-1} where ω\omega and cc are angular frequency and velocity of electromagnetic wave respectively. The energy contained in a volume of 5×10−4 m35 \times 10^{-4} \mathrm{~m}^{3} will be (Given ε0=8.85×10−12C2/Nm2\varepsilon_{0}=8.85 \times 10^{-12} \mathrm{C}^{2} / \mathrm{Nm}^{2} )
A 17⋅7×10−13 J17 \cdot 7 \times 10^{-13} \mathrm{~J}
B 28⋅5×10−13 J28 \cdot 5 \times 10^{-13} \mathrm{~J}
C 8⋅85×10−13 J8 \cdot 85 \times 10^{-13} \mathrm{~J}
D 88⋅5×10−13 J88 \cdot 5 \times 10^{-13} \mathrm{~J}
Correct Answer
Option C
Solution

To find the energy contained in a volume of an electromagnetic wave, we need to calculate the energy density and then multiply it by the volume.

For an electromagnetic wave, the energy density

uu

is given by:

u=12ε0E2u = \frac{1}{2} \varepsilon_0 E^2

where

ε0\varepsilon_0

is the vacuum permittivity and

EE

is the electric field amplitude. First, let's find the amplitude of the electric field. In the given equation:

E→=20sin⁡ω(t−xc)j→NC−1\overrightarrow{\mathrm{E}}=20 \sin \omega\left(\mathrm{t}-\frac{x}{\mathrm{c}}\right) \overrightarrow{\mathrm{j}} \mathrm{NC}^{-1}

The amplitude

EE

is

20 N/C20 \mathrm{~N/C}

. Now, we can find the energy density:

u=12⋅8.85×10−12C2/Nm2⋅(20 N/C)2u = \frac{1}{2} \cdot 8.85 \times 10^{-12} \mathrm{C}^{2} / \mathrm{Nm}^{2} \cdot (20 \mathrm{~N/C})^2
u=12⋅8.85×10−12⋅400u = \frac{1}{2} \cdot 8.85 \times 10^{-12} \cdot 400
u=1.77×10−9J/m3u = 1.77 \times 10^{-9} \mathrm{J/m^3}

The energy contained in a volume of

5×10−4m35 \times 10^{-4} \mathrm{m^3}

will be:

Etotal=u⋅VE_\text{total} = u \cdot V
Etotal=1.77×10−9J/m3⋅5×10−4m3E_\text{total} = 1.77 \times 10^{-9} \mathrm{J/m^3} \cdot 5 \times 10^{-4} \mathrm{m^3}
Etotal=8.85×10−13JE_\text{total} = 8.85 \times 10^{-13} \mathrm{J}
Q114
A plane electromagnetic wave travels in a medium of relative permeability 1.61 and relative permittivity 6.44. If magnitude of magnetic intensity is 4.5 ×\times 10−-2 Am−-1 at a point, what will be the approximate magnitude of electric field intensity at that point? (Given : Permeability of free space μ\mu0 = 4π\pi ×\times 10−-7 NA−-2, speed of light in vacuum c = 3 ×\times 108 ms−-1)
A 16.96 Vm−-1
B 2.25 ×\times 10−-2 Vm−-1
C 8.48 Vm−-1
D 6.75 ×\times 106 Vm−-1
Correct Answer
Option C
Solution

H = 4.5 ×\times 10−-2 So B = μ\mu0μ\muH Thus

E=cnBE = {c \over n}B

(where n ⇒\Rightarrow refractive index) So

E=3×108×4π×10−7×1.61×4.5×10−21.61×6.44E = {{3 \times {{10}^8} \times 4\pi \times {{10}^{ - 7}} \times 1.61 \times 4.5 \times {{10}^{ - 2}}} \over {\sqrt {1.61 \times 6.44} }}
E=8.48E = 8.48
Q115

Match with : .tg .tg

List - IList - II
(a) Ultraviolet rays (i) Study crystal structure
(b) Microwaves (ii) Greenhouse effect
(c) Infrared rays (iii) Sterilizing surgical instrument
(d) X-rays (iv) Radar system
A (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
B (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)
C (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
D (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
Correct Answer
Option A
Solution

UV rays are used to sterilize surgical material.

Microwaves are used in radar system.

Infrared are used for green house effect and X-rays are used to study crystal structure.

Q116
Which of the following Maxwell's equation is valid for time varying conditions but not valid for static conditions :
A ∮E→⋅dl→=0\oint \overrightarrow{\mathrm{E}} \cdot \overrightarrow{d l}=0
B ∮B⃗⋅dl→=μ0I\oint \vec{B} \cdot \overrightarrow{d l}=\mu_{0} I
C ∮E⃗⋅dl→=−∂ϕB∂t\oint \vec{E} \cdot \overrightarrow{d l}=-\dfrac{\partial \phi_{B}}{\partial t}
D ∮D⃗⋅dA→=Q\oint \vec{D} \cdot \overrightarrow{d A}=Q
Correct Answer
Option C
Solution

Maxwell's equations describe the behavior of electric and magnetic fields.

There are four equations, and each has a specific role.

In the given options, Option C refers to Faraday's Law of Electromagnetic Induction, which is the only equation among the options that is not valid for static conditions.

Option C: Faraday's Law of Electromagnetic Induction:

∮E⃗⋅dl→=−∂ϕB∂t\oint \vec{E} \cdot \overrightarrow{d l}=-\frac{\partial \phi_{B}}{\partial t}

This equation states that a time-varying magnetic field (changing magnetic flux, ϕB\phi_B) induces an electromotive force (EMF) in a closed conducting loop, creating an electric field.

In static conditions, the magnetic field doesn't change over time, and there is no induced EMF.

Therefore, Faraday's Law is valid for time-varying conditions but not for static conditions.

Q117
An electromagnetic wave in vacuum has the electric and magnetic field E→\mathop E\limits^ \to and B→\mathop B\limits^ \to , which are always perpendicular to each other. The direction of polarization is given by X→\mathop X\limits^ \to and that of wave propagation by k→\mathop k\limits^ \to . Then
A X→∣∣B→\mathop X\limits^ \to ||\mathop B\limits^ \to and X→∣∣B→×E→\mathop X\limits^ \to ||\mathop B\limits^ \to \times \mathop E\limits^ \to
B X→∣∣E→\mathop X\limits^ \to ||\mathop E\limits^ \to and k→∣∣E→×B→\mathop k\limits^ \to ||\mathop E\limits^ \to \times \mathop B\limits^ \to
C X→∣∣B→\mathop X\limits^ \to ||\mathop B\limits^ \to and k→∣∣E→×B→\mathop k\limits^ \to ||\mathop E\limits^ \to \times \mathop B\limits^ \to
D X→∣∣E→\mathop X\limits^ \to ||\mathop E\limits^ \to and k→∣∣B→×E→\mathop k\limits^ \to ||\mathop B\limits^ \to \times \mathop E\limits^ \to
Correct Answer
Option B
Solution

as The

E.M.E.M.

wave are transverse in nature i.e.,

=k→×E→μω=H→                 ...(i)= {{\overrightarrow k \times \overrightarrow E } \over {\mu \omega }} = \overrightarrow H \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)

where

H→=B→μ\overrightarrow H = {{\overrightarrow B } \over \mu }

and

k→×H→ωε=−E→                 ...(ii){{\overrightarrow k \times \overrightarrow H } \over {\omega \varepsilon }} = - \overrightarrow E \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)
k→\overrightarrow k

is

⊥  H→\bot \,\,\overrightarrow H

and

k→\overrightarrow k

is also ⊥\bot to

E→\overrightarrow E

or In other words

X→∣∣E→\overrightarrow X ||\overrightarrow E

and

k→∣∣E→×B→\overrightarrow k ||\overrightarrow E \times \overrightarrow B
Q118
The rmsrms value of the electric field of the light coming from the Sun is 720720 N/C.N/C. The average total energy density of the electromagnetic wave is
A 4.58×10−6 J/m34.58 \times {10^{ - 6}}\,J/{m^3}
B 6.37×10−9 J/m36.37 \times {10^{ - 9}}\,J/{m^3}
C 81.35×10−12 J/m381.35 \times {10^{ - 12}}\,J/{m^3}
D 3.3×10−3 J/m33.3 \times {10^{ - 3}}\,J/{m^3}
Correct Answer
Option A
Solution
Erms=720{E_{rms}} = 720

The average total energy density

=12∈0 E02=12∈0[2Erms]2=∈0 Erms2= {1 \over 2}{ \in _0}\,E_0^2 = {1 \over 2}{ \in _0}{\left[ {\sqrt 2 {E_{rms}}} \right]^2} = { \in _0}\,E_{rms}^2
=8.85×10−12×(720)2= 8.85 \times {10^{ - 12}} \times {\left( {720} \right)^2}
=4.58×10−6 J/m3= 4.58 \times {10^{ - 6}}\,J/{m^3}
Q119
A plane electromagnetic wave of frequency 500 MHz is travelling in vacuum along y-direction. At a particular point in space and time, B→\overrightarrow B = 8.0 ×\times 10−-8 z^\widehat zT. The value of electric field at this point is : (speed of light = 3 ×\times 108 ms−-1) x^\widehat x, y^\widehat y, z^\widehat z are unit vectors along x, y and z directions.
A 2.6 x^\widehat x V/m
B −-24 x^\widehat x V/m
C 24 x^\widehat x V/m
D −-2.6 y^\widehat y V/m
Correct Answer
Option B
Solution
E0=B.C{E_0} = B.C
E0=(8×10−8)×(3×108){E_0} = (8 \times {10^{ - 8}}) \times (3 \times {10^8})
⇒E0=24\Rightarrow {E_0} = 24

Direction of wave travelling is in

E→×B→\overrightarrow E \times \overrightarrow B

So,

(−x^)×z^=+y^( - \widehat x) \times \widehat z = + \widehat y

∴\therefore

E^=−24x^\widehat E = -24\widehat x

V/m

Q120
The relative permittivity of distilled water is 81. The velocity of light in it will be : (Given μ\mur = 1)
A 4.33 ×\times 107 m/s
B 2.33 ×\times 107 m/s
C 3.33 ×\times 107 m/s
D 5.33 ×\times 107 m/s
Correct Answer
Option C
Solution

The speed of light in a medium is given by the equation:

v=cεrμrv = \frac{c}{\sqrt{\varepsilon_r \mu_r}}

where: cc is the speed of light in vacuum (approximately 3×1083 \times 10^8 m/s), εr\varepsilon_r is the relative permittivity of the medium (in this case, distilled water, and is given as 81), μr\mu_r is the relative permeability of the medium (given as 1 for distilled water).

Substituting the given values into the equation, we get:

v=3×108 m/s81×1v = \frac{3 \times 10^8 \, \text{m/s}}{\sqrt{81 \times 1}}
v=3×108 m/s9v = \frac{3 \times 10^8 \, \text{m/s}}{9}
v=3.33×107 m/sv = 3.33 \times 10^7 \, \text{m/s}

So, the speed of light in distilled water is 3.33×1073.33 \times 10^7 m/s.

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