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JEE • PHYSICS • ELECTROMAGNETIC WAVES

Electromagnetic waves practice · Medium

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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121 Medium
121 of 226
Understanding +4 | -1
Which one of the following is the property of a monochromatic, plane electromagnetic wave in free space
View answer and explanation
Correct option: B
The energy contribution of both electric and magnetic fields are equal
Option B is correct because in a monochromatic plane electromagnetic wave, the energy is equally distributed between the electric and magnetic fields.
122 Medium
122 of 226
Understanding +4 | -1
If electric field E=E0sin⁡(ωt−kz)i^E=E_0 \sin (\omega t-k z) \hat{i} and magnetic field B=B0sin⁡(ωt−kz)j^,B=B_0 \sin (\omega t-k z) \hat{j}, then electromagnetic wave travels along
View answer and explanation
Correct option: A
+z+z direction
Option A (+z direction) is correct because the direction of propagation of an electromagnetic wave is given by the direction of E⃗×B⃗\vec{E} \times \vec{B}.
123 Medium
123 of 226
Application +4 | -1
An electromagnetic wave is travelling to the east. At one instant at a given point its E vector points straight up. What is the direction of is B vector ?
View answer and explanation
Correct option: C
South
Option C (South) is correct because the direction of propagation of an electromagnetic wave is given by the cross product of the electric field vector and the magnetic field vector (E→×B→\overrightarrow E \times \overrightarrow B).
124 Medium
124 of 226
Understanding +4 | -1
A charged particle can continue to move with a constant velocity in a region wherein,

A. E=0,B≠0E=0, B \neq 0

B. E≠0,B=0E \neq 0, B=0

C. E≠0,B≠0E \neq 0, B \neq 0

D. E=0,B=0E=0, B=0

Select correct alternative:
View answer and explanation
Correct option: C
Only A, C, D
Option C (Only A, C, D) is correct because a charged particle can move with constant velocity when there is no net force, which occurs when E=0, B� 0 (forces cancel), E� 0, B� 0 (forces cancel), or E=0, B=0.
125 Medium
125 of 226
Understanding +4 | -1
Given below are two statements:

Statement I : Electromagnetic waves are not deflected by electric and magnetic field.

Statement II: The amplitude of electric field and the magnetic field in electromagnetic waves are related to each other as E0=μ0ε0B0E_0=\sqrt{\frac{\mu_0}{\varepsilon_0}} B_0.

In the light of the above statements, choose the correct answer from the options given below :
View answer and explanation
Correct option: C
Statement I is true but statement II is false
Option C (Statement I is true but statement II is false) is correct because electromagnetic waves are not deflected by electric or magnetic fields, and the correct relationship is E0=cB0E_0 = cB_0.
126 Medium
126 of 226
Application +4 | -1
Suppose that the electric field amplitude of an electromagnetic wave is E0=120 NC−1E_0=120 \mathrm{~NC}^{-1} and that its frequency is v=50.0 MHzv=50.0 \mathrm{~MHz}. Determine, B0,ω,kB_0, \omega, k and λ\lambda.
View answer and explanation
Correct option: A
4×107 T,3.14×108 rad s−1,1.05 m−1,6.00 m4 \times 10^7 \mathrm{~T}, 3.14 \times 10^8 \mathrm{~rad} \mathrm{~s^{-1}}, 1.05 \mathrm{~m}^{-1}, 6.00 \mathrm{~m}
Option A (4×107 T,3.14×108 rad s−1,1.05 m−1,6.00 m4 \times 10^7 \mathrm{~T}, 3.14 \times 10^8 \mathrm{~rad} \mathrm{~s^{-1}}, 1.05 \mathrm{~m}^{-1}, 6.00 \mathrm{~m}) is correct because it correctly calculates the magnetic field amplitude, angular frequency, wave number, and wavelength based on the given electric field amplitude and frequency.
127 Medium
127 of 226
Application +4 | -1
A cylindrical region of radius 1.2 m1.2 \mathrm{~m} is containing a uniform electric field that is increasing uniformly with time. At t=0\mathrm{t}=0 the field is zero and at t=5.0 s\mathrm{t}=5.0 \mathrm{~s} the field is 200 V/m.200 \mathrm{~V} / \mathrm{m}. The total displacement current through a cross section of the region is:
View answer and explanation
Correct option: D
1.6×10−9 A1.6 \times 10^{-9} \mathrm{~A}
Option D (1.6×10−9 A1.6 \times 10^{-9} \mathrm{~A}) is correct because the displacement current is calculated using the formula id=ε0dϕEdt=ε0AdEdti_d =\varepsilon_0 \frac{d \phi_E}{d t} = \varepsilon_0 {A} \frac{d E}{d t}.
128 Medium
128 of 226
Remembering +4 | -1
Infrared radiation is detected by
View answer and explanation
Correct option: B
Pyrometer
Option B (Pyrometer) is correct because pyrometers are specifically designed to detect infrared radiation, often by measuring thermal radiation.
129 Medium
129 of 226
Understanding +4 | -1
If an electromagnetic wave propagating through vacuum is described by Ey=E0sin⁡(kx−ωt),Bz=B0sin⁡(kx−ωt)E_y=E_0 \sin (k x-\omega t), B_z=B_0 \sin (k x-\omega t), then
View answer and explanation
Correct option: A
E0k=B0ωE_0 k=B_0 \omega
Option A (E0k=B0ωE_0 k=B_0 \omega) is correct because the ratio of electric field amplitude to magnetic field amplitude equals the speed of light, leading to the given relation.
130 Medium
130 of 226
Application +4 | -1
A particle of specific charge ( charge  mass )α\left(\frac{\text { charge }}{\text { mass }}\right) \alpha start moving from origin under action of an electric field E⃗=E0i^\vec{E}=E_0 \hat{i} and magnetic field B⃗=B0k^\vec{B}=B_0 \hat{k}. Its velocity at (x0,y0,0)\left(x_0, y_0, 0\right) is 4i^+3j^4 \hat{i}+3 \hat{j}. The value of x0x_0 is
View answer and explanation
Correct option: C
252αE0\frac{25}{2 \alpha E_0}
Option C (252αE0\frac{25}{2 \alpha E_0}) is correct because it is derived from the work-energy theorem considering the electric field's influence.
131 Medium
131 of 226
Analyzing +4 | -1
The electric field of an electromagnetic wave travelling through vaccum is given by the equation E=E0sin⁡(kx−ωt)E=E_0 \sin (k x-\omega t). The quantity that is independent of wavelength is
View answer and explanation
Correct option: B
kω\frac{k}{\omega}
Option B (kω\frac{k}{\omega}) is correct because the ratio kω\frac{k}{\omega} equals 1c\frac{1}{c}, which is independent of wavelength in a vacuum.
132 Medium
132 of 226
Analyzing +4 | -1
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R
Assertion A : EM waves used for optical communication have longer wavelengths than that of microwave, employed in Radar technology.
Reason R : Infrared EM waves are more energetic than microwaves, (used in Radar)
In the light of given statements, choose the correct answer from the options given below.
View answer and explanation
Correct option: B
A is false but R is true
Option B is correct because the assertion is false (optical communication uses shorter wavelengths than microwaves), while the reason is true (infrared waves are more energetic than microwaves).
133 Medium
133 of 226
Understanding +4 | -1
Statement I: The magnetic flux through a closed surface is zero

Statement II: Gauss's law applies in the case of electric flux only
View answer and explanation
Correct option: B
Statement I is true; statement II is false
Option B (Statement I is true; statement II is false) is correct because Gauss's law applies to both electric and magnetic flux, but magnetic flux through a closed surface is always zero.
134 Medium
134 of 226
Understanding +4 | -1
The direction of propagation of electromagnetic wave is along
View answer and explanation
Correct option: C
E⃗×B⃗\vec{E} \times \vec{B}
Option C (E⃗×B⃗\vec{E} \times \vec{B}) is correct because the direction of propagation of an electromagnetic wave is perpendicular to both the electric and magnetic field vectors, and is given by the Poynting vector, which is proportional to E⃗×B⃗\vec{E} \times \vec{B}.
135 Medium
135 of 226
Understanding +4 | -1
The amplitude of an electromagnetic wave in vacuum 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 true?
View answer and explanation
Correct option: D
None of the above
Option D (None of the above) is correct because the speed, frequency, and wavelength of an electromagnetic wave in a vacuum are independent of its amplitude.
136 Medium
136 of 226
Remembering +4 | -1
The amplitudes of electric and magnetic fields in an electromagnetic wave are related to each other as
View answer and explanation
Correct option: B
E0=cB0E_0=c B_0
Option B (E0=cB0E_0=c B_0) is correct because the relationship between the amplitudes of electric and magnetic fields in an electromagnetic wave is given by E0=cB0E_0 = cB_0, where c is the speed of light.
137 Medium
137 of 226
Understanding +4 | -1
Statement I: Ultra violet radiations being higher frequency waves are dangerous to human beings.

Statement II: Ultraviolet radiations are absorbed by the atmosphere
View answer and explanation
Correct option: B
Statement I is true; statement II is false
Option B (Statement I is true; statement II is false) is correct because UV radiation is high frequency and harmful, but it's mostly absorbed by the ozone layer, not the entire atmosphere.
138 Medium
138 of 226
Understanding +4 | -1
Assertion: Dipole oscillations produce electromagnetic waves

Reason: Accelerated charge produces electromagnetic waves
View answer and explanation
Correct option: A
Both Assertion and reason are true and the reason is correct explanation of the Assertion.
Option A is correct because dipole oscillations involve accelerating charges, which produce electromagnetic waves, making the reason a correct explanation of the assertion.
139 Medium
139 of 226
Understanding +4 | -1
A plane electromagnetic wave is traveling in vacuum in X-direction. The plane in which
magnetic field propagates is
View answer and explanation
Correct option: A
X-Y
Option A (X-Y) is correct because the magnetic field propagates in a plane perpendicular to the direction of propagation (X-direction) and the electric field.
140 Medium
140 of 226
Understanding +4 | -1
Statement I: Electromagnetic waves exert pressure, called radiation pressure.

Statement II: Electromagnetic waves carry energy.
View answer and explanation
Correct option: A
Statement I and statement II are true
Option A is correct because electromagnetic waves do exert radiation pressure and they carry energy, both of which are fundamental properties of these waves.
141 Medium
141 of 226
Remembering +4 | -1
In which one of the following regions of the electromagnetic spectrum will the vibrational motion of molecules give rise to absorption
View answer and explanation
Correct option: B
Microwaves
Option B (Microwaves) is correct because vibrational motions of molecules give rise to absorption in the microwave region of the electromagnetic spectrum.
142 Medium
142 of 226
Application +4 | -1
An electromagnetic wave with frequency ω\omega and wavelength λ\lambda travels in the +y+y direction. Its magnetic field is along +x−+x- axis. The vector equation for the associated electric field (of amplitude E0E_0 ) is
View answer and explanation
Correct option: C
E⃗=E0cos⁡(ωt−2πλy)z^\vec{E}=E_0 \cos \left(\omega t-\frac{2 \pi}{\lambda} y\right) \hat{z}
Option C (E⃗=E0cos⁡(ωt−2πλy)z^\vec{E}=E_0 \cos \left(\omega t-\frac{2 \pi}{\lambda} y\right) \hat{z}) is correct because it represents an electromagnetic wave traveling in the +y direction with the magnetic field along the +x axis, thus the electric field must be along the +z axis and in phase.
143 Medium
143 of 226
Remembering +4 | -1
Which scientist experimentally proved the existence of electromagnetic waves
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Correct option: D
Hertz
Option D (Hertz) is correct because Hertz experimentally proved the existence of electromagnetic waves.
144 Medium
144 of 226
Understanding +4 | -1
If ε0  and  μ0{\mathrm{\varepsilon }_0}\mathrm{\;and\;}{\mathrm{\mu }_0} are the electric permittivity and magnetic permeability of free space and ε  and  μ\mathrm{\varepsilon \;and\;\mu } are the corresponding quantities in the medium, the index of refraction of the medium in terms of above parameter is
View answer and explanation
Correct option: B
(εμε0μ0    )1/2{\left( {\frac{{\mathrm{\varepsilon \mu }}}{{{\mathrm{\varepsilon }_0}{\mathrm{\mu }_0}}}\begin{array}{cccccccccccccccccccc} \mathrm{\;}\\ \mathrm{\;} \end{array}} \right)^{1/2}}
Option B ((εμε0μ0  mathrm  )1/2{\left( {\frac{{\mathrm{\varepsilon \mu }}}{{{\mathrm{\varepsilon }_0}{\mathrm{\mu }_0}}}\begin{array}{cccccccccccccccccccc}\mathrm{\;}\\mathrm{\;}\end{array}} \right)^{1/2}}) is correct because the refractive index is the ratio of the speed of light in vacuum to the speed of light in the medium, which can be expressed in terms of permittivity and permeability.
145 Medium
145 of 226
Understanding +4 | -1
In free space, the energy of electromagnetic wave in electric field is UE{U_E} and in magnetic field is UB{U_B}. Then
View answer and explanation
Correct option: A
UE=UB{U_E} = {U_B}
Option A (UE=UB{U_E} = {U_B}) is correct because the energy is equally distributed between the electric and magnetic fields in an electromagnetic wave.
146 Medium
146 of 226
Understanding +4 | -1
The ratio of amplitude of magnetic field to the amplitude of electric field for an electromagnetic wave propagating in vacuum is equal to :
View answer and explanation
Correct option: B
reciprocal of speed of light in vacuum
Option B (reciprocal of speed of light in vacuum) is correct because the ratio of the magnetic field amplitude to the electric field amplitude in an EM wave is 1/c.
147 Medium
147 of 226
Understanding +4 | -1
During the propagation of electromagnetic waves in a medium
View answer and explanation
Correct option: C
electric energy density is equal to the magnetic energy density
Option C (electric energy density is equal to the magnetic energy density) is correct because in electromagnetic waves, the energy is equally divided between the electric and magnetic fields.
148 Medium
148 of 226
Remembering +4 | -1
Choose the correct option relating wavelengths of different parts of electromagnetic wave spectrum
View answer and explanation
Correct option: C
λradiowaves>λmicrowaves>λvisible>λX−rays{\lambda _{\mathrm{radio waves}}} > {\lambda _{\mathrm{micro waves}}} > {\lambda _{\mathrm{visible}}} > {\lambda _{\mathrm{X - rays}}}
Option C (λradiowaves>λmicrowaves>λvisible>λX−rays{\lambda _{\mathrm{radio waves}}} > {\lambda _{\mathrm{micro waves}}} > {\lambda _{\mathrm{visible}}} > {\lambda _{\mathrm{X - rays}}}) is correct because it accurately represents the order of wavelengths in the electromagnetic spectrum, from longest to shortest.
149 Medium
149 of 226
Remembering +4 | -1
The pressure exerted by an electromagnetic wave of intensity I (Wm−2)\left(\mathrm{Wm}^{-2}\right) on a non-reflecting surface is ( CC is the velocity of light)
View answer and explanation
Correct option: C
I/cI/c
Option C (I/cI/c) is correct because the pressure exerted by an electromagnetic wave on a non-reflecting surface is given by the intensity divided by the speed of light.
150 Medium
150 of 226
Understanding +4 | -1
Displacement current is a source of
View answer and explanation
Correct option: D
time varying magnetic field
Option D (time varying magnetic field) is correct because displacement current, which arises from a time-varying electric field, acts as a source of a time-varying magnetic field, similar to conduction current.
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