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JEE • PHYSICS • ELECTROMAGNETIC WAVES

Electromagnetic waves practice · Mixed difficulty

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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 250
Remembering +4 | -1
The infra-red spectrum lies between
View answer and explanation
Correct option: B
the micro-wave and visible region
Option B (the micro-wave and visible region) is correct because the infrared spectrum lies between the microwave and visible regions of the electromagnetic spectrum.
122 Medium
122 of 250
Understanding +4 | -1
If a free electron is placed in the path of a plane electromagnetic wave, it will start moving along
View answer and explanation
Correct option: D
Electric field
Option D (Electric field) is correct because a free electron experiences a force due to the electric field component of an electromagnetic wave.
123 Medium
123 of 250
Remembering +4 | -1
Which of the following electromagnetic waves have minimum frequency
View answer and explanation
Correct option: B
Audible waves
Option D (Radiowaves) is correct because radiowaves have the lowest frequency among the given electromagnetic waves.
124 Medium
124 of 250
Application +4 | -1
A plane EM wave is propagating along xx direction. It has a wavelength of 4 mm4 \mathrm{~mm}. If electric field is in yy direction with the maximum magnitude of 60 Vm−160 \mathrm{~Vm}^{-1}, the equation for magnetic field is:
View answer and explanation
Correct option: B
BZ=2×10−7sin⁡[π2×103(x−3×108t)]k^ T{B}_Z=2 \times 10^{-7} \sin \left[\frac{\pi}{2} \times 10^3\left(x-3 \times 10^8 t\right)\right] \hat{k} \mathrm{~T}
Option B (BZ=2×10−7sin⁡[π2×103(x−3×108t)]k^ T{B}_Z=2 \times 10^{-7} \sin \left[\frac{\pi}{2} \times 10^3\left(x-3 \times 10^8 t\right)\right] \hat{k} \mathrm{~T}) is correct because it represents the magnetic field component of the EM wave, calculated using the relationship between electric and magnetic field amplitudes and the wave's propagation characteristics.
125 Medium
125 of 250
Understanding +4 | -1
Statement I: Microwaves have more energy than the radio waves

Statement II: Microwaves have greater wavelength than the radiowaves
View answer and explanation
Correct option: B
Statement I is true; statement II is false
Option B is correct because microwaves have higher frequency and therefore more energy than radio waves, and wavelength is inversely proportional to frequency.
126 Easy
126 of 250
Application +4 | -1
A plane electromagnetic wave of frequency 35MHz35 {MHz} travels in free space along the XX-direction. At a particular point (in space and time) E⃗=9.6j^ V/m\vec{E}=9.6 \hat{j} {~V} / {m}. The value of magnetic field at this point is :
View answer and explanation
Correct option: D
3.2×10−8k^ T3.2 \times 10^{-8} \hat{k}\, T
Option D (3.2×10−8k^ T3.2 \times 10^{-8} \hat{k}\, T) is correct because the magnetic field is calculated using the relation E = cB, and the direction is determined by the cross product of the wave's propagation direction and the electric field direction.
127 Medium
127 of 250
Remembering +4 | -1
Which of the following rays is emitted by a human body?
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Correct option: D
IR rays
Option D (IR rays) is correct because the human body emits infrared radiation due to its temperature.
128 Medium
128 of 250
Application +4 | -1
A point source is placed at origin. Its intensity at distance of 2 cm2 \mathrm{~cm} from source is II then intensity at distance 4 cm4 \mathrm{~cm} from the source shall be.
View answer and explanation
Correct option: C
I4\frac{I}{4}
Option C (I4\frac{I}{4}) is correct because the intensity of a point source decreases with the square of the distance from the source.
129 Medium
129 of 250
Remembering +4 | -1
Ozone layer blocks the radiations of wavelength
View answer and explanation
Correct option: B
Equal to 3×10−73 \times {10^{ - 7}}m
Option A (Less than 3×10−73 \times {10^{ - 7}}m) is incorrect. Ozone layer blocks the radiations of wavelength less than 3×10−73 \times {10^{ - 7}}m.
130 Medium
130 of 250
Analyzing +4 | -1
An EM wave from air enters a medium. The electric fields are E1=E01x^cos⁡[2πv(zc−t)]{E_1} = {E_{01}}\hat x\cos \left[ {2\pi v\left( {\frac{z}{c} - t} \right)} \right] in air and E2=E02x^cos⁡[k(2z−ct)]{E_2} = {E_{02}}\hat x\cos \left[ {k\left( {2z - ct} \right)} \right] in medium, where the wave number k and frequency ν refer to their values in air. The medium is non-magnetic.
If εr1{\varepsilon _{{r_1}}} and εr2{\varepsilon _{{r_2}}} refer to relative permittivities of air and medium respectively, which of the following options is correct?
View answer and explanation
Correct option: C
εr1εr2=14\frac{{{\varepsilon _{{r_1}}}}}{{{\varepsilon _{{r_2}}}}} = \frac{1}{4}
Option C (εr1εr2=14\frac{{\varepsilon _{{r_1}}}}{{\varepsilon _{{r_2}}}} = \frac{1}{4}) is correct because the ratio of relative permittivities is inversely proportional to the square of the ratio of speeds.
131 Medium
131 of 250
Understanding +4 | -1
Radio waves and visible light in vacuum have
View answer and explanation
Correct option: A
Same velocity but different wavelength
Option A (Same velocity but different wavelength) is correct because all electromagnetic waves travel at the speed of light in a vacuum, but have different wavelengths and frequencies.
132 Medium
132 of 250
Application +4 | -1
The magnetic field of a beam emerging from a filter facing a floodlight is given by

B0=12×10−8sin⁡(1.20×107z−3.60×1015t)T\mathrm{B}_0=12 \times 10^{-8} \sin ( 1.20 \times 10^7 \mathrm{z} \left.-3.60 \times 10^{15} \mathrm{t}\right) \mathrm{T}.

What is the average intensity of the beam?
View answer and explanation
Correct option: A
1.71
Option A (1.71) is correct because it is calculated using the formula for the average intensity of an electromagnetic wave.
133 Medium
133 of 250
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.
134 Medium
134 of 250
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}.
135 Medium
135 of 250
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).
136 Medium
136 of 250
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.
137 Medium
137 of 250
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.
138 Medium
138 of 250
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.
139 Medium
139 of 250
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}.
140 Medium
140 of 250
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.
141 Medium
141 of 250
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.
142 Medium
142 of 250
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.
143 Medium
143 of 250
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.
144 Medium
144 of 250
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).
145 Medium
145 of 250
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.
146 Easy
146 of 250
Understanding +4 | -1
In the case of a photon :-
View answer and explanation
Correct option: D
Its speed can change but cannot become equal to zero.
Option D (Its speed can change but cannot become equal to zero) is correct because a photon always travels at the speed of light in a given medium, but its speed changes when entering a different medium.
147 Hard
147 of 250
Application +4 | -1
A parallel plate capacitor of plate separation 2 mm is connected in on electric circuit having source voltage 400 volts. It the plate area is 60 cm260\,c{m^2}, then the value of displacement current for 10−6{10^{ - 6}} sec. will be
View answer and explanation
Correct option: B
1.062×10−2amp1.062 \times {10^{ - 2}}\mathrm{ amp}
Option B (1.062×10−2amp1.062 \times {10^{ - 2}}\mathrm{ amp}) is correct because it is the value of displacement current calculated using the formula ID=∈o∂ϕE∂t{I_D} = { \in _\mathrm{o}}\frac{{\partial {\phi _E}}}{{\partial t}}.
148 Medium
148 of 250
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}.
149 Medium
149 of 250
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.
150 Medium
150 of 250
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.
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