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NEET • PHYSICS • ELECTROMAGNETIC WAVES

Electromagnetic waves practice · Mixed difficulty

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Showing 1-30 of 145 questions
1 Medium
1 of 145
Understanding +4 | -1
What is order of energy of X-rays (EX{E_X}), radio waves (ER{E_R}) and microwave (EM{E_M})?
View answer and explanation
Correct option: B
EX  <EM>ER{E_{X\;}} < {E_M} > {E_R}
Option D (EM  <EX>ER{E_{M\;}} < {E_X} > {E_R}) is correct because X-rays have the highest energy, followed by microwaves, and then radio waves, based on their frequencies.
2 Medium
2 of 145
Application +4 | -1
The amplitude of the sinusoidally oscillating electric field of plane wave is 60  Vm60\,\,\frac{V}{m}. Then the amplitude of the magnetic field is
View answer and explanation
Correct option: D
2×10−7T2 \times {10^{ - 7}}T
Option D (2×10−7T2 \times {10^{ - 7}}T) is correct because it is calculated using the relationship between electric and magnetic field amplitudes in an electromagnetic wave, where E = cB.
3 Medium
3 of 145
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 refractive index of the medium is 2, and the relative permittivity is related to the square of the refractive index.
4 Medium
4 of 145
Application +4 | -1
Electric field in a plane electromagnetic wave is given by E=50sin⁡ (500x−10×1010t) v/m.E = 50\sin \,(500x - 10 \times {10^{10}}t)\,v/m. The velocity of electromagnetic wave in this medium is
(Given, cc = speed of light in vacuum)
View answer and explanation
Correct option: C
23 c\frac{2}{3}\,\mathrm{c}
Option C (23 c\frac{2}{3}\,\mathrm{c}) is correct because the velocity of an electromagnetic wave is given by ωk\frac{\omega}{k}, where ω\omega is the angular frequency and kk is the wave number.
5 Medium
5 of 145
Remembering +4 | -1
If an electromagnetic wave is propagation in a medium with permeability&nbsp;ε\mathrm{\varepsilon } and permittivity μ\mathrm{\mu }, then με\sqrt {\frac{\mathrm{\mu }}{\mathrm{\varepsilon }}} is the
View answer and explanation
Correct option: A
Intrinsic impedance of the medium
Option A (Intrinsic impedance of the medium) is correct because the ratio of the square root of permeability to permittivity defines the intrinsic impedance of a medium.
6 Medium
6 of 145
Understanding +4 | -1
In an E.M. wave the average energy density is associated with:-
View answer and explanation
Correct option: C
electric & magnetic fields equally
Option C (electric & magnetic fields equally) is correct because the average energy density in an electromagnetic wave is equally distributed between the electric and magnetic fields.
7 Medium
7 of 145
Application +4 | -1
A plane electromagnetic wave of frequency 500 MHz is travelling in vacuum along y−y -direction. At a particular point in space and time, B→=8.0×10−8z^T\overrightarrow B = 8.0 \times {10^{ - 8}}\widehat zT. The value of electric field at this point is:
(speed of light =3×108ms−1= 3 \times {10^8}m{s^{ - 1}}). x^, y^, z^\widehat x,\widehat {\,y},\widehat {\,z} are unit vectors along x, yx,\,y and zz directions.
View answer and explanation
Correct option: D
−24  x^  V/m- 24\,\,\widehat x\,\,\mathrm{V/m}
Option D (−24  x^  V/m-24\,\,\widehat x\,\,\mathrm{V/m}) is correct because the electric field is perpendicular to both the magnetic field and the direction of propagation, and its magnitude is related to the magnetic field by E=cBE = cB.
8 Easy
8 of 145
Understanding +4 | -1
An electromagnetic wave is propagating along the +z+\mathrm{z}-axis. Direction of electric and the magnetic field of this wave can be represented by
View answer and explanation
Correct option: A
E⃗=E0i^,B⃗=B0j^\vec{E}=E_0 \hat{i}, \vec{B}=B_0 \hat{j}
Option A (E⃗=E0i^,B⃗=B0j^\vec{E}=E_0 \hat{i}, \vec{B}=B_0 \hat{j}) is correct because the electric and magnetic fields are perpendicular to each other and to the direction of propagation, which is the z-axis.
9 Medium
9 of 145
Remembering +4 | -1
According to Maxwell’s equation the velocity of light in any medium is expressed as
View answer and explanation
Correct option: B
1με\frac{1}{{\sqrt {\mathrm{\mu \varepsilon }} }}
Option B (1με\frac{1}{{\sqrt {\mathrm{\mu \varepsilon }} }}) is correct because it represents the speed of light in a medium with permeability μ and permittivity ε, as derived from Maxwell's equations.
10 Medium
10 of 145
Understanding +4 | -1
If μ0\mu_0 be the peremeability and ε0\varepsilon_0 be the permittivity of a medium, then its refractive index is given by
View answer and explanation
Correct option: C
μ0ε0\sqrt{\mu_0 \varepsilon_0}
Option C (μ0ε0\sqrt{\mu_0 \varepsilon_0}) is correct. The refractive index is the square root of the product of permeability and permittivity.
11 Medium
11 of 145
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 kω\frac{k}{\omega} represents the inverse of the wave's velocity, which depends on the medium and is independent of the wavelength in a vacuum.
12 Medium
12 of 145
Application +4 | -1
The magnetic field in travelling EM ware has a peak value of 20. The peak value of electric field strength is [in V/m]
View answer and explanation
Correct option: B
6
Option B (6) is correct because it is calculated using the relationship between electric and magnetic field amplitudes in an electromagnetic wave, where E = cB, and B is given in micro Tesla.
13 Easy
13 of 145
Application +4 | -1
Calculate the momentum transferred to a surface when a radiation of energy E falls normally on it. Assume that the reflectivity of the surface is unity.
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Correct option: B
2Ec\frac{2 E}{c}
Option B (2Ec\frac{2 E}{c}) is correct because the momentum transferred to a surface with reflectivity of unity is twice the energy divided by the speed of light.
14 Medium
14 of 145
Application +4 | -1
In a plane electromagnetic wave propagating in space has an electric field of amplitude 9×103  Vm−19 \times {10^3}\mathrm{\;V}{\mathrm{m}^{ - 1}}, then the amplitude of the magnetic field is
View answer and explanation
Correct option: D
3.0×10−5  T3.0 \times {10^{ - 5}}\mathrm{\;T}
Option D (3.0×10−5  T3.0 \times {10^{ - 5}}{\;\mathrm{T}}) is correct because the amplitude of the magnetic field is calculated by dividing the electric field amplitude by the speed of light.
15 Medium
15 of 145
Understanding +4 | -1
For an electromagnetic wave travelling in free space, the relation between average energy densities due to electric (Ue)({U_e}) and magnetic (Um)({U_m}) fields is:
View answer and explanation
Correct option: B
Ue=Um{U_e} = {U_m}
Option B (Ue=Um{U_e} = {U_m}) is correct because in an electromagnetic wave, the average energy density due to electric and magnetic fields are equal.
16 Medium
16 of 145
Remembering +4 | -1
Which of the following rays is emitted by a human body?
View answer and explanation
Correct option: D
IR rays
Option D is correct because the human body emits infrared (IR) radiation due to its temperature.
17 Medium
17 of 145
Understanding +4 | -1
Identify the correct statements from the following descriptions of various properties of electromagnetic waves.
(A) In a plane electromagnetic wave electric field and magnetic field must be perpendicular to each other and direction of propagation of wave should be along electric field or magnetic field.
(B) The energy in electromagnetic wave is divided equally between electric and magnetic fields.
(C) Both electric field and magnetic field are parallel to each other and perpendicular to the direction of propagation of wave.
(D) The electric field, magnetic field and direction of propagation of wave must be perpendicular to each other.
(E) The ratio of amplitude of magnetic field to the amplitude of electric field is equal to speed of light.
Choose the most appropriate answer from the options given below
View answer and explanation
Correct option: B
(B) and (D) only
Option B ((B) and (D) only) is correct. The energy in an electromagnetic wave is divided equally between electric and magnetic fields, and the electric field, magnetic field, and direction of propagation are mutually perpendicular.
18 Easy
18 of 145
Application +4 | -1
Radiations of intensity 0.5 Wm−2W{m^{ - 2}}are striking a metal plate. The pressure on the plate is
View answer and explanation
Correct option: A
0.166×10−8  Nm−20.166 \times {10^{ - 8}}\;\mathrm{N}{\mathrm{m}^{ - 2}}
Option A (0.166×10−8  Nm−20.166 \times {10^{ - 8}}\;\mathrm{N}{\mathrm{m}^{ - 2}}) is correct because the pressure is calculated as intensity divided by the speed of light when the radiation is fully absorbed.
19 Medium
19 of 145
Analyzing +4 | -1
An electron moving with an initial velocity V=V0i^{V}={V}_0 \hat{{i}} in a magnetic field B=B0j^{B}={B}_0 \hat{{j}}. Then, its de- Broglie wavelength
View answer and explanation
Correct option: A
remains constant
Option A is correct because the magnetic field only changes the direction of the electron's velocity, not its speed. Since the de Broglie wavelength depends on momentum (and thus speed), it remains constant.
20 Medium
20 of 145
Application +4 | -1
50W/m250\mathrm{ W/}{\mathrm{m}^2} energy density of sunlight is normally incident on the surface of a solar panel. Some part of incident energy (25%) is reflected from the surface and the rest is absorbed. The force exerted on 1m2\mathrm{1 }{\mathrm{m}^\mathrm{2}} surface area will be close to (c=3×108  m/s)(c = 3 \times {10^8}{\rm\,\,{m/s)}}
View answer and explanation
Correct option: A
20×10−8N20 \times {10^{ - 8}}\mathrm{N}
Option A (20×10−8N20 \times {10^{ - 8}}\mathrm{N}) is correct because it is calculated by considering both the absorbed and reflected portions of the energy.
21 Easy
21 of 145
Understanding +4 | -1
The wavelength of the short radio waves, micro waves, ultraviolet waves are λ1,  λ2  and  λ3{\lambda _1},\,\,{\lambda _2}\,\,\mathrm{and}\,\,{\lambda _3} respectively. Arrange them in decreasing order
View answer and explanation
Correct option: B
λ1, λ2,λ3{\lambda _1},\,{\lambda _2},{\lambda _3}
Option D (λ2,λ1,λ3{\lambda _2},{\lambda _1},{\lambda _3}) is correct because it arranges the wavelengths in decreasing order, knowing that microwaves have the longest wavelength, followed by short radio waves, and then ultraviolet waves.
22 Medium
22 of 145
Application +4 | -1
A plane EMW propagating in the X-direction has a wave length of 60 mm. The electric field is in the y-direction and its maximum magnitude is 33 Vm−133\,V{m^{ - 1}}. The equation for the electric field as function of xx and tt is
View answer and explanation
Correct option: B
33sin⁡π×10+10[t−xc]33\sin \pi \times {10^{ + 10}}\left[ {t - \frac{x}{c}} \right]
Option B is correct because it represents the electric field with the correct amplitude, wave number, and angular frequency derived from the given wavelength and maximum magnitude.
23 Medium
23 of 145
Application +4 | -1
The amplitude of the magnetic field part of a harmonic Electromagnetic Wave in vacuum is B0{B_0}=510 nT. What is the amplitude of the electric field part of the wave?
View answer and explanation
Correct option: B
153NC−1153\mathrm{ }N{C^{ - 1}}
Option B (153  NC−1153{\;N}{C^{ - 1}}) is correct because the amplitude of the electric field is calculated by multiplying the magnetic field amplitude by the speed of light.
24 Easy
24 of 145
Remembering +4 | -1
Part of electromagnetic spectrum that has largest penetrating power is
View answer and explanation
Correct option: A
γ\gamma-rays
Option A is correct because gamma rays have the largest penetrating power due to their high energy and short wavelength in the electromagnetic spectrum.
25 Hard
25 of 145
Application +4 | -1
Calculate the electric and magnetic fields produced by the radiation coming from a 100 watt bulb at a distance of 3 m3 \mathrm{~m}. Assume that the efficiency of the bulb is 2.5%2.5 \% and it is a point source.
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Correct option: A
2.9 Vm−1,1.4×10−7 T2.9 \mathrm{~Vm}^{-1}, 1.4 \times 10^{-7} \mathrm{~T}
Option A (2.9 Vm−1,1.4×10−7 T2.9 \mathrm{~Vm}^{-1}, 1.4 \times 10^{-7} \mathrm{~T}) is correct because it is calculated using the efficiency of the bulb to find the power radiated, then using that power to find the intensity, and finally the electric and magnetic fields.
26 Medium
26 of 145
Understanding +4 | -1
X-ray are not used for radar purpose, because they are not
View answer and explanation
Correct option: A
Reflected by target
Option A (Reflected by target) is correct because radar relies on the reflection of electromagnetic waves from the target, and X-rays are not effectively reflected.
27 Medium
27 of 145
Analyzing +4 | -1
If a charged particle goes unaccelerated in a region containing electric and magnetic fields
View answer and explanation
Correct option: B
V→\overrightarrow V must be perpendicular to  E→\,\overrightarrow E
The correct answer is that V→\overrightarrow V must be perpendicular to E→\overrightarrow E because the net force on the charged particle must be zero for it to remain unaccelerated.
28 Medium
28 of 145
Understanding +4 | -1
In free space, an electromagnetic wave of 3 GHz frequency strikes over the edge of an object of size λ100,\frac{\lambda }{{100}}, where λ\lambda is the wavelength of the wave in free space. The phenomenon, which happens there will be:
View answer and explanation
Correct option: D
Scattering
Option D (Scattering) is correct. When the size of the object is much smaller than the wavelength of the wave, scattering is the dominant phenomenon.
29 Medium
29 of 145
Understanding +4 | -1
Identify the mismatched pair in electromagnetic wave spectrum
View answer and explanation
Correct option: C
Microwaves: To destroy cancer cells
Option C (Microwaves: To destroy cancer cells) is correct because microwaves are typically used for heating and communication, not for destroying cancer cells (which is more commonly done with radiation therapy using X-rays or gamma rays).
30 Medium
30 of 145
Analyzing +4 | -1
An electromagnetic wave, going through vacuum is described by E=E0sin⁡(kx−ωt)E = {E_0}\sin \left( {kx - \omega t} \right). Which of the following is independent of wavelength
View answer and explanation
Correct option: C
k/ωk/\omega
Option B is correct because the amount of energy transmitted is the difference between the incident energy and the sum of reflected and absorbed energy.
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  • ✓ This set contains 145 single-correct-answer MCQs on ELECTROMAGNETIC WAVES.
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