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NEET • PHYSICS • ELECTROMAGNETIC WAVES

Electromagnetic waves practice · Mixed difficulty

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91 Medium
91 of 145
Application +4 | -1
The amplitude of electric field in a parallel beam of light of intensity 4  Wm−24\mathrm{\;W}{\mathrm{m}^{ - 2}} is
View answer and explanation
Correct option: D
55.5  NC−155.5\mathrm{\;N}{\mathrm{C}^{ - 1}}
Option D (55.5  NC−155.5\mathrm{\;N}{\mathrm{C}^{ - 1}}) is correct because the amplitude of the electric field is calculated using the intensity formula I=12cϵ0E02I = \frac{1}{2}c\epsilon_0 E_0^2.
92 Medium
92 of 145
Analyzing +4 | -1
If a charge particle of charge 2μC2 \mu {C} and mass 1gm1 {gm} is released at origin in gravity free space having E⃗=E0j^\vec{E}=E_0 \hat{j} and B⃗=−B0j^\vec{B}=-B_0 \hat{j}

(E0=103 N/C\left(E_0=10^3 {~N} / {C}\right. and B0=5 T)\left.B_0=5 {~T}\right), then its speed
View answer and explanation
Correct option: B
Depends only on YY co-ordinate
Option B is correct because the electric field is along the y-axis, causing acceleration in the y-direction. The magnetic field, also along the y-axis, does not exert a force on the particle since the velocity is initially along the y-axis. Therefore, the speed depends only on the y-coordinate.
93 Medium
93 of 145
Application +4 | -1
Light with energy flux of 24 w/m224\,w/{m^2} is incident on a well polished disc of radios 3.5 cm for one hour. The momentum transferred to the disc is [in kgm s−1{s^{ - 1}}]
View answer and explanation
Correct option: B
2.2×10−62.2 \times {10^{ - 6}}
Option B (2.2×10−62.2 \times {10^{ - 6}}) is correct because it is the momentum transferred calculated using the formula p=(2UA)/cp = (2UA)/c for a reflecting surface.
94 Medium
94 of 145
Understanding +4 | -1
Radio wave diffract around building although light waves do not. The reason is that radio waves
View answer and explanation
Correct option: B
Have much larger wavelength than light
Option B (Have much larger wavelength than light) is correct because diffraction is more pronounced when the wavelength of the wave is comparable to the size of the obstacle.
95 Medium
95 of 145
Application +4 | -1
In an EMW in free space the root mean square value of the electric field is Erms=6 Vm{E_{rms}} = 6\,\frac{V}{m}. The peak value of the magnetic field is [in Tesla]
View answer and explanation
Correct option: B
2.83×10−82.83 \times {10^{ - 8}}
Option B (2.83×10−82.83 \times {10^{ - 8}}) is correct because it is calculated using the relationship between electric and magnetic field amplitudes in an electromagnetic wave and the given RMS electric field.
96 Easy
96 of 145
Application +4 | -1
In a place e.m. wave, the electric field oscillates sinusoidally at a frequency of 2.5×1010 Hz2.5 \times 10^{10} {~Hz} and amplitude 480 V/m480 {~V} / {m}. The amplitude of oscillating magnetic field will be :-
View answer and explanation
Correct option: C
1.6×10−6 Wb/m21.6 \times 10^{-6} {~Wb} / {m}^2
Option C (1.6×10−6 Wb/m21.6 \times 10^{-6} {~Wb} / {m}^2) is correct because the amplitude of the oscillating magnetic field is calculated using the formula B0=E0cB_0 = \frac{E_0}{c}, where c is the speed of light.
97 Medium
97 of 145
Analyzing +4 | -1
Assertion: The microwaves are better carriers of signals than radio waves.

Reason: The electromagnetic waves do not require any medium for propagation.
View answer and explanation
Correct option: B
If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion
Option B (If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion) is correct because both statements are true, but the reason doesn't explain why microwaves are better carriers.
98 Easy
98 of 145
Remembering +4 | -1
Rays in electromagnetic spectrum that has shortest wavelength are
View answer and explanation
Correct option: D
X-rays
Option D is correct because X-rays have the shortest wavelength among the given options in the electromagnetic spectrum.
99 Hard
99 of 145
Application +4 | -1
Light with energy flux of 18Wcm218\frac{W}{{c{m^2}}} falls on a non-reflecting surface of area 20  cm220\,\,c{m^2} at normal in cadence the momentum delivered in 30 minutes is [in kgm/s]
View answer and explanation
Correct option: B
2.16×10−32.16 \times {10^{ - 3}}
Option B (2.16×10−32.16 \times {10^{ - 3}}) is correct because it is calculated by finding the momentum delivered by the light to the surface over the given time.
100 Medium
100 of 145
Remembering +4 | -1
For an electromagnetic wave traveling in free space
View answer and explanation
Correct option: A
E0=cB0E_0=c B_0
Option A (E0=cB0E_0=c B_0) is correct because the electric field amplitude is equal to the speed of light times the magnetic field amplitude in an electromagnetic wave.
101 Easy
101 of 145
Remembering +4 | -1
Displacement Current is:
View answer and explanation
Correct option: D
Due to time Varying Electric field
Option D is correct because displacement current arises from a time-varying electric field, as described by Maxwell's equations.
102 Easy
102 of 145
Remembering +4 | -1
The S.I. unit of displacement current is:
View answer and explanation
Correct option: C
Ampere
Option C (Ampere) is correct because displacement current, like regular current, is measured in Amperes.
103 Medium
103 of 145
Analyzing +4 | -1
An electromagnetic wave travelling along z{z}-axis is given as: E⃗=E⃗0cos⁡(kz−ωt)\vec{E}=\vec{E}_0 \cos (k z-\omega t). Choose the INCORRECT options from the following;
View answer and explanation
Correct option: D
k^×E⃗=0→,k^×B⃗=0→\hat{k} \times \vec{E}=\overrightarrow{0}, \hat{k} \times \vec{B}=\overrightarrow{0}
Option D is the INCORRECT option because for an electromagnetic wave traveling along the z-axis, the electric and magnetic fields are perpendicular to the direction of propagation, meaning their cross product with the direction vector is non-zero.
104 Medium
104 of 145
Application +4 | -1
A plane electromagnetic wave is propagating along the direction i^+j^2\frac{{\hat i + \hat j}}{{\sqrt 2 }} with its polarization along the direction k^.\hat k. The correct form of the magnetic field of the wave would be (here B0{B_0} is an appropriate constant)
View answer and explanation
Correct option: C
B0i^−j^2cos⁡(ωt−ki^+j^2){B_0}\frac{{\hat i - \hat j}}{{\sqrt 2 }}\cos \left( {\omega t - k\frac{{\hat i + \hat j}}{{\sqrt 2 }}} \right)
Option C (B0i^−j^2cos⁡(ωt−ki^+j^2){B_0}\frac{{\hat i - \hat j}}{{\sqrt 2 }}\cos \left( {\omega t - k\frac{{\hat i + \hat j}}{{\sqrt 2 }}} \right)) is correct because the magnetic field must be perpendicular to both the direction of propagation and the polarization direction of the electric field.
105 Medium
105 of 145
Understanding +4 | -1
The magnetic field BB and electric field EE in an electromagnetic wave are:
View answer and explanation
Correct option: C
mutually perpendicular but with 0∘0^{\circ} phase difference
Option C (mutually perpendicular but with 0∘0^{\circ} phase difference) is correct because the electric and magnetic fields in an electromagnetic wave are perpendicular to each other and oscillate in phase.
106 Medium
106 of 145
Remembering +4 | -1
In an Electromagnetic Wave, direction of propagation is in the direction of
View answer and explanation
Correct option: C
E  ×  B\mathbf{E}\; \times \;\mathbf{B}
Option C (E  ×  B\mathbf{E}\; \times \;\mathbf{B}) is correct because the direction of propagation of an electromagnetic wave is given by the Poynting vector, which is proportional to the cross product of the electric and magnetic field vectors.
107 Medium
107 of 145
Remembering +4 | -1
In a medium of dielectric constant KK, the electric field is E\mathbf{E}. If ε0{\varepsilon _0} is permittivity of the free space, the electric displacement vector is
View answer and explanation
Correct option: D
Kε0EK{\varepsilon _0}\mathbf{E}
Option D (Kε0EK{\varepsilon _0}\mathbf{E}) is correct because the electric displacement vector in a medium is given by the product of the dielectric constant, permittivity of free space, and the electric field.
108 Medium
108 of 145
Understanding +4 | -1
If a long hollow copper pipe carries a direct current, the magnetic field associated with the current will be
View answer and explanation
Correct option: B
Only outside the pipe
Option B (Only outside the pipe) is correct because, according to Ampere's law, the magnetic field inside a hollow conductor carrying a direct current is zero, and the field exists only outside.
109 Easy
109 of 145
Understanding +4 | -1
Let the velocities of light in vacuum and glass are represented by vv and vgv_{\mathrm{g}}. When electrons move through glass with a velocity vev_e, there emitted a visible radiation in glass if
View answer and explanation
Correct option: A
ve>vgv_e > v_g
Option A (ve>vgv_e > v_g) is correct because Cherenkov radiation is emitted when a charged particle moves through a dielectric medium faster than the speed of light in that medium.
110 Medium
110 of 145
Understanding +4 | -1
The average value of electric energy density in an Electromagnetic Waves is (E0{E_0} is peak value)
View answer and explanation
Correct option: D
14ε0E02\frac{1}{4}{\varepsilon _0}E_0^2
Option D (14ε0E02\frac{1}{4}{\varepsilon _0}E_0^2) is correct because the average electric energy density is half of the product of permittivity and the square of the RMS electric field, and the RMS electric field is E0/2E_0/\sqrt{2}.
111 Medium
111 of 145
Understanding +4 | -1
The energy of an electromagnetic wave contained in a small volume oscillates with
View answer and explanation
Correct option: A
Double the frequency of the wave
Option A (Double the frequency of the wave) is correct because the energy density is proportional to the square of the electric or magnetic field, and squaring a sinusoidal function doubles its frequency.
112 Medium
112 of 145
Remembering +4 | -1
Phase difference between electric field and magnetic field associated with a plane
polarized light is
View answer and explanation
Correct option: A
zero
Option A (zero) is correct because the electric and magnetic fields in a plane polarized light wave are in phase.
113 Easy
113 of 145
Analyzing +4 | -1
A proton moving with a constant velocity passes through a region of space without any change in its velocity. If E⃗\vec{E} and B⃗\vec{B} represent the electric and magnetic fields respectively, then the region of space may have:

(A) E=0,B=0E=0, B=0

(B) E=0,B≠0E=0, B \neq 0

(C) E≠0,B=0E \neq 0, B=0

(D) E≠0,B≠0E \neq 0, B \neq 0

Choose the most appropriate answer from the options given below:
View answer and explanation
Correct option: C
(A), (B) and (D) only
Option C ((A), (B) and (D) only) is correct because a proton can move with constant velocity if there are no fields, only a magnetic field (velocity parallel to B), or both electric and magnetic fields such that their forces cancel out.
114 Medium
114 of 145
Understanding +4 | -1
The ratio of average electric energy density and total average energy density of electromagnetic wave is
View answer and explanation
Correct option: D
12\frac{1}{2}
Option D (12\frac{1}{2}) is correct because the average electric energy density is half of the total average energy density in an electromagnetic wave, as the energy is equally distributed between the electric and magnetic fields.
115 Hard
115 of 145
Application +4 | -1
An alternating voltage (in volts) =1002sin⁡100t=100 \sqrt{2} \sin 100 t is connected to one microfarad capacitor. The reading of the ammeter shall be
View answer and explanation
Correct option: D
10 mA10 \mathrm{~mA}
Option D is correct because the current is calculated using the formula I=V/XcI = V/X_c where Xc=1/(ωC)X_c = 1/(\omega C).
116 Medium
116 of 145
Understanding +4 | -1
A parallel plate capacitor is charged to 60μC60\mathrm{\mu C}. Due to a radioactive source, the plate loss charge at the rate of 1.8  ×  10−8  Cs−1.1.8\mathrm{\;} \times \mathrm{\;}{10^{ - 8}}\mathrm{\;C}{\mathrm{s}^{ - 1}}. The magnitude of displacement current is
View answer and explanation
Correct option: A
1.8×10−8  Cs−11.8 \times {10^{ - 8}}\mathrm{\;C}{\mathrm{s}^{ - 1}}
Option A (1.8×10−8  Cs−11.8 \times {10^{ - 8}}\mathrm{\;C}{\mathrm{s}^{ - 1}}) is correct because the displacement current is equal to the rate of change of charge on the capacitor plates.
117 Easy
117 of 145
Understanding +4 | -1
If E→  and  B→\overrightarrow E \,\,\mathrm{and}\,\,\overrightarrow B are the electric and magnetic field vectors of E.M waves then the direction of propagation of E.M wave is along the direction of
View answer and explanation
Correct option: C
E→×B→\overrightarrow E \times \overrightarrow B
Option C, E→×B→\overrightarrow E \times \overrightarrow B, is correct because the direction of propagation of an electromagnetic wave is given by the Poynting vector, which is proportional to the cross product of the electric and magnetic field vectors.
118 Medium
118 of 145
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, electric field, and magnetic field are mutually perpendicular, and following the right-hand rule, if E is up and the wave travels East, B must be South.
119 Medium
119 of 145
Application +4 | -1
The wave function (in S.I units) for an electromagnetic wave is given as −ψ(X,t)=103sin⁡π(3×106X−9×1014t)- \psi \left( {X,t} \right) = {10^3}\sin \pi \left( {3 \times {{10}^6}X - 9 \times {{10}^{14}}t} \right). The speed of wave is :
View answer and explanation
Correct option: B
3×108 m/s3 \times {10^8}\,m/s
Option B (3×108 m/s3 \times {10^8}\,m/s) is correct because the speed of the wave is calculated by dividing the coefficient of time by the coefficient of position in the wave equation.
120 Easy
120 of 145
Application +4 | -1
A plane electromagnetic wave travels in a medium of relative permittivity 2 and relative permeability 8.0.8.0. If the magnitude of magnetic intensity is 6π×10−2 A/m\frac{6}{\pi} \times 10^{-2} \mathrm{~A} / \mathrm{m} at a point, then the electric field at that point in ___ V/m\mathrm{V} / \mathrm{m}.
View answer and explanation
Correct option: A
1.801.80
Option A (1.80) is correct because the electric field is calculated using the formula E=vBE = vB, where v=cμrϵrv = \frac{c}{\sqrt{\mu_r \epsilon_r}} is the speed of light in the medium.
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